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Difficulty: 1 – 5

5161184

Find the rule that changes each top number into the number below it. Then complete the table. <table> <tr> <td>\(1200\)</td><td>\(5000\)</td><td>\(15{,}000\)</td><td>\(100{,}000\)</td><td>\(240{,}000\)</td><td>\(600{,}000\)</td> </tr> <tr> <td>\(3700\)</td><td>\(7500\)</td><td>\(17{,}500\)</td><td>\(102{,}500\)</td><td></td><td></td> </tr> </table>

Hints

- Compare a known top number with the number below it. - Determine whether the same amount is added each time. - Test your rule on every completed column before using it.

Solution

1. Compare the first pair: \(3700 - 1200 = 2500\). 2. The rule \(+2500\) works for the other known pairs: \(5000 + 2500 = 7500\), \(15{,}000 + 2500 = 17{,}500\), and \(100{,}000 + 2500 = 102{,}500\). 3. Apply the rule to the missing entries: \(240{,}000 + 2500 = 242{,}500\) and \(600{,}000 + 2500 = 602{,}500\).

Answer

Rule: add \(2500\) Missing numbers: \(242{,}500\) and \(602{,}500\)

5161194

Find the rule that changes each top number into the number below it. Then complete the table. <table> <tr> <td>\(15\)</td><td>\(60\)</td><td>\(200\)</td><td>\(1000\)</td><td>\(3000\)</td><td>\(12{,}000\)</td> </tr> <tr> <td>\(45\)</td><td>\(180\)</td><td>\(600\)</td><td>\(3000\)</td><td></td><td></td> </tr> </table>

Hints

- Compare a top number with the number below it. - Find the factor that changes \(15\) into \(45\). - Check that the same multiplication rule works for the larger numbers.

Solution

1. Compare the first pair: \(15 \times 3 = 45\). 2. The rule \(\times 3\) works for the other known pairs: \(60 \times 3 = 180\), \(200 \times 3 = 600\), and \(1000 \times 3 = 3000\). 3. Apply the rule to the missing entries: \(3000 \times 3 = 9000\) and \(12{,}000 \times 3 = 36{,}000\).

Answer

Rule: multiply by \(3\) Missing numbers: \(9000\) and \(36{,}000\)

5161964

Continue the multiplication pattern with two more equations. Calculate every product, and describe how the products change. \(4\times 70=\square\) \(8\times 70=\square\) \(12\times 70=\square\) \(16\times 70=\square\)

Hints

- Notice how the first factor changes. - Check whether the second factor changes. - Compare consecutive products. - Connect the change in the factor to the change in the product.

Solution

1. The first four products are \(280,560,840,1120\). 2. The first factor increases by \(4\), while the second factor stays \(70\). 3. The next equations are \(20\times 70=1400\) and \(24\times 70=1680\). 4. Each product increases by \(4\times 70=280\).

Answer

\(20\times 70=1400\) \(24\times 70=1680\) The products increase by \(280\) each time.

5162324

Calculate each row. What pattern do you notice among the quotients within each row? a) \(240\div 4\), \(120\div 4\), \(60\div 4\) b) \(600\div 6\), \(300\div 6\), \(150\div 6\) c) \(800\div 8\), \(400\div 8\), \(200\div 8\)

Hints

- Track how the dividend changes within each row. - Check whether the divisor changes. - Compare each quotient with the previous quotient. - State exactly how the quotients change.

Solution

1. Part a gives \(60,30,15\). 2. Part b gives \(100,50,25\). 3. Part c gives \(100,50,25\). 4. In each row, the dividend is halved while the divisor stays fixed, so the quotient is also halved.

Answer

a) \(60,30,15\) b) \(100,50,25\) c) \(100,50,25\) In each row, halving the dividend halves the quotient.

5162444

Study the division pattern and write the next equation. \(24{,}000 \div 6 = 4000\) \(30{,}000 \div 6 = 5000\) \(36{,}000 \div 6 = 6000\) \(42{,}000 \div 6 = 7000\)

Hints

- Track how the dividend changes from one equation to the next. - Compare the quotients. - Notice which number stays the same.

Solution

1. The divisor remains \(6\). 2. Each dividend increases by \(6000\), so the next dividend is \(42{,}000 + 6000 = 48{,}000\). 3. Each quotient increases by \(1000\), so the next quotient is \(8000\). 4. The next equation is \(48{,}000 \div 6 = 8000\).

Answer

\(48{,}000 \div 6 = 8000\)

5162564

Continue the number pattern with three more numbers and describe the rule. \(125{,}400, 126{,}100, 126{,}800, 127{,}500, \ldots\)

Hints

- Find the difference between the first two terms. - Check whether the same difference appears again. - Apply the repeated change three times.

Solution

1. The difference between consecutive terms is \(700\). 2. Add \(700\) to get \(127{,}500 + 700 = 128{,}200\). 3. Continue: \(128{,}200 + 700 = 128{,}900\). 4. Continue again: \(128{,}900 + 700 = 129{,}600\).

Answer

Next numbers: \(128{,}200, 128{,}900, 129{,}600\) Rule: Add \(700\) each time.

5163584

Continue counting by \(500\) until you reach \(351{,}000\). \(348{,}000, 348{,}500, \ldots, 351{,}000\)

Hints

- Add \(500\) to each term. - Watch how the hundreds and thousands digits change at each full thousand. - Stop when you reach the target number.

Solution

1. Add \(500\) repeatedly. 2. The missing terms are \(349{,}000\), \(349{,}500\), \(350{,}000\), and \(350{,}500\). 3. One more step gives \(350{,}500 + 500 = 351{,}000\).

Answer

\(348{,}000, 348{,}500, 349{,}000, 349{,}500, 350{,}000, 350{,}500, 351{,}000\)

5163594

Find the step size and direction in each pattern. Then write the next three terms. a) \(456{,}700, 456{,}800, 456{,}900, \ldots\) b) \(820{,}000, 810{,}000, 800{,}000, \ldots\)

Hints

- Compare the first two terms in each pattern. - Decide whether the pattern increases or decreases. - Pay attention to changes across a full thousand or hundred thousand.

Solution

1. For a), each term increases by \(100\). 2. The next three terms are \(457{,}000\), \(457{,}100\), and \(457{,}200\). 3. For b), each term decreases by \(10{,}000\). 4. The next three terms are \(790{,}000\), \(780{,}000\), and \(770{,}000\).

Answer

a) Step: \(+100\); next terms: \(457{,}000, 457{,}100, 457{,}200\) b) Step: \(-10{,}000\); next terms: \(790{,}000, 780{,}000, 770{,}000\)

5163604

Count backward from \(1{,}000{,}000\) by \(1000\) until you reach \(994{,}000\). Write the complete sequence.

Hints

- Remove one thousand at each step. - Pay attention to the first change from \(1{,}000{,}000\) to a six-digit number. - Stop at the stated target.

Solution

1. Begin at \(1{,}000{,}000\). 2. Subtract \(1000\) repeatedly to get \(999{,}000\), \(998{,}000\), \(997{,}000\), \(996{,}000\), \(995{,}000\), and \(994{,}000\).

Answer

\(1{,}000{,}000, 999{,}000, 998{,}000, 997{,}000, 996{,}000, 995{,}000, 994{,}000\)

5163714

Fill in the missing numbers in this counting pattern. \(599{,}996, \square, \square, 599{,}999, \square, 600{,}001, \square\)

Hints

- Determine the distance between the given terms. - Adding \(1\) to a number ending in several nines changes multiple places. - Identify the number exactly between \(599{,}999\) and \(600{,}001\).

Solution

1. The pattern counts forward by \(1\). 2. The first two missing numbers are \(599{,}997\) and \(599{,}998\). 3. The number between \(599{,}999\) and \(600{,}001\) is \(600{,}000\). 4. The number after \(600{,}001\) is \(600{,}002\).

Answer

\(599{,}997, 599{,}998, 600{,}000, 600{,}002\)

5163894

Count forward by \(100\) until you reach the next multiple of \(10{,}000\). a) \(359{,}400, 359{,}500, \ldots, 360{,}000\) b) \(819{,}600, 819{,}700, \ldots, 820{,}000\)

Hints

- Add \(100\) at each step. - Watch what happens when the hundreds digit passes \(9\). - Count how many steps remain before the target.

Solution

1. For a), add \(100\) repeatedly. The missing terms are \(359{,}600\), \(359{,}700\), \(359{,}800\), and \(359{,}900\). 2. For b), the missing terms are \(819{,}800\) and \(819{,}900\).

Answer

a) \(359{,}600, 359{,}700, 359{,}800, 359{,}900\) b) \(819{,}800, 819{,}900\)

5165154

Count forward by ones. What three numbers come after \(399{,}999\)? \(399{,}997, 399{,}998, 399{,}999, \ldots\)

Hints

- Pay close attention to the transition after a number ending in several nines. - Determine how many place values change when \(1\) is added to \(399{,}999\).

Solution

1. \(399{,}999 + 1 = 400{,}000\). 2. \(400{,}000 + 1 = 400{,}001\). 3. \(400{,}001 + 1 = 400{,}002\).

Answer

\(400{,}000, 400{,}001, 400{,}002\)

5170694

Multiply \(505\) by \(3\), \(5\), \(7\), and \(9\). 1. Calculate the four products. 2. Describe the digit pattern in the products.

Hints

- Calculate all four products carefully. - Compare the first two digits with the last two digits in each product. - Relate the repeated block to the multiplication facts for \(5\).

Solution

1. The products are \(505\times 3=1515\), \(505\times 5=2525\), \(505\times 7=3535\), and \(505\times 9=4545\). 2. Each product consists of the same two-digit block repeated twice. 3. That two-digit block equals \(5\) times the one-digit factor: \(5\times 3=15\), \(5\times 5=25\), \(5\times 7=35\), and \(5\times 9=45\).

Answer

1. \(1515,2525,3535,4545\) 2. Each product repeats the same two-digit block twice. The block is \(5\) times the factor.

5170724

Multiply \(198\) by \(2\), \(3\), \(4\), and \(5\). a) Calculate the four products. b) What do you notice about the tens digit of every product? c) Add the hundreds digit and ones digit of each product. What do you notice?

Hints

- Calculate each product carefully. - Compare the middle digit of the four products. - Add the first and last digit of each product.

Solution

1. The products are \(198\times 2=396\), \(198\times 3=594\), \(198\times 4=792\), and \(198\times 5=990\). 2. The tens digit is \(9\) in every product. 3. The outer digits also add to \(9\): \(3+6=9\), \(5+4=9\), \(7+2=9\), and \(9+0=9\).

Answer

a) \(396,594,792,990\) b) The tens digit is always \(9\). c) The hundreds digit and ones digit always add to \(9\).

5170844

Continue the sequence using this rule: each new term is the sum of the two previous terms. Write the sequence through the sixth term. Starting terms: \(2400\) and \(3700\)

Hints

- Add the two most recent terms to find the next term. - Begin by adding the two starting terms. - After each step, use the new term as one of the next two addends.

Solution

1. The third term is \(2400 + 3700 = 6100\). 2. The fourth term is \(3700 + 6100 = 9800\). 3. The fifth term is \(6100 + 9800 = 15{,}900\). 4. The sixth term is \(9800 + 15{,}900 = 25{,}700\).

Answer

\(2400, 3700, 6100, 9800, 15{,}900, 25{,}700\)

5173514

Continue the number pattern with the next five terms. Then describe the rule. \(92; 84; 76; 68; 60; \ldots\)

Hints

- Find the difference between each pair of neighboring terms. - Check whether the same change repeats. - Apply the rule one term at a time.

Solution

1. The difference between consecutive terms is \(8\): \(92-84=8\), \(84-76=8\), and \(76-68=8\). 2. Subtract \(8\) each time: \(60-8=52\), \(52-8=44\), \(44-8=36\), \(36-8=28\), and \(28-8=20\).

Answer

Next five terms: \(52; 44; 36; 28; 20\) Rule: Subtract \(8\) each time.

5173634

A number pattern starts with \(2\) and \(3\). Each new term is the sum of the two terms immediately before it. Find the first ten terms.

Hints

- Write the first two terms. - Add them to find the third term. - Keep using the two most recent terms until you have ten terms.

Solution

1. Start with \(2, 3\). 2. Add the two preceding terms each time: \(2+3=5\), \(3+5=8\), \(5+8=13\), \(8+13=21\), \(13+21=34\), \(21+34=55\), \(34+55=89\), and \(55+89=144\).

Answer

\(2; 3; 5; 8; 13; 21; 34; 55; 89; 144\)

5182114

A number pattern starts with \(10\). To find each new term, double the previous term and then subtract \(5\). Write the first six terms.

Hints

- “Double” means multiply by \(2\). - Subtract \(5\) after doubling. - Use each new term to create the next term.

Solution

1. Start with \(10\). 2. Apply the rule repeatedly: \(10\times2-5=15\), \(15\times2-5=25\), \(25\times2-5=45\), \(45\times2-5=85\), and \(85\times2-5=165\).

Answer

\(10, 15, 25, 45, 85, 165\)

5188644

Write the next three terms in each number pattern. a) \(60{,}200, 60{,}100, 60{,}000, \ldots\) b) \(899{,}997, 899{,}998, 899{,}999, \ldots\) c) \(120{,}000, 110{,}000, 100{,}000, \ldots\)

Hints

- Find the difference between the first two terms in each pattern. - Decide whether each pattern increases or decreases. - Watch for changes across a major place-value boundary.

Solution

1. For a), subtract \(100\) each time. The next terms are \(59{,}900\), \(59{,}800\), and \(59{,}700\). 2. For b), add \(1\) each time. The next terms are \(900{,}000\), \(900{,}001\), and \(900{,}002\). 3. For c), subtract \(10{,}000\) each time. The next terms are \(90{,}000\), \(80{,}000\), and \(70{,}000\).

