- Read the bar values in order. - Look for a change that repeats. - Check the pattern using more than one pair of bars.

1. The bar values are \(10, 15, 20, 25,\) and \(30\). 2. Each value is \(5\) greater than the value before it.

The number of math puzzles solved increases by \(5\) each day.

- Read the bar values in order. - Find the change that repeats from day to day. - Apply the same change to Thursday’s value.

1. The values increase by \(4\) each day: \(12, 16, 20, 24\). 2. The next value is \(24+4=28\).

Friday’s bar should have a value of \(28\).

- Compare neighboring numbers. - Use the same change to keep the pattern going.

1. The numbers increase by \(4\) each time. 2. The next two numbers are \(22\) and \(26\).

Rule: add \(4\). Next numbers: \(22,26\).

- Decide whether the numbers go up or down. - Find the same-sized change between the numbers shown.

1. Each number is \(5\) less than the one before it. 2. The next numbers are \(15\) and \(10\).

Rule: subtract \(5\). Missing numbers: \(15,10\).

- Trace the shape from the bottom to the blank brick. - Should the top number be greater or less than each positive number beneath it? Check your result.

1. The two bottom bricks are \(3\) and \(4\). 2. Their sum is \(3+4=7\).

\(7\).

- Compare the last digit of each number in both sequences. - Subtract one pair of numbers in the same position. - Compare the hundreds places in the two sequences.

1. Every number in both sequences has \(7\) in the ones place. 2. Subtract numbers in the same position: for example, \(207-7=200\) and \(217-17=200\). 3. Each number in Sequence B is \(200\) greater than the number in the same position in Sequence A.

a) Every number has \(7\) in the ones place. b) Each number in Sequence B is \(200\) greater than the number in the same position in Sequence A.

- Find the difference between consecutive terms. - Determine how much must be added to \(225\) to reach \(250\). - Check whether that difference stays the same. - Continue using the same change until you reach \(375\).

1. The difference between the first two terms is \(250-225=25\), and \(275-250=25\). 2. The rule is to add \(25\) each time. 3. The missing terms are \(275+25=300\), \(300+25=325\), and \(325+25=350\). 4. The next term is \(350+25=375\), which matches the given endpoint.

The missing numbers are \(300, 325,\) and \(350\). Rule: add \(25\) each time.

- Compare the first factors in each pair. - How can the known product help you find the other product? - What happens to a product when one factor is doubled or halved and the other factor stays the same?

1. \(3 \times 4 = 12\). Doubling the first factor from \(3\) to \(6\) doubles the product, so \(6 \times 4 = 24\). 2. \(10 \times 7 = 70\). Halving the first factor from \(10\) to \(5\) halves the product, so \(5 \times 7 = 35\). 3. \(4 \times 8 = 32\). Doubling the first factor from \(4\) to \(8\) doubles the product, so \(8 \times 8 = 64\).

a) \(3 \times 4 = 12\) and \(6 \times 4 = 24\); the first factor and the product are doubled. b) \(10 \times 7 = 70\) and \(5 \times 7 = 35\); the first factor and the product are halved. c) \(4 \times 8 = 32\) and \(8 \times 8 = 64\); the first factor and the product are doubled.

- Find how much each term decreases. - Think about the number that is \(2\) less than \(700\). - Check that you wrote exactly eight new terms.

1. Each term is \(2\) less than the term before it. 2. Starting from \(708\), subtract \(2\) eight times. 3. The sequence passes from \(700\) to \(698\).

\(706, 704, 702, 700, 698, 696, 694, 692\)

- Find the amount you count by in each part. - Do not include the starting or ending number. - Keep adding the same amount until the next number would reach or pass the end.

1. Count by \(100\) from \(100\) through \(900\). 2. Count by \(10\) from \(470\) through \(530\). 3. Count by \(50\) from \(750\) through \(900\).

a) \(100, 200, 300, 400, 500, 600, 700, 800, 900\) b) \(470, 480, 490, 500, 510, 520, 530\) c) \(750, 800, 850, 900\)

- Compare the first two terms in each sequence. - Decide whether the terms increase or decrease. - Find the amount of the repeated change. - In b), pay attention to the tens and hundreds transition.