Answer

a) \(59{,}900, 59{,}800, 59{,}700\) b) \(900{,}000, 900{,}001, 900{,}002\) c) \(90{,}000, 80{,}000, 70{,}000\)

5199904

Write the five numbers that come immediately before \(300{,}003\) in the counting sequence.

Hints

- Count backward by ones. - Pay attention when subtracting \(1\) from a number ending in zeros. - Present the five numbers in their normal increasing order.

Solution

1. Count backward by \(1\): \(300{,}002\), \(300{,}001\), \(300{,}000\), \(299{,}999\), and \(299{,}998\). 2. Write those five numbers in increasing order: \(299{,}998, 299{,}999, 300{,}000, 300{,}001, 300{,}002\).

Answer

\(299{,}998, 299{,}999, 300{,}000, 300{,}001, 300{,}002\)

5199914

Count forward by tens. What four numbers come immediately after \(699{,}980\)?

Hints

- Add \(10\) at each step. - Track the tens digit. - Pay attention to the transition after \(699{,}990\).

Solution

1. \(699{,}980 + 10 = 699{,}990\). 2. \(699{,}990 + 10 = 700{,}000\). 3. Continue by tens: \(700{,}010\) and \(700{,}020\).

Answer

\(699{,}990, 700{,}000, 700{,}010, 700{,}020\)

5213094

Continue the number pattern. The first two terms are \(999{,}800\) and \(999{,}850\). Each term increases by the same amount. Write the next three terms.

Hints

- Find how much the pattern increases from the first term to the second. - Add that same amount for each new term. - Pay attention to the place-value change after \(999{,}950\).

Solution

1. Find the difference between the first two terms: \(999{,}850 - 999{,}800 = 50\). 2. Add \(50\) repeatedly: \(999{,}850 + 50 = 999{,}900\), \(999{,}900 + 50 = 999{,}950\), and \(999{,}950 + 50 = 1{,}000{,}000\).

Answer

\(999{,}900, 999{,}950, 1{,}000{,}000\)

5217304

Start at \(8970\) and list the next five numbers in each pattern. a) Count forward by tens. b) Count backward by hundreds. c) Count forward by fives.

Hints

- Decide whether each pattern increases or decreases. - Change the correct place value at every step. - Pay attention when the pattern crosses a multiple of \(1000\).

Solution

1. a) Add \(10\) five times: \(8980, 8990, 9000, 9010, 9020\). 2. b) Subtract \(100\) five times: \(8870, 8770, 8670, 8570, 8470\). 3. c) Add \(5\) five times: \(8975, 8980, 8985, 8990, 8995\).

Answer

a) \(8980, 8990, 9000, 9010, 9020\) b) \(8870, 8770, 8670, 8570, 8470\) c) \(8975, 8980, 8985, 8990, 8995\)

5373894

Panels a), b), and c) show \(2\), \(4\), and \(6\) rows of dots, with \(7\) dots in each row. Describe the rule relating the number of rows to the number of dots. How many dots would there be in \(9\) rows?

Figure for problem 537389

Hints

- Compare the pairs \(2 \rightarrow 14\), \(4 \rightarrow 28\), and \(6 \rightarrow 42\). - The number of dots in each row stays the same.

Solution

1. Multiply the number of rows by \(7\) to find the number of dots. 2. For \(9\) rows: \(9 \times 7 = 63\) dots.

Answer

Rule: Multiply the number of rows by \(7\). Nine rows contain \(63\) dots.

5157044

Continue the sequence and explain the rule. \(320, 321, 323, 326, 330, \ldots, \ldots, \ldots, 356\)

Hints

- Find how much is added from one term to the next. - Compare the successive differences. - Determine how the differences themselves change.

Solution

1. The changes are \(321-320=1\), \(323-321=2\), \(326-323=3\), and \(330-326=4\). 2. The amount added increases by \(1\) each time. 3. Continue by adding \(5\), \(6\), and \(7\): \(330+5=335\), \(335+6=341\), and \(341+7=348\). 4. Adding \(8\) gives \(348+8=356\), which matches the final term.

Answer

The missing numbers are \(335, 341,\) and \(348\). Rule: add \(1, 2, 3, 4, 5, \ldots\), increasing the amount added by \(1\) each time.

5157054

Find the rule and fill in the missing terms. \(650, 660, 655, 665, 660, \ldots, \ldots, 675\)

Hints

- Compare two consecutive changes rather than only one. - Consider whether two different operations alternate. - Test an “add, then subtract” pattern.

Solution

1. The first changes are \(+10\), then \(-5\). 2. The same pair of changes repeats: \(655+10=665\) and \(665-5=660\). 3. Continue the alternating pattern: \(660+10=670\), then \(670-5=665\). 4. The next \(+10\) step gives \(665+10=675\), which matches the final term.

Answer

The missing numbers are \(670\) and \(665\). Rule: alternate between adding \(10\) and subtracting \(5\).

5161154

Study this pattern of addition equations: \(245{,}100 + 312{,}400 = 557{,}500\) \(255{,}200 + 322{,}500 = 577{,}700\) \(265{,}300 + 332{,}600 = 597{,}900\) a) Continue the pattern by writing the fourth equation. b) By how much does the sum increase from one equation to the next? Explain using the addends.

Hints

- Find how the first addend changes from one row to the next. - Find how the second addend changes. - Combine the two changes to determine the change in the sum.

Solution

1. Each addend increases by \(10{,}100\). The next addends are \(265{,}300 + 10{,}100 = 275{,}400\) and \(332{,}600 + 10{,}100 = 342{,}700\). 2. The fourth equation is \(275{,}400 + 342{,}700 = 618{,}100\). 3. Because both addends increase by \(10{,}100\), the sum increases by \(10{,}100 + 10{,}100 = 20{,}200\) each time.

Answer

a) \(275{,}400 + 342{,}700 = 618{,}100\) b) The sum increases by \(20{,}200\) because each addend increases by \(10{,}100\).

5161164

One equation does not follow the pattern. 1) \(980{,}000 - 450{,}000 = 530{,}000\) 2) \(950{,}000 - 420{,}000 = 530{,}000\) 3) \(920{,}000 - 390{,}000 = 530{,}000\) 4) \(890{,}000 - 350{,}000 = 540{,}000\) a) Which equation breaks the pattern? b) Correct the fourth equation so that it follows the pattern.

Hints

- Compare the differences in the first three equations. - Track how both numbers in each subtraction change. - If both numbers decrease by the same amount, the difference stays the same.

Solution

1. In the first three equations, both the minuend and subtrahend decrease by \(30{,}000\), so the difference remains \(530{,}000\). 2. The fourth minuend correctly decreases to \(890{,}000\), but the subtrahend should decrease from \(390{,}000\) to \(360{,}000\). 3. The corrected equation is \(890{,}000 - 360{,}000 = 530{,}000\).

Answer

a) Equation 4 b) \(890{,}000 - 360{,}000 = 530{,}000\)

5161174

Build an addition pattern using these rules: - The first equation is \(300{,}000 + 200{,}000 = 500{,}000\). - In each new equation, the first addend increases by \(50{,}000\). - In each new equation, the second addend decreases by \(20{,}000\). Find the fifth equation and its sum.

Hints

- You may write the second, third, and fourth equations before finding the fifth. - The change is applied four times after the first equation. - Each step changes the total by the increase in one addend minus the decrease in the other.

Solution

1. From the first equation to the fifth, each change happens \(4\) times. 2. The first addend becomes \(300{,}000 + 4 \times 50{,}000 = 500{,}000\). 3. The second addend becomes \(200{,}000 - 4 \times 20{,}000 = 120{,}000\). 4. The fifth equation is \(500{,}000 + 120{,}000 = 620{,}000\).

Answer

\(500{,}000 + 120{,}000 = 620{,}000\)

5161514

Choose two consecutive digits from \(1\) through \(9\), such as \(3\) and \(4\). a) Use the digits to form the two four-digit numbers with patterns \(abba\) and \(baab\). For \(3\) and \(4\), the numbers are \(3443\) and \(4334\). b) Subtract the smaller number from the larger number. c) Repeat with a different pair of consecutive digits. What pattern do you notice?

Hints

- Consecutive digits differ by \(1\). - Check each subtraction carefully. - Compare the results from parts b and c.

Solution

1. With \(3\) and \(4\), the numbers are \(3443\) and \(4334\), and \(4334-3443=891\). 2. With \(7\) and \(8\), the numbers are \(7887\) and \(8778\), and \(8778-7887=891\). 3. The difference is always \(891\) when the two selected digits differ by \(1\).

Answer

The difference is always \(891\).

5161534

Investigate three-digit numbers and their reversed numbers. a) Choose a three-digit number that does not end in \(0\) and whose hundreds digit is exactly \(3\) greater than its ones digit. One example is \(653\). b) Reverse the digits and subtract the reversed number from your chosen number. c) Repeat with two other numbers that meet the same condition. What do you notice?

Hints

- The tens digit may be any digit. - Reverse the digit order exactly. - Subtract one place at a time and compare your results.

Solution

1. For \(653\), the reversed number is \(356\), and \(653-356=297\). 2. For \(411\), the reversed number is \(114\), and \(411-114=297\). 3. For \(906\), the reversed number is \(609\), and \(906-609=297\). 4. The difference is always \(297\) when the hundreds digit is \(3\) greater than the ones digit.

Answer

The difference is always \(297\).

5161944

Examine the factors in this number pattern. One equation breaks the pattern. Which equation is it? 1. \(10\times 100=1000\) 2. \(20\times 90=1800\) 3. \(30\times 80=2400\) 4. \(40\times 70=2800\) 5. \(50\times 50=2500\)

Hints

- Track how the first factor changes from one equation to the next. - Then track how the second factor changes. - Check whether the last equation follows both rules.

Solution

1. The first factors are \(10,20,30,40,50\), increasing by \(10\). 2. The second factors should be \(100,90,80,70,60\), decreasing by \(10\). 3. Therefore, the fifth equation should use \(60\), not \(50\). 4. The corrected equation is \(50\times 60=3000\). The products then increase by \(800,600,400,200\).

Answer

Equation 5 breaks the pattern. It should be \(50\times 60=3000\).

5161974

Study this multiplication pattern. Continue it with three more equations, and describe what happens to the products. \(10\times 90=\square\) \(20\times 80=\square\) \(30\times 70=\square\) \(40\times 60=\square\)

Hints

- Track both factors at the same time. - Calculate the products for the continued equations. - Compare later products with earlier products for symmetry. - Notice how the differences between consecutive products change.

Solution

1. The given products are \(900,1600,2100,2400\). 2. The first factor increases by \(10\), and the second factor decreases by \(10\). 3. The next equations are \(50\times 50=2500\), \(60\times 40=2400\), and \(70\times 30=2100\). 4. The products increase to a maximum of \(2500\) and then decrease. The products repeat in reverse order around \(50\times 50\).

Answer

\(50\times 50=2500\) \(60\times 40=2400\) \(70\times 30=2100\) The products rise to \(2500\) and then fall, repeating earlier values in reverse order.

5162054

Continue the division pattern with the next two equations and calculate their quotients. \(900\div 9\) \(720\div 8\) \(560\div 7\) \(420\div 6\) \(300\div 5\)

Hints

- Notice how the divisors change. - Calculate the existing quotients and identify their pattern. - Predict the next quotient. - Use multiplication to find a dividend from a divisor and quotient.

Solution

1. The divisors decrease by \(1\): \(9,8,7,6,5\). The next divisors are \(4\) and \(3\). 2. The quotients are \(100,90,80,70,60\), decreasing by \(10\). The next quotients are \(50\) and \(40\). 3. Find the next dividends: \(50\times 4=200\) and \(40\times 3=120\). 4. The next equations are \(200\div 4=50\) and \(120\div 3=40\).

Answer

\(200\div 4=50\) \(120\div 3=40\)

5162064

Continue the pattern by writing the next division equation. \(216\div 40=5\text{ R }16\) \(175\div 40=4\text{ R }15\) \(134\div 40=3\text{ R }14\) \(93\div 40=2\text{ R }13\) \(52\div 40=1\text{ R }12\)

Hints

- Compare consecutive dividends. - Track the quotient and remainder patterns separately. - Identify the value that stays constant. - Use all three patterns to write the next equation.

Solution

1. The dividends decrease by \(41\), so the next dividend is \(52-41=11\). 2. The divisor remains \(40\). 3. The quotients decrease by \(1\), so the next quotient is \(0\). 4. The remainders decrease by \(1\), so the next remainder is \(11\). 5. Therefore, \(11\div 40=0\text{ R }11\).

Answer

\(11\div 40=0\text{ R }11\)

5162074

Find the rule in this division pattern and write the next two equations. \(50\div 5=10\) \(110\div 10=11\) \(180\div 15=12\) \(260\div 20=13\) \(350\div 25=14\)

Hints

- Examine how the divisors change. - Examine how the quotients change. - Choose the next divisor and quotient first. - Use multiplication to find the matching dividend.

Solution

1. The divisors increase by \(5\), so the next divisors are \(30\) and \(35\). 2. The quotients increase by \(1\), so the next quotients are \(15\) and \(16\). 3. Use multiplication to find the dividends: \(15\times 30=450\) and \(16\times 35=560\). 4. The next equations are \(450\div 30=15\) and \(560\div 35=16\).

Answer

\(450\div 30=15\) \(560\div 35=16\)

5162314

Determine which division expressions have the same quotient. Make a prediction from the relationships among the numbers, explain it, and then verify each quotient. a) \(744\div 6\) b) \(372\div 3\) c) \(744\div 3\) d) \(248\div 2\)

Hints

- Compare \(744\) with \(372\) and \(6\) with \(3\). - Look for a similar relationship between parts a and d. - Verify the predicted quotients with division.

Solution

1. In part b, both the dividend and divisor from part a are halved, so the quotient remains the same. 2. In part d, both the dividend and divisor from part a are divided by \(3\), so the quotient again remains the same. 3. Verify: \(744\div 6=124\), \(372\div 3=124\), \(744\div 3=248\), and \(248\div 2=124\). 4. Therefore, parts a, b, and d have the same quotient.

Answer

a), b), and d) have the same quotient, \(124\). Part c equals \(248\).

5162334

Calculate and compare the quotients. How are the changes in the dividends related to each divisor? a) \(560\div 7\), \(567\div 7\), \(553\div 7\) b) \(420\div 6\), \(426\div 6\), \(414\div 6\) c) \(320\div 4\), \(328\div 4\), \(312\div 4\)

Hints

- Calculate the first quotient in each row. - Compare the dividends within a row. - Express each dividend difference as a multiple of the divisor. - Use that relationship to predict the other quotients.

Solution

1. Part a gives \(80,81,79\). Adding or subtracting one divisor changes the quotient by \(1\). 2. Part b gives \(70,71,69\). Again, adding or subtracting one divisor changes the quotient by \(1\). 3. Part c gives \(80,82,78\). Adding or subtracting \(8\), which is twice the divisor, changes the quotient by \(2\). 4. In general, changing the dividend by a multiple of the divisor changes the quotient by that same multiple.

Answer

a) \(80,81,79\) b) \(70,71,69\) c) \(80,82,78\) Changing the dividend by a multiple of the divisor changes the quotient by the same multiple.

5162344

Calculate each group. What do you notice about the quotients? a) \(640\div 8\), \(320\div 4\), \(160\div 2\) b) \(420\div 6\), \(210\div 3\), \(70\div 1\) c) \(240\div 6\), \(120\div 3\), \(40\div 1\)

Hints

- Compare how the dividend and divisor change from one expression to the next. - Calculate the first quotient in each group. - Predict the other quotients before calculating. - State a rule that explains why the quotient stays the same.

Solution

1. Part a gives \(80,80,80\). 2. Part b gives \(70,70,70\). 3. Part c gives \(40,40,40\). 4. Within each group, the dividend and divisor are divided by the same number. This keeps the quotient unchanged.

Answer

a) \(80,80,80\) b) \(70,70,70\) c) \(40,40,40\) Dividing the dividend and divisor by the same nonzero number does not change the quotient.

5162454

Study the division pattern and write the next equation. \(64{,}000 \div 8 = 8000\) \(56{,}000 \div 7 = 8000\) \(48{,}000 \div 6 = 8000\) \(40{,}000 \div 5 = 8000\)

Hints

- Notice what stays the same in every equation. - Track the changes in both the dividend and divisor. - Use the constant quotient to check the next equation.

Solution

1. The quotient remains \(8000\). 2. The divisor decreases by \(1\) each time, so the next divisor is \(4\). 3. The dividend decreases by \(8000\) each time, so the next dividend is \(40{,}000 - 8000 = 32{,}000\). 4. The next equation is \(32{,}000 \div 4 = 8000\).

Answer

\(32{,}000 \div 4 = 8000\)

5162464

Continue the pattern by writing the next equation. \(21{,}000 \div 3 = 7000\) \(32{,}000 \div 4 = 8000\) \(45{,}000 \div 5 = 9000\) \(60{,}000 \div 6 = 10{,}000\)

Hints

- Find the pattern in the divisors. - Find the pattern in the quotients. - Multiply the new divisor by the new quotient to find the dividend.

Solution

1. The divisor increases by \(1\), so the next divisor is \(7\). 2. The quotient increases by \(1000\), so the next quotient is \(11{,}000\). 3. Multiply to find the next dividend: \(7 \times 11{,}000 = 77{,}000\). 4. The next equation is \(77{,}000 \div 7 = 11{,}000\).

Answer

\(77{,}000 \div 7 = 11{,}000\)

5162574

Continue the number pattern with three more numbers and explain the rule. \(850{,}000, 800{,}000, 825{,}000, 775{,}000, 800{,}000, \ldots\)

Hints

- The direction changes from one step to the next. - Examine the first two changes separately. - Look for a repeating two-step rule.

Solution

1. The first change is \(-50{,}000\), and the second change is \(+25{,}000\). 2. These two changes repeat: \(825{,}000 - 50{,}000 = 775{,}000\) and \(775{,}000 + 25{,}000 = 800{,}000\). 3. Continue with \(-50{,}000\): \(800{,}000 - 50{,}000 = 750{,}000\). 4. Then add \(25{,}000\): \(750{,}000 + 25{,}000 = 775{,}000\). 5. Then subtract \(50{,}000\): \(775{,}000 - 50{,}000 = 725{,}000\).

Answer

Next numbers: \(750{,}000, 775{,}000, 725{,}000\) Rule: Alternate subtracting \(50{,}000\) and adding \(25{,}000\).

5162614

Find the rule and complete each number pattern. a) \(200{,}000, 350{,}000, 500{,}000, \square, \square, 950{,}000\) b) \(900{,}000, 820{,}000, 740{,}000, \square, \square, \square, 420{,}000\) c) \(125{,}000, 250{,}000, 500{,}000, \square\)

Hints

- Ignore the zeros at first and compare the leading values. - Find the difference between consecutive terms in a) and b). - In c), compare each term with the one before it.

Solution

1. For a), add \(150{,}000\) each time. The missing terms are \(650{,}000\) and \(800{,}000\), followed by the given term \(950{,}000\). 2. For b), subtract \(80{,}000\) each time. The missing terms are \(660{,}000\), \(580{,}000\), and \(500{,}000\), followed by the given term \(420{,}000\). 3. For c), multiply by \(2\) each time, so the missing term is \(1{,}000{,}000\).

Answer

a) Rule: add \(150{,}000\); missing terms: \(650{,}000, 800{,}000\) b) Rule: subtract \(80{,}000\); missing terms: \(660{,}000, 580{,}000, 500{,}000\) c) Rule: multiply by \(2\); missing term: \(1{,}000{,}000\)

5162724

Two terms are missing from this number pattern. Fill in the blanks. \(100{,}000, 120{,}000, 150{,}000, \square, 240{,}000, 300{,}000, \square\)

Hints

- Find the differences between consecutive known terms. - Examine how the differences themselves change. - Use the pattern of differences on both sides of each blank. - Check that every step follows the same rule.

Solution

1. The first differences are \(20{,}000\) and \(30{,}000\). 2. The amount added increases by \(10{,}000\) each time. 3. Add \(40{,}000\) to \(150{,}000\) to get \(190{,}000\). 4. Then the additions are \(50{,}000\), \(60{,}000\), and \(70{,}000\), so the final missing term is \(300{,}000 + 70{,}000 = 370{,}000\).

Answer

\(190{,}000\) and \(370{,}000\)

5164004

How many steps of the given size are needed to reach the next thousand? a) Start: \(342{,}400\); step size: \(200\) b) Start: \(712{,}750\); step size: \(50\) c) Start: \(899{,}920\); step size: \(20\)

Hints

- First identify the next thousand after each starting number. - Find the distance from the start to that target. - Divide the distance by the step size.

Solution

1. For a), the next thousand is \(343{,}000\). The distance is \(600\), and \(600 \div 200 = 3\) steps. 2. For b), the next thousand is \(713{,}000\). The distance is \(250\), and \(250 \div 50 = 5\) steps. 3. For c), the next thousand is \(900{,}000\). The distance is \(80\), and \(80 \div 20 = 4\) steps.

Answer

a) \(3\) steps b) \(5\) steps c) \(4\) steps

5164024

Complete the table so that each row lands exactly on the target, which is the next thousand. <table> <tr><th>Start</th><th>Step size</th><th>Target</th><th>Number of steps</th></tr> <tr><td>\(451{,}200\)</td><td>\(200\)</td><td>\(452{,}000\)</td><td>?</td></tr> <tr><td>\(123{,}750\)</td><td>?</td><td>\(124{,}000\)</td><td>\(5\)</td></tr> <tr><td>?</td><td>\(10\)</td><td>\(888{,}000\)</td><td>\(3\)</td></tr> </table>

Hints

- A different value is missing in each row. - To find a start, work backward from the target. - To find a step size, divide the total distance by the number of steps. - Check each completed row by counting forward.

Solution

1. Row 1: The distance is \(452{,}000 - 451{,}200 = 800\). Then \(800 \div 200 = 4\) steps. 2. Row 2: The distance is \(124{,}000 - 123{,}750 = 250\). Then \(250 \div 5 = 50\), so the step size is \(50\). 3. Row 3: Three steps of \(10\) cover \(3 \times 10 = 30\). Count backward from the target: \(888{,}000 - 30 = 887{,}970\).

Answer

Row 1: \(4\) steps Row 2: step size \(50\) Row 3: start \(887{,}970\)

5164314

In this number pattern, each step adds \(100{,}000 + 10{,}000 + 1000 + 100 + 10 + 1\). a) Continue with two more numbers: \(123{,}456 \to 234{,}567 \to \square \to \square\) b) Find the next number: \(489{,}000 \to 600{,}111 \to \square\) c) How many times must \(111{,}111\) be added to move from \(222{,}222\) to \(888{,}888\)?

Hints

- Combine the place-value amounts to find one complete step. - Add the same amount at every step. - For part c, track repeated additions from the starting number to the target.

Solution

1. The amount added each time is \(111{,}111\). 2. For a), \(234{,}567 + 111{,}111 = 345{,}678\), and \(345{,}678 + 111{,}111 = 456{,}789\). 3. For b), \(600{,}111 + 111{,}111 = 711{,}222\). 4. For c), repeated addition gives \(333{,}333\), \(444{,}444\), \(555{,}555\), \(666{,}666\), \(777{,}777\), and \(888{,}888\), so the amount is added \(6\) times.

Answer

a) \(345{,}678\) and \(456{,}789\) b) \(711{,}222\) c) \(6\) times

5164704

Evaluate each set of division expressions. Then describe the pattern in the quotients. a) \(1{,}000{,}000 \div 1\) \(1{,}000{,}000 \div 10\) \(1{,}000{,}000 \div 100\) \(1{,}000{,}000 \div 1000\) b) \(100{,}000 \div 2\) \(100{,}000 \div 4\) \(100{,}000 \div 8\)

Hints

- In part a, compare the number of zeros in each quotient. - In part b, compare how the divisors change and what happens to the quotients.

Solution

1. For a), the quotients are \(1{,}000{,}000\), \(100{,}000\), \(10{,}000\), and \(1000\). 2. Each divisor in a) is \(10\) times the previous divisor, so each quotient is one-tenth of the previous quotient. 3. For b), the quotients are \(50{,}000\), \(25{,}000\), and \(12{,}500\). 4. Each divisor in b) doubles, so each quotient is half the previous quotient.

Answer

a) \(1{,}000{,}000, 100{,}000, 10{,}000, 1000\); each quotient is one-tenth of the previous quotient. b) \(50{,}000, 25{,}000, 12{,}500\); each quotient is half the previous quotient.

5165814

Find the step size in each pattern and write the next three terms. a) \(0, 150{,}000, 300{,}000, \ldots\) b) \(1{,}000{,}000, 925{,}000, 850{,}000, \ldots\) c) \(500{,}000, 575{,}000, 650{,}000, \ldots\)

Hints

- Find the difference between two consecutive terms. - Decide whether each pattern increases or decreases. - Apply the same step three more times.

Solution

1. For a), the step is \(+150{,}000\). The next terms are \(450{,}000\), \(600{,}000\), and \(750{,}000\). 2. For b), the step is \(-75{,}000\). The next terms are \(775{,}000\), \(700{,}000\), and \(625{,}000\). 3. For c), the step is \(+75{,}000\). The next terms are \(725{,}000\), \(800{,}000\), and \(875{,}000\).

Answer

a) Step: \(+150{,}000\); next terms: \(450{,}000, 600{,}000, 750{,}000\) b) Step: \(-75{,}000\); next terms: \(775{,}000, 700{,}000, 625{,}000\) c) Step: \(+75{,}000\); next terms: \(725{,}000, 800{,}000, 875{,}000\)

5165824

Luke and Maya count within \(1{,}000{,}000\). Luke counts forward by \(250{,}000\), starting at \(0\). Maya counts backward by \(100{,}000\), starting at \(1{,}000{,}000\). Which numbers appear in both counting patterns?

Hints

- Write each counting pattern completely. - Compare the two lists. - Mark every number that appears in both.

Solution

1. Luke's pattern is \(0, 250{,}000, 500{,}000, 750{,}000, 1{,}000{,}000\). 2. Maya's pattern contains every multiple of \(100{,}000\) from \(1{,}000{,}000\) down to \(0\). 3. The numbers in both patterns are \(0\), \(500{,}000\), and \(1{,}000{,}000\).

Answer

\(0\), \(500{,}000\), and \(1{,}000{,}000\)

5166504

Study the pattern. Continue it for two more rows, and describe the sums. \(12{,}121+21{,}212=33{,}333\) \(23{,}232+21{,}212=44{,}444\) \(34{,}343+21{,}212=55{,}555\)

Hints

- Compare the first addends from one row to the next. - Use the same increase to predict the next first addend and sum. - Look at the digits in each sum. - Describe what all the sums have in common.

Solution

1. The first addend increases by \(11{,}111\) in each row, so the sum increases by \(11{,}111\). 2. The next two rows are \(45{,}454+21{,}212=66{,}666\) and \(56{,}565+21{,}212=77{,}777\). 3. Each sum has the same digit in all five places.

Answer

\(45{,}454+21{,}212=66{,}666\) \(56{,}565+21{,}212=77{,}777\) Every sum has one digit repeated in all five places.

5166514

Study the pattern. Write the next equation before calculating it. Then explain why all the sums are equal. \(12{,}345+87{,}654=99{,}999\) \(23{,}456+76{,}543=99{,}999\) \(34{,}567+65{,}432=99{,}999\)

Hints

- Compare the first addends in consecutive rows. - Compare the second addends in consecutive rows. - Check whether the two changes have the same size. - Use those changes to write the next equation.

Solution

1. The first addend increases by \(11{,}111\), while the second addend decreases by \(11{,}111\). 2. The next equation is \(45{,}678+54{,}321=99{,}999\). 3. Because one addend increases by exactly the amount the other decreases, the sum remains unchanged.

Answer

\(45{,}678+54{,}321=99{,}999\) Each time, one addend increases by \(11{,}111\) and the other decreases by \(11{,}111\), so the sum stays \(99{,}999\).

5166644

Study the pattern. Continue it for two more rows, and describe how the digits change. \(4321-2889=1432\) \(5432-2889=2543\) \(6543-2889=3654\) \(7654-2889=4765\)

Hints

- Compare consecutive minuends and consecutive differences. - Use the common increase to predict the next two rows. - Compare the digit order in each minuend and its difference. - Describe the repeated digit movement.

Solution

1. Each minuend increases by \(1111\), so each difference also increases by \(1111\). 2. The next two rows are \(8765-2889=5876\) and \(9876-2889=6987\). 3. In each row, the last digit of the minuend becomes the first digit of the difference, and the other three digits shift one place to the right.

Answer

\(8765-2889=5876\) \(9876-2889=6987\) The last digit of each minuend moves to the front of the difference, while the other digits shift one place to the right.

5167104

Complete each set of equations. Then describe how the quotients change. a) \(48{,}000\div 2=\square\) \(48{,}000\div 4=\square\) \(48{,}000\div 6=\square\) \(48{,}000\div 8=\square\) b) \(12{,}000\div 3=\square\) \(24{,}000\div 3=\square\) \(36{,}000\div 3=\square\) \(48{,}000\div 3=\square\)

Hints

- Identify which number stays fixed in each part. - Use basic division facts, then account for the zeros. - Compare consecutive dividends, divisors, and quotients. - Describe each pattern separately.

Solution

1. In part a, the dividend stays \(48{,}000\) while the divisor increases. The quotients are \(24{,}000,12{,}000,8000,6000\). 2. In part b, the divisor stays \(3\) while the dividend increases by \(12{,}000\). The quotients are \(4000,8000,12{,}000,16{,}000\). 3. In part a, larger divisors produce smaller quotients. In part b, each increase of \(12{,}000\) in the dividend increases the quotient by \(4000\).

Answer

a) \(24{,}000,12{,}000,8000,6000\) b) \(4000,8000,12{,}000,16{,}000\) In part a, the quotients decrease as the divisors increase. In part b, the quotients increase by \(4000\) each time.

5169304

Continue the sequence with one more equation, calculate all the quotients, and describe the pattern. \(1112\div 8\) \(2224\div 8\) \(3336\div 8\) \(\square\)

Hints

- Determine how the dividend changes from one equation to the next. - Use the same change to write the next dividend. - Compare consecutive quotients. - Relate the change in the quotients to \(1112\div 8\).

Solution

1. The first three quotients are \(1112\div 8=139\), \(2224\div 8=278\), and \(3336\div 8=417\). 2. The dividend increases by \(1112\), so the next equation is \(4448\div 8=556\). 3. Because \(1112\div 8=139\), each increase of \(1112\) in the dividend increases the quotient by \(139\).

Answer

The quotients are \(139,278,417\). The next equation is \(4448\div 8=556\). The quotient increases by \(139\) each time.

5170454

Study the equation pattern. \(1 \times 9 + 2 = 11\) \(12 \times 9 + 3 = 111\) \(123 \times 9 + 4 = 1111\) a) Write and evaluate the fourth equation. b) Write and evaluate the fifth equation.

Hints

- Notice which digit is appended to the first factor in each row. - Track how the added number changes. - Use the pattern in the results to check your calculations.

Solution

1. Each new first factor appends the next digit: \(1, 12, 123, 1234, 12{,}345\). The added number also increases by \(1\). 2. The fourth equation is \(1234 \times 9 + 5\). Since \(1234 \times 9 = 11{,}106\), the result is \(11{,}111\). 3. The fifth equation is \(12{,}345 \times 9 + 6\). Since \(12{,}345 \times 9 = 111{,}105\), the result is \(111{,}111\).

Answer

a) \(1234 \times 9 + 5 = 11{,}111\) b) \(12{,}345 \times 9 + 6 = 111{,}111\)

5170464

Study the number pattern. \(6 + 6 = 12\) \(66 + 66 = 132\) \(666 + 666 = 1332\) \(6666 + 6666 = 13{,}332\) a) What is the next equation and result? b) Describe how the result changes when one more digit \(6\) is added to each addend.

Hints

- Calculate the next equation to test your prediction. - Identify which digits appear in every result. - Compare the number of \(6\)s in an addend with the number of \(3\)s in the result.

Solution

1. The next equation is \(66{,}666 + 66{,}666 = 133{,}332\). 2. Every result begins with \(1\), ends with \(2\), and has only \(3\)s between them. 3. Each added \(6\) in the addends creates one additional \(3\) in the middle of the result.

Answer

a) \(66{,}666 + 66{,}666 = 133{,}332\) b) The result begins with \(1\) and ends with \(2\); each additional \(6\) adds another \(3\) between them.

5170664

Choose two different digits from \(1\) through \(9\), and call them \(a\) and \(b\). a) Form the two four-digit palindromes \(abba\) and \(baab\). For example, the digits \(3\) and \(8\) form \(3883\) and \(8338\). b) Subtract the smaller palindrome from the larger one. c) Repeat the process with two other pairs of digits. d) Find the sum of the digits in each of your three differences. What do you notice?

Hints

- Keep the outside digits equal and the inside digits equal when forming each palindrome. - Align the numbers by place value before subtracting. - Add every digit in each difference. - Test digit pairs with both small and large differences.

Solution

1. Using \(3\) and \(8\), \(8338-3883=4455\), and \(4+4+5+5=18\). 2. Using \(1\) and \(2\), \(2112-1221=891\), and \(8+9+1=18\). 3. Using \(4\) and \(9\), \(9449-4994=4455\), and the sum of the digits is \(18\). 4. For every valid pair, the sum of the digits in the difference is \(18\).

Answer

The sum of the digits in each difference is always \(18\).

5170684

Use two different digits to form the four-digit palindromes \(abba\) and \(baab\). a) Find the sums for the digit pairs \((2,5)\) and \((3,4)\). What do you notice? b) Which two different digits from \(1\) through \(9\) make the sum exactly \(8888\)? Find all three pairs.

Hints

- Add corresponding digits in the two palindromes. - Compare the repeated result digit with the two chosen digits. - For part b, list pairs of different positive digits whose sum is \(8\).

Solution

1. For \((2,5)\), \(2552+5225=7777\). 2. For \((3,4)\), \(3443+4334=7777\). 3. In each place, the result digit is the sum of the two chosen digits, provided that sum is less than \(10\). 4. To obtain \(8888\), the two digits must add to \(8\). 5. The three pairs of different positive digits are \((1,7)\), \((2,6)\), and \((3,5)\).

Answer

a) Both sums are \(7777\). Each result has the same digit in all four places. b) The pairs are \((1,7)\), \((2,6)\), and \((3,5)\).

5170704

A researcher multiplied \(606\) by several one-digit numbers and obtained \(1212\), \(1818\), \(2424\), and \(3030\). 1. Which one-digit factor produced each result? 2. Predict \(606\times 7\) using the pattern.

Hints

- Split each result into two equal two-digit blocks. - Relate each block to a multiplication fact for \(6\). - Use the same relationship for a factor of \(7\).

Solution

1. Each result repeats a two-digit block. Dividing the block values \(12,18,24,30\) by \(6\) gives the factors \(2,3,4,5\). 2. For a factor of \(7\), \(6\times 7=42\). 3. Repeating the block \(42\) gives \(4242\), so \(606\times 7=4242\).

Answer

1. The factors are \(2,3,4,5\), respectively. 2. \(606\times 7=4242\).

5170734

Multiply \(495\) by \(2\), \(3\), \(4\), and \(5\). a) Calculate the four products. b) Find the sum of the digits in each product. What do you notice? c) For each four-digit product, add the first and third digits, and then add the second and fourth digits. What do you notice?

Hints

- Add every digit in each product for part b). - For part c), pair digits that are two places apart. - Compare the two pair sums within each four-digit product.

Solution

1. The products are \(495\times 2=990\), \(495\times 3=1485\), \(495\times 4=1980\), and \(495\times 5=2475\). 2. The sums of the digits are \(9+9+0=18\), \(1+4+8+5=18\), \(1+9+8+0=18\), and \(2+4+7+5=18\). 3. In \(1485\), \(1+8=9\) and \(4+5=9\). The same relationship holds for \(1980\) and \(2475\).

Answer

a) \(990,1485,1980,2475\) b) The sum of the digits is always \(18\). c) In each four-digit product, the first and third digits add to \(9\), and the second and fourth digits also add to \(9\).

5170754

A hundreds chart lists the numbers from \(1\) through \(100\) in ten rows. <table><tbody><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td></tr><tr><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td></tr><tr><td>21</td><td>22</td><td>23</td><td>24</td><td>25</td><td>26</td><td>27</td><td>28</td><td>29</td><td>30</td></tr><tr><td>31</td><td>32</td><td>33</td><td>34</td><td>35</td><td>36</td><td>37</td><td>38</td><td>39</td><td>40</td></tr><tr><td>41</td><td>42</td><td>43</td><td>44</td><td>45</td><td>46</td><td>47</td><td>48</td><td>49</td><td>50</td></tr><tr><td>51</td><td>52</td><td>53</td><td>54</td><td>55</td><td>56</td><td>57</td><td>58</td><td>59</td><td>60</td></tr><tr><td>61</td><td>62</td><td>63</td><td>64</td><td>65</td><td>66</td><td>67</td><td>68</td><td>69</td><td>70</td></tr><tr><td>71</td><td>72</td><td>73</td><td>74</td><td>75</td><td>76</td><td>77</td><td>78</td><td>79</td><td>80</td></tr><tr><td>81</td><td>82</td><td>83</td><td>84</td><td>85</td><td>86</td><td>87</td><td>88</td><td>89</td><td>90</td></tr><tr><td>91</td><td>92</td><td>93</td><td>94</td><td>95</td><td>96</td><td>97</td><td>98</td><td>99</td><td>100</td></tr></tbody></table> 1. Find \(1+100\). 2. Find \(2+99\). 3. Find \(3+98\). 4. What pattern do you notice? How many such pairs can be formed from all \(100\) numbers? 5. Use the pattern to find the sum of all the numbers from \(1\) through \(100\).

Hints

- Pair the smallest unused number with the largest unused number. - Compare the sums of the first few pairs. - Divide the total number of entries by \(2\) to count the pairs. - Multiply the number of pairs by the common pair sum.

Solution

1. The first pair has sum \(1+100=101\). 2. The second pair has sum \(2+99=101\). 3. The third pair has sum \(3+98=101\). 4. Every smallest-largest pair has sum \(101\). The \(100\) numbers form \(50\) pairs. 5. Therefore, the total is \(50\times 101=50\times 100+50=5050\).

Answer

1. \(101\) 2. \(101\) 3. \(101\) 4. Every pair sums to \(101\), and there are \(50\) pairs. 5. The total is \(5050\).

5170784

A four-digit repeating-block number has the form \(abab\). Swapping the digits gives \(baba\). a) Use the digits \(3\) and \(7\) to form \(3737\) and \(7373\). Subtract the smaller number from the larger number. b) Use two other digits that also differ by \(4\), such as \(1\) and \(5\). Find the new difference. What do you notice? c) Use the digits \(2\) and \(3\). How many times as large is the result from part a) as this new difference?

Hints

- Keep the digit order \(abab\) or \(baba\) when forming each number. - Compare the difference between the two chosen digits. - Relate \(909\) to \(3636\) using multiplication.

Solution

1. Part a gives \(7373-3737=3636\). 2. Part b gives \(5151-1515=3636\). Equal digit differences produce equal number differences. 3. Part c gives \(3232-2323=909\). 4. Since \(909\times 4=3636\), the result from part a is four times the result from part c.

Answer

a) \(3636\) b) \(3636\); the differences are equal. c) The new difference is \(909\), and \(3636\) is \(4\) times as large.

5170804

Investigate sums of four-digit repeating-block numbers of the forms \(abab\) and \(baba\). a) Use \(a=2\) and \(b=5\). Find \(2525+5252\). b) Use \(a=1\) and \(b=8\). Find \(1818+8181\). c) A student claims, “If \(a+b=10\), then \(abab+baba\) is always \(11{,}110\).” Test the claim using the digit pairs \((3,7)\) and \((4,6)\).

Hints

- Align the numbers by place value. - Compare the repeated result digit with \(a+b\). - Watch for regrouping when \(a+b=10\). - Test both required digit pairs before deciding about the claim.

Solution

1. Part a gives \(2525+5252=7777\). 2. Part b gives \(1818+8181=9999\). 3. For \((3,7)\), \(3737+7373=11{,}110\). 4. For \((4,6)\), \(4646+6464=11{,}110\). 5. Both tests support the claim. In general, the sum is \(1111(a+b)\), so when \(a+b=10\), the sum is \(11{,}110\).

Answer

a) \(7777\) b) \(9999\) c) The claim is supported: \(3737+7373=11{,}110\) and \(4646+6464=11{,}110\).

5170834

Find the sum of all four-digit palindromes that begin and end with \(1\): \(1001,1111,1221,1331,1441,1551,1661,1771,1881,1991\)

Hints

- Pair the first number with the last number. - Continue pairing numbers from the outside inward. - Check whether all pair sums are equal. - Multiply the common pair sum by the number of pairs.

Solution

1. Pair the first and last numbers, then work inward: \(1001+1991=2992\), \(1111+1881=2992\), \(1221+1771=2992\), \(1331+1661=2992\), and \(1441+1551=2992\). 2. There are \(5\) pairs. 3. The total is \(5\times 2992=14{,}960\).

Answer

\(14{,}960\)

5170854

In a sequence, each term is the sum of the two previous terms. The third term is \(12{,}500\), and the fourth term is \(20{,}000\). Find the first, second, and fifth terms.

Hints

- Use subtraction to work backward from a known sum. - The fifth term can be found by applying the rule directly. - Each earlier term equals a later term minus the term beside it.

Solution

1. The fifth term is \(12{,}500 + 20{,}000 = 32{,}500\). 2. The second term is \(20{,}000 - 12{,}500 = 7500\), because the second and third terms add to the fourth. 3. The first term is \(12{,}500 - 7500 = 5000\), because the first and second terms add to the third.

Answer

First term: \(5000\) Second term: \(7500\) Fifth term: \(32{,}500\)

5170904

Each sequence follows the rule that every new term is the sum of the two previous terms. The second term is always \(20{,}000\), while the first term increases by \(1000\) from one sequence to the next. Sequence 1: \(10{,}000, 20{,}000, \ldots\) Sequence 2: \(11{,}000, 20{,}000, \ldots\) Sequence 3: \(12{,}000, 20{,}000, \ldots\) Sequence 4: \(13{,}000, 20{,}000, \ldots\) Find the fifth term of each sequence. How do the fifth terms change?

Hints

- Continue each sequence to its fifth term. - Compare the four fifth terms. - Relate the change in the starting term to the change in the fifth term.

Solution

1. Sequence 1 continues \(30{,}000, 50{,}000, 80{,}000\), so its fifth term is \(80{,}000\). 2. Sequence 2 continues \(31{,}000, 51{,}000, 82{,}000\). 3. Sequence 3 continues \(32{,}000, 52{,}000, 84{,}000\). 4. Sequence 4 continues \(33{,}000, 53{,}000, 86{,}000\). 5. The fifth terms increase by \(2000\) from one sequence to the next.

Answer

Fifth terms: \(80{,}000, 82{,}000, 84{,}000, 86{,}000\) They increase by \(2000\) each time.

5170914

Continue each sequence through the fifth term. Every new term is the sum of the two previous terms. The first term is always \(50{,}000\), while the second term increases by \(10{,}000\). a) \(50{,}000, 10{,}000, \ldots\) b) \(50{,}000, 20{,}000, \ldots\) c) \(50{,}000, 30{,}000, \ldots\) d) \(50{,}000, 40{,}000, \ldots\) By how much does the fifth term increase from one sequence to the next?

Hints

- Add the two most recent terms to find each new term. - Compare only the fifth terms after completing all four sequences. - Determine how the change in the second starting term affects the fifth term.

Solution

1. Sequence a) is \(50{,}000, 10{,}000, 60{,}000, 70{,}000, 130{,}000\). 2. Sequence b) is \(50{,}000, 20{,}000, 70{,}000, 90{,}000, 160{,}000\). 3. Sequence c) is \(50{,}000, 30{,}000, 80{,}000, 110{,}000, 190{,}000\). 4. Sequence d) is \(50{,}000, 40{,}000, 90{,}000, 130{,}000, 220{,}000\). 5. The fifth terms differ by \(30{,}000\) each time.

Answer

The fifth terms are \(130{,}000, 160{,}000, 190{,}000\), and \(220{,}000\). They increase by \(30{,}000\) each time.

5173524

Continue the number pattern with the next five terms. Explain the repeating rule. \(5; 15; 10; 30; 25; 75; \ldots\)

Hints

- The same operation does not happen at every step. - Compare the first-to-second change with the second-to-third change. - Look for two operations that repeat in order.

Solution

1. The pattern alternates between multiplying by \(3\) and subtracting \(5\): \(5\times3=15\), \(15-5=10\), \(10\times3=30\), \(30-5=25\), and \(25\times3=75\). 2. Continue the rule: \(75-5=70\), \(70\times3=210\), \(210-5=205\), \(205\times3=615\), and \(615-5=610\).

Answer

Next five terms: \(70; 210; 205; 615; 610\) Rule: Alternate subtracting \(5\) and multiplying by \(3\).

5175964

Create a number pattern using these rules: - If the current number is odd, multiply it by \(3\) and add \(1\). - If the current number is even, divide it by \(2\). Start with \(13\). List every term until the pattern first reaches \(1\). How many operations are needed?

Hints

- Decide whether each current term is odd or even before choosing a rule. - Record every new term so none are skipped. - Count operations, not terms.

Solution

1. Apply the rule to each term: \(13\times3+1=40\), \(40\div2=20\), \(20\div2=10\), \(10\div2=5\), \(5\times3+1=16\), \(16\div2=8\), \(8\div2=4\), \(4\div2=2\), and \(2\div2=1\). 2. The sequence is \(13, 40, 20, 10, 5, 16, 8, 4, 2, 1\). 3. There are \(9\) operations between the starting term and \(1\).

Answer

Sequence: \(13, 40, 20, 10, 5, 16, 8, 4, 2, 1\) Operations: \(9\)

5175974

Create a number pattern using these rules: - If the number is less than \(20\), double it. - If the number is \(20\) or greater, subtract \(7\). Start with \(5\). Continue until a number appears for the second time. List the terms through that repetition. Which number repeats first?

Hints

- Use the second rule when the term is exactly \(20\). - Keep a complete list of the terms. - Compare each new term with every earlier term.

Solution

1. Apply the correct rule to each term: \(5\times2=10\), \(10\times2=20\), \(20-7=13\), \(13\times2=26\), \(26-7=19\), \(19\times2=38\), \(38-7=31\), \(31-7=24\), \(24-7=17\), \(17\times2=34\), \(34-7=27\), and \(27-7=20\). 2. The sequence is \(5, 10, 20, 13, 26, 19, 38, 31, 24, 17, 34, 27, 20\). 3. The first repeated number is \(20\).

Answer

Sequence: \(5, 10, 20, 13, 26, 19, 38, 31, 24, 17, 34, 27, 20\) First repeated number: \(20\)

5175984

A number machine uses these rules: - If a number ends in \(0\), remove the final zero. This is the same as dividing by \(10\). - If a number does not end in \(0\), add \(6\). a) Starting with \(14\), list the first eight terms. b) Describe the pattern you notice.

Hints

- Count the starting value as the first term. - Apply exactly one rule to create each next term. - Look for the first term that appears again.

Solution

1. Start with \(14\). Apply the rules to get \(14+6=20\), \(20\div10=2\), \(2+6=8\), \(8+6=14\), \(14+6=20\), \(20\div10=2\), and \(2+6=8\). 2. The first eight terms are \(14, 20, 2, 8, 14, 20, 2, 8\). 3. The four-term cycle \(14, 20, 2, 8\) repeats.

Answer

a) \(14, 20, 2, 8, 14, 20, 2, 8\) b) The pattern repeats every four terms.

5185114

Study the multiplication pattern. \(9 \times 7\) \(9 \times 77\) \(9 \times 777\) a) Calculate the three products. b) Describe the digit pattern in the products. c) Without using the standard multiplication algorithm, write the fifth equation and product in the pattern.

Hints

- Compare the first and last digits of the products. - Count the \(7\)s in each factor and the \(9\)s in each product. - Extend the digit rule through the fourth equation before writing the fifth.

Solution

1. The first three products are \(9 \times 7 = 63\), \(9 \times 77 = 693\), and \(9 \times 777 = 6993\). 2. Each product begins with \(6\), ends with \(3\), and has \(9\)s between them. The number of \(9\)s is one less than the number of \(7\)s in the second factor. 3. The fifth second factor is \(77{,}777\), so the fifth equation is \(9 \times 77{,}777 = 699{,}993\).

Answer

a) \(63, 693, 6993\) b) Each product begins with \(6\), ends with \(3\), and gains one more \(9\) in the middle. c) \(9 \times 77{,}777 = 699{,}993\)

5185294

Study this sequence of sums: 1. \(1\) 2. \(1+2\) 3. \(1+2+3\) 4. \(1+2+3+4\) a) Find the results for rows \(1\) through \(6\). b) What is the result in row \(10\)? c) In which row is the result \(55\)?

Hints

- Each new row adds one more whole number. - Use the previous result instead of starting the sum again. - For row \(10\), pair numbers from the beginning and end when helpful.

Solution

1. a) The results are \(1, 3, 6, 10, 15\), and \(21\). Each row adds the next whole number to the previous result. 2. b) Row \(10\) is \(1+2+3+4+5+6+7+8+9+10=55\). 3. c) Since row \(10\) has a result of \(55\), the answer is row \(10\).

Answer

a) \(1, 3, 6, 10, 15, 21\) b) \(55\) c) Row \(10\)

5193034

A number machine multiplies each digit by itself and then adds the products. For example, \(13\) becomes \(1\times1+3\times3=10\), and \(10\) becomes \(1\times1+0\times0=1\). Start with \(49\). Repeat the rule until you reach \(1\). List the number chain and state how many steps it takes.

Hints

- Apply the rule to every digit in the current number. - Record each result before repeating the rule. - Count the operations between numbers, not the number of terms.

Solution

1. \(4\times4+9\times9=16+81=97\). 2. \(9\times9+7\times7=81+49=130\). 3. \(1\times1+3\times3+0\times0=1+9+0=10\). 4. \(1\times1+0\times0=1\). 5. The chain is \(49\rightarrow97\rightarrow130\rightarrow10\rightarrow1\), which takes \(4\) steps.

Answer

Chain: \(49\rightarrow97\rightarrow130\rightarrow10\rightarrow1\) Steps: \(4\)

5193044

Use this number-machine rule: multiply each digit by itself and add the products. Start with \(20\) and apply the rule eight times. List the chain. What do you notice after the eighth step?

Hints

- Apply the rule to each digit, including any zero. - Keep the terms in order. - Compare the eighth result with the starting number.

Solution

1. Apply the rule repeatedly: \(2\times2+0\times0=4\), \(4\times4=16\), \(1\times1+6\times6=37\), \(3\times3+7\times7=58\), \(5\times5+8\times8=89\), \(8\times8+9\times9=145\), \(1\times1+4\times4+5\times5=42\), and \(4\times4+2\times2=20\). 2. The chain is \(20\rightarrow4\rightarrow16\rightarrow37\rightarrow58\rightarrow89\rightarrow145\rightarrow42\rightarrow20\). 3. After eight steps, the chain returns to \(20\), so the cycle repeats.

Answer

Chain: \(20\rightarrow4\rightarrow16\rightarrow37\rightarrow58\rightarrow89\rightarrow145\rightarrow42\rightarrow20\) The chain returns to \(20\) after eight steps and repeats.

5205534

Investigate products with \(25\) and identify a pattern. a) Calculate: \(4 \times 25 = \dots\) \(8 \times 25 = \dots\) \(12 \times 25 = \dots\) b) What happens to the product each time the first factor increases by \(4\)? c) Use the pattern from part b to find \(16 \times 25\) and \(20 \times 25\).

Hints

- Compare consecutive products and determine their difference. - Relate that difference to \(4 \times 25\). - Continue the same change for the next two first factors.

Solution

1. Part a: \(4 \times 25 = 100\), \(8 \times 25 = 200\), and \(12 \times 25 = 300\). 2. Part b: Each time the first factor increases by \(4\), the product increases by \(4 \times 25 = 100\). 3. Part c: Continue the pattern: \(16 \times 25 = 300 + 100 = 400\), and \(20 \times 25 = 400 + 100 = 500\).

Answer

a) \(100\), \(200\), \(300\) b) The product increases by \(100\) each time. c) \(16 \times 25 = 400\) and \(20 \times 25 = 500\)

5217314

Complete the table. Starting from each number, count five steps in the stated direction. <table> <tr> <th>Starting number</th> <th>Direction</th> <th>Step size</th> <th>Step 1</th> <th>Step 2</th> <th>Step 3</th> <th>Step 4</th> <th>Step 5</th> </tr> <tr> <td>\(24{,}500\)</td> <td>forward</td> <td>\(250\)</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> <tr> <td>\(10{,}200\)</td> <td>backward</td> <td>\(40\)</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </table>

Hints

- Use each result as the starting point for the next step. - Add when counting forward and subtract when counting backward. - Check that the difference between consecutive entries is constant.

Solution

1. For the first row, add \(250\) at each step: \(24{,}750, 25{,}000, 25{,}250, 25{,}500, 25{,}750\). 2. For the second row, subtract \(40\) at each step: \(10{,}160, 10{,}120, 10{,}080, 10{,}040, 10{,}000\).

Answer

First row: \(24{,}750, 25{,}000, 25{,}250, 25{,}500, 25{,}750\) Second row: \(10{,}160, 10{,}120, 10{,}080, 10{,}040, 10{,}000\)

5328294

The diagrams show a pattern of cube buildings inside increasingly large boxes. a) How many cubes are used in Building \(1\) and Building \(2\)? b) If the pattern continues by adding the same number of cubes, how many cubes will the next building have inside a \(4 \times 4 \times 4\) box?

Figure for problem 532829

Hints

- Count the cubes in each of the two buildings. - Compare the two totals to find how many cubes were added. - Add the same number of cubes once more.

Solution

1. Building \(1\) has \(2 + 1 + 1 = 4\) cubes. 2. Building \(2\) has \(3 + 1 + 1 + 1 + 1 = 7\) cubes. 3. The number of cubes increases by \(7 - 4 = 3\). Continuing that rule gives \(7 + 3 = 10\) cubes for the next building.

Answer

a) Building \(1\) has \(4\) cubes, and Building \(2\) has \(7\) cubes. b) The next building has \(10\) cubes.

5329484

Lucas builds a staircase from small cubes each day. On Day \(1\), the staircase has \(1\) cube. On Day \(2\), he adds a new step, so the staircase has \(3\) cubes. On Day \(3\), he adds another step, so the staircase has \(6\) cubes. If Lucas continues the pattern, how many cubes will the staircase have in all on Day \(5\)?

Figure for problem 532948

Hints

- Determine how many new cubes are added each day. - Look for a pattern in the numbers being added. - Extend the totals through Days \(4\) and \(5\).

Solution

1. The totals begin \(1, 3, 6\). The numbers of cubes added are \(2\) on Day \(2\) and \(3\) on Day \(3\). 2. Continue the pattern by adding \(4\) cubes on Day \(4\): \(6+4=10\). 3. Add \(5\) cubes on Day \(5\): \(10+5=15\).

Answer

On Day \(5\), the staircase has \(15\) cubes in all.

5352514

Some bricks are missing from these number walls. Which walls can be completed using whole numbers? Select them and find the missing numbers.

Figure for problem 535251

Hints

- The top brick equals the two outside bottom bricks plus twice the middle bottom brick. - Compare the top brick with the sum of the outside bottom bricks. - Subtract the outside values first, then interpret the remainder.

Solution

1. Wall a): The top brick is the sum of the two outside bottom bricks plus twice the middle bottom brick. Thus \(80 - 10 - 20 = 50\), and the middle bottom brick is \(50 \div 2 = 25\). The middle row is \(35\) and \(45\). 2. Wall b): The outside bottom bricks already sum to \(45 + 40 = 85\), which is greater than the top brick \(80\). It cannot be completed with whole numbers. 3. Wall c): \(80 - 25 - 25 = 30\), so the middle bottom brick is \(30 \div 2 = 15\). The middle row is \(40\) and \(40\). 4. Wall d): The outside bottom bricks already sum to \(41 + 42 = 83\), which is greater than \(80\). It cannot be completed with whole numbers.

Answer

Walls a) and c) can be completed. a) The middle bottom brick is \(25\), and the middle row is \(35\), \(45\). c) The middle bottom brick is \(15\), and the middle row is \(40\), \(40\). Walls b) and d) cannot be completed using whole numbers.

5353244

Compare the two number walls. If every bottom-row value is doubled, what happens to the top value?

Figure for problem 535324

Hints

- Calculate the top value in each wall. - Compare corresponding bottom-row values in a) and b). - Check whether the same relationship holds for the top values.

Solution

1. In wall a), the second row is \(10 + 25 = 35\) and \(25 + 15 = 40\), so the top is \(35 + 40 = 75\). 2. In wall b), the second row is \(20 + 50 = 70\) and \(50 + 30 = 80\), so the top is \(70 + 80 = 150\). 3. Since \(150 = 2 \times 75\), doubling every bottom-row value doubles the top value.

Answer

The top value also doubles, from \(75\) to \(150\).

5353284

Complete the number wall. Then add \(2\) to every bottom-row value. By how much does the top value increase?

Figure for problem 535328

Hints

- First calculate the original top value. - Build the new wall with \(7\) in each bottom brick. - Compare the two top values.

Solution

1. The original wall has second row \(10, 10, 10\), third row \(20, 20\), and top value \(40\). 2. After adding \(2\), the bottom row becomes \(7, 7, 7, 7\). The new rows are \(14, 14, 14\), then \(28, 28\), with \(56\) at the top. 3. The increase is \(56 - 40 = 16\). 4. Equivalently, increasing each of four bottom bricks by \(1\) increases the top by \(8\), so an increase of \(2\) raises the top by \(2 \times 8 = 16\).

Answer

The top value increases by \(16\).

5353594

Fill the bottom row so that the top brick is \(100\). The middle bottom brick must be \(40\). Find one possible solution.

Figure for problem 535359

Hints

- Determine how many times the middle bottom value contributes to the top. - Find the required sum of the two outside bottom bricks; many pairs are possible.

Solution

1. The middle bottom brick contributes twice to the top, so it contributes \(2 \times 40 = 80\). 2. The two outside bottom bricks must therefore have a sum of \(100 - 80 = 20\). 3. One possible pair is \(12\) and \(8\), giving the bottom row \(12\), \(40\), \(8\). 4. Check: the middle row is \(52\) and \(48\), and \(52 + 48 = 100\).

Answer

One possible bottom row is \(12\), \(40\), \(8\). Any two outside numbers with a sum of \(20\) also work.

5353634

Jordan says, “If I increase the bottom-left brick of a four-brick number wall by \(10\), the top value also increases by exactly \(10\).” Is Jordan correct? Calculate wall b) to test the claim.

Figure for problem 535363

Hints

- Complete wall b) from the bottom row upward. - Compare its top value with \(500\).

Solution

1. Wall a) has top value \(500\). 2. In wall b), the second row is \(110 + 50 = 160\), \(50 + 50 = 100\), and \(50 + 100 = 150\). 3. The third row is \(160 + 100 = 260\) and \(100 + 150 = 250\). 4. The new top is \(260 + 250 = 510\). 5. Since \(510 - 500 = 10\), Jordan is correct.

Answer

Yes. The top value changes from \(500\) to \(510\), an increase of \(10\).

5353844

Complete both number walls and compare their top values. What happens when one inner bottom brick is increased by \(1\)?

Figure for problem 535384

Hints

- Calculate the top value of each wall. - Trace how often the changed value contributes to bricks above it.

Solution

1. In wall a), the second row is \(20, 20, 20\), the third row is \(40, 40\), and the top is \(80\). 2. In wall b), the second row is \(21, 21, 20\), the third row is \(42, 41\), and the top is \(83\). 3. The top increases by \(83 - 80 = 3\). The changed inner brick contributes along three paths to the top.

Answer

The top value increases by \(3\), from \(80\) to \(83\).

5157064

This sequence decreases. Find the pattern and fill in the three missing terms. \(880, 878, 874, 868, \ldots, \ldots, \ldots, 824\)

Hints

- Since the terms decrease, examine how much is subtracted each time. - Compare the successive decreases. - Consider whether the amounts subtracted follow the even-number pattern.

Solution

1. The decreases are \(880-878=2\), \(878-874=4\), and \(874-868=6\). 2. The sequence subtracts consecutive even numbers: \(2, 4, 6, 8, \ldots\). 3. Continue with \(868-8=860\), \(860-10=850\), and \(850-12=838\). 4. The next step is \(838-14=824\), which matches the final term.

Answer

The missing numbers are \(860, 850,\) and \(838\). Rule: subtract consecutive even numbers \(2, 4, 6, 8, 10, 12, 14, \ldots\).

5162584

Find the rule and continue the number pattern with three more terms. \(2500, 5000, 4000, 8000, 7000, \ldots\)

Hints

- Compare the first term with the second. - Then compare the second term with the third. - Look for a repeating two-step rule.

Solution

1. The first step is \(2500 \times 2 = 5000\). 2. The second step is \(5000 - 1000 = 4000\). 3. The same two-step rule repeats: multiply by \(2\), then subtract \(1000\). 4. Continue: \(7000 \times 2 = 14{,}000\), \(14{,}000 - 1000 = 13{,}000\), and \(13{,}000 \times 2 = 26{,}000\).

Answer

Next terms: \(14{,}000, 13{,}000, 26{,}000\) Rule: Alternate multiplying by \(2\) and subtracting \(1000\).

5164324

A repeated operation adds \(1\), \(10\), \(100\), \(1000\), \(10{,}000\), and \(100{,}000\) to a starting number. a) Apply the complete operation twice to \(345{,}123\). What is the final number? b) Starting at \(100{,}000\), how many complete operations can you apply without going above \(1{,}000{,}000\)?

Hints

- First find the total amount added by one complete operation. - Applying the operation twice adds twice that amount. - For part b, add the same amount repeatedly and check against \(1{,}000{,}000\).

Solution

1. One complete operation adds \(1 + 10 + 100 + 1000 + 10{,}000 + 100{,}000 = 111{,}111\). 2. For a), two operations add \(2 \times 111{,}111 = 222{,}222\). Then \(345{,}123 + 222{,}222 = 567{,}345\). 3. For b), after \(8\) operations the number is \(100{,}000 + 888{,}888 = 988{,}888\). 4. A ninth operation would give \(1{,}099{,}999\), which is greater than \(1{,}000{,}000\). Therefore, the operation can be applied \(8\) times.

Answer

a) \(567{,}345\) b) \(8\) times

5166494

Study the pattern. Continue it for two more rows, and explain what you notice. \(10{,}203+19{,}998=30{,}201\) \(20{,}304+19{,}998=40{,}302\) \(30{,}405+19{,}998=50{,}403\)

Hints

- Compare corresponding digits in the first addends. - Find the amount added from one row to the next. - Use the same change to predict the next sums. - Compare the digit order in each first addend and its sum.

Solution

1. The first addend increases by \(10{,}101\) in each row, so the sum also increases by \(10{,}101\). 2. The next two rows are \(40{,}506+19{,}998=60{,}504\) and \(50{,}607+19{,}998=70{,}605\). 3. In every row, the sum contains the digits of the first addend in reverse order.

Answer

\(40{,}506+19{,}998=60{,}504\) \(50{,}607+19{,}998=70{,}605\) The digits of each sum are the digits of the first addend in reverse order.

5166664

Study the pattern. Calculate the sums, write the next equation, and determine how much the sum increases each time. \(12{,}345+23{,}456\) \(23{,}456+34{,}567\) \(34{,}567+45{,}678\)

Hints

- Calculate each sum carefully using place value. - Compare the first addends in consecutive rows. - Compare the second addends in consecutive rows. - Combine the two changes to predict how the sums change.

Solution

1. The sums are \(12{,}345+23{,}456=35{,}801\), \(23{,}456+34{,}567=58{,}023\), and \(34{,}567+45{,}678=80{,}245\). 2. Each addend increases by \(11{,}111\), so the next equation is \(45{,}678+56{,}789=102{,}467\). 3. Since both addends increase by \(11{,}111\), the sum increases by \(22{,}222\) each time.

Answer

The sums are \(35{,}801\), \(58{,}023\), and \(80{,}245\). The next equation is \(45{,}678+56{,}789=102{,}467\). The sum increases by \(22{,}222\) each time.

5167834

A number pattern starts at \(10\). Multiply in this repeating order: by \(10\), then by \(5\), then by \(2\). How many multiplication steps are needed to obtain a number greater than \(1{,}000{,}000\)?

Hints

- The target must be exceeded, not merely reached. - Keep the repeating factor order visible as you work. - Count each multiplication as one step.

Solution

1. Apply the repeating factors: \(10 \times 10 = 100\), \(100 \times 5 = 500\), and \(500 \times 2 = 1000\). 2. Continue: \(1000 \times 10 = 10{,}000\), \(10{,}000 \times 5 = 50{,}000\), and \(50{,}000 \times 2 = 100{,}000\). 3. The seventh step gives \(100{,}000 \times 10 = 1{,}000{,}000\), which is not greater than \(1{,}000{,}000\). 4. The eighth step gives \(1{,}000{,}000 \times 5 = 5{,}000{,}000\), so \(8\) steps are needed.

Answer

\(8\) steps

5170474

Find the rule and continue the equation pattern. \(9 \times 9 + 7 = 88\) \(98 \times 9 + 6 = 888\) \(987 \times 9 + 5 = 8888\) a) Write and evaluate the next two equations. b) Without calculating, how many digits are in the result of the fifth equation? Explain.

Hints

- Track how the digits in the first factor change. - Track how the added number changes. - Count the \(8\)s in the results and relate that count to the row number.

Solution

1. The first factor appends the next lower digit each time: \(9, 98, 987, 9876, 98{,}765\). The added number decreases by \(1\). 2. The fourth equation is \(9876 \times 9 + 4 = 88{,}888\). 3. The fifth equation is \(98{,}765 \times 9 + 3 = 888{,}888\). 4. The fifth result has \(6\) digits because each new row adds one more digit \(8\) to the result.

Answer

a) \(9876 \times 9 + 4 = 88{,}888\) and \(98{,}765 \times 9 + 3 = 888{,}888\) b) \(6\) digits

5170674

Two four-digit palindromes of the forms \(abba\) and \(baab\), made from the same two digits, differ by exactly \(2673\). Which two digits could have been used? Find three different digit pairs.

Hints

- Start with two digits that differ by \(1\) and find the palindrome difference. - Compare \(2673\) with that smallest difference. - List digit pairs from \(1\) through \(9\) with the required difference. - Check three pairs by subtraction.

Solution

1. When the two digits differ by \(1\), the two palindromes differ by \(891\). 2. Since \(3\times 891=2673\), the two chosen digits must differ by \(3\). 3. Possible pairs are \((1,4)\), \((2,5)\), \((3,6)\), \((4,7)\), \((5,8)\), and \((6,9)\). 4. For example, \(4114-1441=2673\), \(5225-2552=2673\), and \(6336-3663=2673\).

Answer

Any three digit pairs that differ by \(3\) are valid. Examples are \((1,4)\), \((2,5)\), and \((3,6)\).

5170764

A hundreds chart has ten numbers in each column. <table><tbody><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td></tr><tr><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td></tr><tr><td>21</td><td>22</td><td>23</td><td>24</td><td>25</td><td>26</td><td>27</td><td>28</td><td>29</td><td>30</td></tr><tr><td>31</td><td>32</td><td>33</td><td>34</td><td>35</td><td>36</td><td>37</td><td>38</td><td>39</td><td>40</td></tr><tr><td>41</td><td>42</td><td>43</td><td>44</td><td>45</td><td>46</td><td>47</td><td>48</td><td>49</td><td>50</td></tr><tr><td>51</td><td>52</td><td>53</td><td>54</td><td>55</td><td>56</td><td>57</td><td>58</td><td>59</td><td>60</td></tr><tr><td>61</td><td>62</td><td>63</td><td>64</td><td>65</td><td>66</td><td>67</td><td>68</td><td>69</td><td>70</td></tr><tr><td>71</td><td>72</td><td>73</td><td>74</td><td>75</td><td>76</td><td>77</td><td>78</td><td>79</td><td>80</td></tr><tr><td>81</td><td>82</td><td>83</td><td>84</td><td>85</td><td>86</td><td>87</td><td>88</td><td>89</td><td>90</td></tr><tr><td>91</td><td>92</td><td>93</td><td>94</td><td>95</td><td>96</td><td>97</td><td>98</td><td>99</td><td>100</td></tr></tbody></table> 1. Find the sum of the first column: \(1,11,21,31,41,51,61,71,81,91\). 2. Find the sum of the second column: \(2,12,22,32,42,52,62,72,82,92\). 3. How much greater is the second column sum? Explain why. 4. Find the sum of the tenth column, \(10,20,\dots,100\), without adding all ten numbers individually. 5. Add the ten column sums to find the total of the hundreds chart.

Hints

- Compare corresponding entries in the first two columns. - Use the repeated change to predict later column sums. - Count how many steps separate the first and tenth columns. - Pair the first and last column sums when finding the final total.

Solution

1. The first column sum is \(1+11+21+31+41+51+61+71+81+91=460\). 2. The second column sum is \(2+12+22+32+42+52+62+72+82+92=470\). 3. The second sum is \(10\) greater because each of its ten entries is \(1\) greater. 4. Each column sum is \(10\) greater than the previous one. The tenth column sum is \(460+9\times 10=550\). 5. The column sums are \(460,470,480,490,500,510,520,530,540,550\), and their total is \(5050\).

Answer

1. \(460\) 2. \(470\) 3. It is \(10\) greater because all ten entries are \(1\) greater. 4. \(550\) 5. \(5050\)

5170774

Use the hundreds chart. <table><tbody><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td></tr><tr><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td></tr><tr><td>21</td><td>22</td><td>23</td><td>24</td><td>25</td><td>26</td><td>27</td><td>28</td><td>29</td><td>30</td></tr><tr><td>31</td><td>32</td><td>33</td><td>34</td><td>35</td><td>36</td><td>37</td><td>38</td><td>39</td><td>40</td></tr><tr><td>41</td><td>42</td><td>43</td><td>44</td><td>45</td><td>46</td><td>47</td><td>48</td><td>49</td><td>50</td></tr><tr><td>51</td><td>52</td><td>53</td><td>54</td><td>55</td><td>56</td><td>57</td><td>58</td><td>59</td><td>60</td></tr><tr><td>61</td><td>62</td><td>63</td><td>64</td><td>65</td><td>66</td><td>67</td><td>68</td><td>69</td><td>70</td></tr><tr><td>71</td><td>72</td><td>73</td><td>74</td><td>75</td><td>76</td><td>77</td><td>78</td><td>79</td><td>80</td></tr><tr><td>81</td><td>82</td><td>83</td><td>84</td><td>85</td><td>86</td><td>87</td><td>88</td><td>89</td><td>90</td></tr><tr><td>91</td><td>92</td><td>93</td><td>94</td><td>95</td><td>96</td><td>97</td><td>98</td><td>99</td><td>100</td></tr></tbody></table> Choose any \(3\times 3\) block of nine numbers. 1. Add the nine numbers. 2. Multiply the center number by \(9\). 3. Compare the two results. 4. Test another \(3\times 3\) block. 5. Explain why the pattern always works.

Hints

- Write all nine numbers in your chosen block. - Compare numbers directly opposite each other across the center. - Add each opposite pair before adding all nine entries. - Express each pair in terms of the center number.

Solution

1. For the block with rows \(14,15,16\), \(24,25,26\), and \(34,35,36\), the sum is \(225\). 2. The center is \(25\), and \(9\times 25=225\). 3. For another block centered at \(58\), the nine-number sum is \(522\), and \(9\times 58=522\). 4. Opposite numbers balance around the center: \((m-1)+(m+1)=2m\), \((m-10)+(m+10)=2m\), \((m-11)+(m+11)=2m\), and \((m-9)+(m+9)=2m\). 5. The four opposite pairs contribute \(8m\), and the center contributes \(m\), for a total of \(9m\).

Answer

1. Answers will vary. For example, the block centered at \(25\) has sum \(225\). 2. For that block, \(9\times 25=225\). 3. The two results are equal. 4. Answers will vary. For example, a block centered at \(58\) has sum \(522\), and \(9\times 58=522\). 5. The sum of any \(3\times 3\) block is \(9\) times its center number because opposite entries pair to twice the center.

5170794

For four-digit repeating-block numbers of the forms \(abab\) and \(baba\), the difference depends on the difference between digits \(a\) and \(b\). a) Find the number difference when \(a=6\) and \(b=1\). b) Find every ordered digit pair \((a,b)\), with both digits from \(1\) through \(9\) and \(a>b\), for which the number difference is \(1818\). c) What is the smallest possible positive difference between two such numbers?

Hints

- First test a digit difference of \(1\). - Compare \(1818\) with the resulting number difference. - List all ordered digit pairs with the required difference. - Use the smallest possible positive digit difference for part c).

Solution

1. For \(a=6\) and \(b=1\), \(6161-1616=4545\). 2. A digit difference of \(1\) produces a number difference of \(909\). Since \(1818=2\times 909\), the digits must differ by \(2\). 3. The ordered pairs are \((3,1)\), \((4,2)\), \((5,3)\), \((6,4)\), \((7,5)\), \((8,6)\), and \((9,7)\). 4. The smallest positive digit difference is \(1\), so the smallest positive number difference is \(909\).

Answer

a) \(4545\) b) \((3,1),(4,2),(5,3),(6,4),(7,5),(8,6),(9,7)\) c) \(909\)

5170864

A sequence begins with \(6500\) and \(9000\). Each new term is the sum of the two previous terms. What is the first term greater than \(100{,}000\), and what position does it have in the sequence?

Hints

- Number the terms as you build the sequence. - Add the two most recent terms each time. - Stop only when a term is greater than \(100{,}000\).

Solution

1. Continue the sequence: \(6500, 9000, 15{,}500, 24{,}500, 40{,}000, 64{,}500, 104{,}500\). 2. The sixth term, \(64{,}500\), is below \(100{,}000\). 3. The seventh term, \(104{,}500\), is greater than \(100{,}000\), so it is the first term that meets the condition.

Answer

The seventh term is the first one greater than \(100{,}000\). It is \(104{,}500\).

5170924

Two students build sequences in which each new term is the sum of the two previous terms. Starting with \(100{,}000\) and \(100{,}000\), the fifth term is \(500{,}000\). Now each student increases exactly one starting term by \(1\): Luke: \(100{,}001, 100{,}000, \ldots\) Mia: \(100{,}000, 100{,}001, \ldots\) Who gets the greater fifth term? Calculate both sequences to justify your answer.

Hints

- Continue both sequences carefully through the fifth term. - Track how often each starting term contributes to later terms. - Compare the two fifth terms only after both sequences are complete.

Solution

1. Luke's sequence is \(100{,}001, 100{,}000, 200{,}001, 300{,}001, 500{,}002\). 2. Mia's sequence is \(100{,}000, 100{,}001, 200{,}001, 300{,}002, 500{,}003\). 3. Since \(500{,}003 > 500{,}002\), Mia gets the greater fifth term.

Answer

Mia gets the greater fifth term: \(500{,}003\), compared with Luke's \(500{,}002\).

5173534

Study the number pattern and find the next five terms. \(2; 3; 6; 11; 18; \ldots\)

Hints

- Write the difference between each pair of neighboring terms. - Look for a pattern in those differences. - Continue the difference pattern before finding the next terms.

Solution

1. Find the differences: \(3-2=1\), \(6-3=3\), \(11-6=5\), and \(18-11=7\). 2. The differences are consecutive odd numbers. The next differences are \(9, 11, 13, 15\), and \(17\). 3. Add them in order: \(18+9=27\), \(27+11=38\), \(38+13=51\), \(51+15=66\), and \(66+17=83\).

Answer

Next five terms: \(27; 38; 51; 66; 83\) Rule: Add consecutive odd numbers: \(1, 3, 5, 7, 9, \ldots\).

5173654

In a number pattern, each term is the sum of the two terms before it. Two consecutive terms are \(7\) and \(11\). Find the two terms immediately before \(7\) and the two terms immediately after \(11\).

Hints

- Add \(7\) and \(11\) to move forward. - Use subtraction to reverse the addition rule. - Check every group of three consecutive terms.

Solution

1. Find the next terms by adding: \(7+11=18\), then \(11+18=29\). 2. Work backward using subtraction. The term before \(7\) is \(11-7=4\). The term before \(4\) is \(7-4=3\). 3. Check the pattern: \(3; 4; 7; 11; 18; 29\). Each term after the first two is the sum of the previous two.

Answer

Before \(7\): \(3\) and \(4\) After \(11\): \(18\) and \(29\)

5185244

Two students play “Reach \(32\).” They begin at \(0\) and take turns adding any whole number from \(1\) through \(5\) to the running total. The player who reaches exactly \(32\) wins. a) Leo goes first and adds \(2\). Explain why this is a strong first move. b) If the other player adds \(x\), what should Leo add next to guarantee a win? c) List the running totals Leo should reach after each of his turns.

Hints

- Work backward from \(32\). - Find a total that two consecutive moves can always make. - Look for equally spaced target totals.

Solution

1. The two players’ additions can always be paired to total \(6\), because if the opponent adds \(x\), Leo can add \(6-x\). 2. Work backward from \(32\) by subtracting \(6\): \(32, 26, 20, 14, 8, 2\). 3. Starting at \(2\) lets Leo reach each of these totals. Since \(1\leq x\leq5\), the response \(6-x\) is also from \(1\) through \(5\). 4. Leo’s target totals are \(2, 8, 14, 20, 26\), and \(32\).

Answer

a) \(2\) is the first total in a winning pattern that increases by \(6\). b) Leo should add \(6-x\). c) \(2, 8, 14, 20, 26, 32\)

5185254

There are \(21\) pieces of candy in a bowl. Two players take turns removing either \(1\) or \(2\) pieces. The player who takes the last piece wins. If both players use the best strategy, which player can force a win? Describe the winning strategy.

Hints

- Try to make each pair of turns remove the same total number of pieces. - List the multiples of \(3\) below \(21\). - Decide which player can always leave a multiple of \(3\).

Solution

1. A pair of moves can always remove \(3\) pieces: if the first player removes \(1\), the second removes \(2\); if the first removes \(2\), the second removes \(1\). 2. Since \(21\) is a multiple of \(3\), the second player can keep leaving \(18, 15, 12, 9, 6, 3\), and then \(0\) pieces. 3. Therefore, the second player can force a win by making each pair of moves remove exactly \(3\) pieces.

Answer

The second player can force a win. After the first player removes \(x\) pieces, the second player removes \(3-x\) pieces.

5185304

Study this sequence of expressions: 1. \(2-1\) 2. \(4-3+2-1\) 3. \(6-5+4-3+2-1\) 4. \(8-7+6-5+4-3+2-1\) a) Find the results of the first four rows. Describe the pattern. b) Write the expression for row \(6\) and find its result. c) Which row has a result of \(25\)?

Hints

- Group each even number with the odd number immediately after it. - Each pair has the same value. - Count the number of pairs in each row.

Solution

1. a) Group consecutive pairs: \((2-1)=1\), \((4-3)+(2-1)=2\), \((6-5)+(4-3)+(2-1)=3\), and \((8-7)+(6-5)+(4-3)+(2-1)=4\). The result equals the row number. 2. b) Row \(6\) is \(12-11+10-9+8-7+6-5+4-3+2-1\). It contains six pairs that each equal \(1\), so the result is \(6\). 3. c) Because the result equals the row number, row \(25\) has a result of \(25\).

Answer

a) \(1, 2, 3, 4\); the result equals the row number. b) \(12-11+10-9+8-7+6-5+4-3+2-1=6\) c) Row \(25\)

5185314

Study this sequence of calculations: 1. \(100-1=99\) 2. \(100-(1+2)=97\) 3. \(100-(1+2+3)=94\) 4. \(100-(1+2+3+4)=90\) a) Find the results for rows \(5\) and \(6\). b) In which row is the result less than \(80\) for the first time? c) What is the result in row \(10\)?

Hints

- Each row adds one more number inside the parentheses. - Use the previous sum to find the next sum. - Compare each result with \(80\).

Solution

1. a) Row \(5\) subtracts \(1+2+3+4+5=15\), so the result is \(100-15=85\). Row \(6\) subtracts \(15+6=21\), so the result is \(100-21=79\). 2. b) Row \(5\) gives \(85\), and row \(6\) gives \(79\). Therefore, row \(6\) is the first result less than \(80\). 3. c) The sum from \(1\) through \(10\) is \(55\), so row \(10\) gives \(100-55=45\).

Answer

a) Row \(5\): \(85\); row \(6\): \(79\) b) Row \(6\) c) \(45\)

5193054

A “happy number” reaches \(1\) after repeatedly multiplying each digit by itself and adding the products. Test \(13\), \(14\), and \(15\). Which are happy numbers? Show each chain until it reaches \(1\) or repeats a number.

Hints

- Work with one starting number at a time. - Record every term in each chain. - Stop when the chain reaches \(1\) or when a term appears for the second time.

Solution

1. For \(13\): \(13\rightarrow10\rightarrow1\). Therefore, \(13\) is happy. 2. For \(14\): \(14\rightarrow17\rightarrow50\rightarrow25\rightarrow29\rightarrow85\rightarrow89\rightarrow145\rightarrow42\rightarrow20\rightarrow4\rightarrow16\rightarrow37\rightarrow58\rightarrow89\). The term \(89\) repeats, so the chain enters a cycle and never reaches \(1\). 3. For \(15\): \(15\rightarrow26\rightarrow40\rightarrow16\rightarrow37\rightarrow58\rightarrow89\rightarrow145\rightarrow42\rightarrow20\rightarrow4\rightarrow16\). The term \(16\) repeats, so this chain also enters a cycle. 4. Only \(13\) is a happy number.

Answer

Only \(13\) is happy: \(13\rightarrow10\rightarrow1\). The chains for \(14\) and \(15\) repeat before reaching \(1\).

5217324

A number pattern starts at \(15{,}000\). After five equal steps forward, it reaches \(17{,}500\). a) Find the size of one step. b) List the five numbers reached by moving forward. c) List the five numbers reached by moving backward from \(15{,}000\) using the same step size.

Hints

- Find the total distance from the starting number to the ending number. - Divide that distance into five equal parts. - Add the step size to move forward and subtract it to move backward.

Solution

1. a) The total change is \(17{,}500-15{,}000=2500\). Divide by the five equal steps: \(2500\div5=500\). 2. b) Add \(500\) repeatedly: \(15{,}500, 16{,}000, 16{,}500, 17{,}000, 17{,}500\). 3. c) Subtract \(500\) repeatedly: \(14{,}500, 14{,}000, 13{,}500, 13{,}000, 12{,}500\).

Answer

a) \(500\) b) \(15{,}500, 16{,}000, 16{,}500, 17{,}000, 17{,}500\) c) \(14{,}500, 14{,}000, 13{,}500, 13{,}000, 12{,}500\)

5319684

Zoe and Ben each build a number wall. Every brick is the sum of the two bricks directly below it. Zoe’s bottom row is \(60\), \(110\), \(80\), and \(50\), from left to right. Ben uses the same numbers except that he increases the second brick from \(110\) to \(120\). a) Find the top brick in Zoe’s wall. b) Find the top brick in Ben’s wall. c) How much greater is Ben’s top brick? Explain why increasing that one bottom brick by \(10\) creates this exact change at the top.

Figure for problem 531968

Hints

- Complete both walls from the bottom upward. - Compare corresponding bricks in the two walls. - Track how many times the second bottom brick contributes to the top.

Solution

1. For Zoe, the second row is \(170\), \(190\), and \(130\). The third row is \(360\) and \(320\), so the top is \(680\). 2. For Ben, the second row is \(180\), \(200\), and \(130\). The third row is \(380\) and \(330\), so the top is \(710\). 3. The difference is \(710 - 680 = 30\). The changed bottom brick contributes to two bricks in the second row and three total contributions to the top. Therefore, increasing it by \(10\) increases the top by \(3 \times 10 = 30\).

Answer

a) \(680\) b) \(710\) c) Ben’s top brick is \(30\) greater. The changed bottom brick contributes three times to the top, so the increase is \(3 \times 10 = 30\).

5320024

Mia and Jonas compare two number walls. In wall a), the bottom row is \(120\), \(150\), and \(180\). Mia creates wall b) by increasing the middle bottom brick from \(150\) to \(160\). Jonas claims, “Because the bottom brick increased by \(10\), the top brick will also increase by exactly \(10\).” Is Jonas correct? Find the top brick in each wall and explain your answer.

Figure for problem 532002

Hints

- Complete wall a) first. - Complete wall b) next. - Compare the two top bricks. - Track how the middle bottom brick contributes to both bricks above it.

Solution

1. In wall a), the second row is \(120 + 150 = 270\) and \(150 + 180 = 330\). The top is \(270 + 330 = 600\). 2. In wall b), the second row is \(120 + 160 = 280\) and \(160 + 180 = 340\). The top is \(280 + 340 = 620\). 3. The top increases by \(620 - 600 = 20\), not \(10\). The middle bottom brick contributes to both bricks in the second row, so its change contributes twice to the top.

Answer

No. Wall a) has a top brick of \(600\), and wall b) has a top brick of \(620\). The top increases by \(20\) because the middle bottom brick contributes twice to the top.

5320224

Complete both large number walls. Every brick is the sum of the two bricks directly below it.

Figure for problem 532022

Hints

- Start where addition or subtraction gives a missing value directly. - In wall b), subtract the known outside bottom values from \(465\). The remaining amount contains the unknown bottom value twice. - Use a variable or question mark to organize the repeated unknown if helpful.

Solution

1. Wall a): \(205 - 85 = 120\), \(45 + 120 = 165\), \(165 + 205 = 370\), \(770 - 370 = 400\), \(400 - 205 = 195\), and \(195 - 85 = 110\). 2. Wall b): \(205 - 70 = 135\). In the right third-row brick, the two outside bottom values contribute once and the unknown bottom value contributes twice. Thus \(465 - 70 - 95 = 300\), so the unknown is \(300 \div 2 = 150\). Then \(70 + 150 = 220\), \(150 + 95 = 245\), \(205 + 220 = 425\), and \(425 + 465 = 890\).

Answer

a) Bottom row: \(45\), \(120\), \(85\), \(110\); second row: \(165\), \(205\), \(195\); third row: \(370\), \(400\); top: \(770\) b) Bottom row: \(135\), \(70\), \(150\), \(95\); second row: \(205\), \(220\), \(245\); third row: \(425\), \(465\); top: \(890\)

5352494

Complete both number walls. Decide where to begin in each wall.

Figure for problem 535249

Hints

- Add when finding an upper brick from two lower bricks. - Subtract when finding a lower brick from an upper brick and its neighbor. - In wall b), think about how the middle bottom brick contributes to both bricks above it.

Solution

1. Wall a): The right brick in the second row is \(210 + 150 = 360\). The left brick in the second row is \(800 - 360 = 440\). The left bottom brick is \(440 - 210 = 230\). 2. Wall b): The outside bottom bricks contribute \(145 + 145 = 290\) to the top. The middle bottom brick contributes twice, so \(600 - 290 = 310\) and \(310 \div 2 = 155\). The two bricks in the second row are both \(145 + 155 = 300\).

Answer

a) Bottom left: \(230\); second row: \(440\), \(360\) b) Bottom middle: \(155\); second row: \(300\), \(300\)

5353094

This number wall has five rows. Find every missing value, and use the symmetry you notice to help you.

Figure for problem 535309

Hints

- Look for mirror symmetry around the center of the wall. - Express the middle \(120\) using the three bottom bricks beneath it. - After finding the bottom row, add upward to check the pattern.

Solution

1. The first bottom brick is \(60 - 40 = 20\). 2. Let the middle bottom brick be \(x\). The middle brick in the third row is made from \(40 + x\) and \(x + 40\), so \(40 + x + x + 40 = 120\). Thus \(2 \times x = 40\), and \(x = 20\). 3. The two middle bricks in the second row are each \(40 + 20 = 60\). 4. The right brick in the third row is \(240 - 120 = 120\). Therefore the right brick in the second row is \(120 - 60 = 60\), and the last bottom brick is \(60 - 40 = 20\). 5. The remaining rows are \(120, 120, 120\), then \(240, 240\), with \(480\) at the top.

Answer

Bottom row: \(20\), \(40\), \(20\), \(40\), \(20\) Second row: \(60\), \(60\), \(60\), \(60\) Third row: \(120\), \(120\), \(120\) Fourth row: \(240\), \(240\) Top: \(480\)

5353114

Complete the five-row number wall. The values are large, but the rule is unchanged: each brick is the sum of the two bricks below it.

Figure for problem 535311

Hints

- Start at the top, where two connected values are already known. - Use the wall's symmetry to check matching values on the left and right. - Keep place values aligned when subtracting three-digit numbers.

Solution

1. The right brick in the fourth row is \(920 - 460 = 460\). 2. The left and right bricks in the third row are each \(460 - 240 = 220\). 3. The left brick in the second row is \(220 - 120 = 100\), so the second bottom brick is \(100 - 45 = 55\). 4. The next brick in the second row is \(240 - 120 = 120\), so the middle bottom brick is \(120 - 55 = 65\). 5. The right brick in the second row is \(220 - 120 = 100\). This gives \(120 - 65 = 55\) for the fourth bottom brick and \(100 - 55 = 45\) for the last one.

Answer

Bottom row: \(45\), \(55\), \(65\), \(55\), \(45\) Second row: \(100\), \(120\), \(120\), \(100\) Third row: \(220\), \(240\), \(220\) Fourth row: \(460\), \(460\) Top: \(920\)

5353264

The top value must become \(800\). Add the same number to each bottom-row value. What will the new bottom row be?

Figure for problem 535326

Hints

- First find how much the top value must increase. - Try adding a small equal amount to all three bottom bricks and track how much the top changes.

Solution

1. The current top value is \(400\), so it must increase by \(800 - 400 = 400\). 2. In a three-brick base, adding \(k\) to each bottom brick increases the two middle bricks by \(2 \times k\) each. The top therefore increases by \(4 \times k\). 3. Solve \(4 \times k = 400\): \(k = 400 \div 4 = 100\). 4. Add \(100\) to each original bottom value: \(100 + 100 = 200\).

Answer

The new bottom row is \(200\), \(200\), \(200\).

5353274

Subtract the same amount from each bottom-row value so that the top value becomes exactly \(120\). What is the new bottom row?

Figure for problem 535327

Hints

- Find how much the top value must decrease. - Determine how many times the common bottom-row decrease appears in the top value. - Subtract that common amount from each original bottom brick.

Solution

1. The top must decrease by \(260 - 120 = 140\). 2. In a three-brick base, subtracting \(k\) from each bottom brick decreases the top by \(4 \times k\). 3. Solve \(4 \times k = 140\): \(k = 140 \div 4 = 35\). 4. Subtract \(35\) from each bottom value: \(60 - 35 = 25\), \(80 - 35 = 45\), and \(40 - 35 = 5\).

Answer

The new bottom row is \(25\), \(45\), \(5\).

5353644

What happens to the top value of a four-brick number wall when the second bottom brick is increased by \(10\)? Compare the two walls and explain why.

Figure for problem 535364

Hints

- Complete wall b) and compare its top with \(500\). - Trace how many times the second bottom brick contributes to the rows above.

Solution

1. Wall a) has top value \(500\). 2. For wall b), the second row is \(160, 110, 150\). 3. The third row is \(160 + 110 = 270\) and \(110 + 150 = 260\). 4. The top is \(270 + 260 = 530\), so it increased by \(530 - 500 = 30\). 5. The second bottom brick contributes to three paths leading to the top. Therefore an increase of \(10\) contributes \(3 \times 10 = 30\) to the top.

Answer

The top value increases by \(30\). The second bottom brick is counted three times in the top value, so increasing it by \(10\) increases the top by \(3 \times 10 = 30\).

5354114

Complete the number wall to the top. Use the mirror symmetry in the bottom row.

Figure for problem 535411

Hints

- Begin with the given \(100\) in the third row. - Use the symmetry to see that the two bricks directly below \(100\) are equal. - Then work row by row both upward and downward.

Solution

1. Let the middle bottom brick be \(x\). The two bricks beneath the middle \(100\) are both \(20 + x\). 2. Therefore \((20 + x) + (x + 20) = 100\), so \(2x + 40 = 100\), \(2x = 60\), and \(x = 30\). 3. The bottom row is \(10, 20, 30, 20, 10\), and the second row is \(30, 50, 50, 30\). 4. Adding upward gives \(80, 100, 80\), then \(180, 180\), and finally \(360\).

Answer

Bottom row: \(10\), \(20\), \(30\), \(20\), \(10\) Second row: \(30\), \(50\), \(50\), \(30\) Third row: \(80\), \(100\), \(80\) Fourth row: \(180\), \(180\) Top: \(360\)

5354124

The number wall is symmetric. Use backward reasoning to complete it.

Figure for problem 535412

Hints

- First subtract downward from the top where only one value is missing. - Use symmetry to connect matching positions on the left and right. - Express the left \(8\) in terms of the missing bottom value.

Solution

1. The missing brick in the fourth row is \(36 - 18 = 18\). 2. The outside bricks in the third row are each \(18 - 10 = 8\), so that row is \(8, 10, 8\). 3. Let the second bottom brick be \(x\). The left \(8\) equals \(1 + 2x + 3\), so \(2x + 4 = 8\), \(2x = 4\), and \(x = 2\). 4. By symmetry, the bottom row is \(1, 2, 3, 2, 1\). 5. Adding upward gives \(3, 5, 5, 3\), then \(8, 10, 8\), then \(18, 18\), and \(36\) at the top.

Answer

Bottom row: \(1\), \(2\), \(3\), \(2\), \(1\) Second row: \(3\), \(5\), \(5\), \(3\) Third row: \(8\), \(10\), \(8\) Fourth row: \(18\), \(18\) Top: \(36\)

5354154

Complete the number wall and describe the pattern in the bottom row.

Figure for problem 535415

Hints

- Start with a blank that can be found by subtracting from \(100\). - Watch for repeated values. - Look for mirror symmetry between the left and right sides.

Solution

1. The third bottom brick is \(100 - 50 = 50\). 2. The right brick in the second row is \(50 + 50 = 100\), so the right brick in the third row is \(100 + 100 = 200\). 3. The left brick in the third row is \(400 - 200 = 200\). 4. The left brick in the second row is \(200 - 100 = 100\), so the first bottom brick is \(100 - 50 = 50\). 5. Every bottom brick is \(50\).

Answer

Bottom row: \(50\), \(50\), \(50\), \(50\) Second row: \(100\), \(100\), \(100\) Third row: \(200\), \(200\) Top: \(400\) The bottom row has the constant pattern \(50, 50, 50, 50\).

5363284

Compare the three product walls. Find each top value. Which walls have the same result, and why is wall c) different?

Figure for problem 536328

Hints

- Calculate all three top values. - Compare the middle bottom value in each wall. - Trace which bottom position is used in both second-row products.

Solution

1. Wall a) has second row \(2 \times 3 = 6\) and \(3 \times 4 = 12\), so the top is \(6 \times 12 = 72\). 2. Wall b) has second row \(4 \times 3 = 12\) and \(3 \times 2 = 6\), so the top is \(12 \times 6 = 72\). 3. Wall c) has second row \(3 \times 2 = 6\) and \(2 \times 4 = 8\), so the top is \(6 \times 8 = 48\). 4. Walls a) and b) have equal top values. The middle bottom value is used in both products in the second row, so it contributes twice. Wall c) puts the smallest value, \(2\), in that influential middle position.

Answer

Walls a) and b) both have top value \(72\). Wall c) has top value \(48\) because its smallest bottom value is in the middle and is used in both products above it.

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