1. Sequence a) increases by \(10\), giving \(700, 710, 720, 730, 740,\) and \(750\). 2. Sequence b) decreases by \(2\), giving \(306, 304, 302, 300, 298,\) and \(296\). 3. Sequence c) increases by \(50\), giving \(600, 650, 700, 750,\) and \(800\).

a) \(670, 680, 690, 700, 710, 720, 730, 740, 750\) b) \(312, 310, 308, 306, 304, 302, 300, 298, 296\) c) \(450, 500, 550, 600, 650, 700, 750, 800\)

- Find the change between two adjacent known terms. - Check whether the same change works throughout the sequence. - Verify each missing term using the next known term.

1. Sequence a) increases by \(25\). The missing terms are \(150+25=175\) and \(225+25=250\). 2. Sequence b) decreases by \(20\). The missing terms are \(820-20=800\) and \(780-20=760\).

a) \(175\) and \(250\) b) \(800\) and \(760\)

- Find the change from the first term to the second term. - Check whether the same change works from the second term to the third. - State the repeated rule in words. - Count the terms carefully so that each sequence has six terms.

1. In a), each term is \(15\) greater than the previous term. Continue with \(60, 75,\) and \(90\). 2. In b), each term is \(30\) less than the previous term. Continue with \(820, 790,\) and \(760\).

a) Rule: add \(15\). Sequence: \(15, 30, 45, 60, 75, 90\) b) Rule: subtract \(30\). Sequence: \(910, 880, 850, 820, 790, 760\)

- Calculate all three differences. - Compare the numbers being subtracted. - Relate the change in the number being subtracted to the change in the difference.

1. \(1000 - 20 = 980\). 2. \(1000 - 40 = 960\). 3. \(1000 - 60 = 940\). 4. The number being subtracted increases by \(20\) each time, so the difference decreases by \(20\) each time.

a) \(980\) b) \(960\) c) \(940\) The differences decrease by \(20\) each time.

- Increase the first factor by \(1\) in each new expression. - Compare expressions with the same first factor across the two sequences. - Relate \(80\) to \(40\).

1. Sequence A continues with \(6 \times 40 = 240\), \(7 \times 40 = 280\), and \(8 \times 40 = 320\). The products increase by \(40\). 2. Sequence B continues with \(6 \times 80 = 480\), \(7 \times 80 = 560\), and \(8 \times 80 = 640\). The products increase by \(80\). 3. Each product in Sequence B is twice the matching product in Sequence A because \(80\) is twice \(40\).

Sequence A: \(6 \times 40 = 240\), \(7 \times 40 = 280\), \(8 \times 40 = 320\) Sequence B: \(6 \times 80 = 480\), \(7 \times 80 = 560\), \(8 \times 80 = 640\) Each product in Sequence B is twice the matching product in Sequence A.

- Counting backward means each term is smaller than the term before it. - Subtract the stated amount each time.

1. In a), subtract \(5\) repeatedly after \(710\): \(705, 700, 695, 690, 685, 680\). 2. In b), subtract \(50\) repeatedly after \(450\): \(400, 350, 300, 250, 200\).

a) \(720, 715, 710, 705, 700, 695, 690, 685, 680\) b) \(550, 500, 450, 400, 350, 300, 250, 200\)

- Compare each first number with the one before it. - The second number stays the same. How should the answer change? - Find how much greater \(400\) is than \(370\).

1. The first number increases by \(10\) each time. 2. The second number stays the same, so the answer also increases by \(10\) each time. 3. From \(370\) to \(400\) is an increase of \(30\). 4. Increase the known answer by \(30\): \(490+30=520\).

a) The first number increases by \(10\) each time. b) The answer increases by \(10\) each time. c) \(400+120=520\)

- Test each change with an equation such as \(100 - 20 = 80\). - What happens when the amount you start with increases? - What happens when the amount you subtract increases?

1. Increasing the first number by \(65\) increases the distance between the two numbers by \(65\), so the difference increases by \(65\). 2. Increasing the number being subtracted by \(65\) means \(65\) more is subtracted, so the difference decreases by \(65\).

a) The difference increases by \(65\). b) The difference decreases by \(65\).

- Identify the number being subtracted. - Think about what happens when more is subtracted. - Test the pattern with a simple equation such as \(10 - 2 = 8\).

1. When the number being subtracted increases, more is subtracted, so the difference decreases. 2. The difference decreases by the same amount, \(60\). 3. The new difference is \(350 - 60 = 290\).

The difference decreases by \(60\), so its new value is \(290\).

- Calculate the three differences first. - Compare consecutive results as the first number increases. - In part c), first find the new number being subtracted after increasing \(100\) by \(50\). - Decide whether subtracting more makes the difference greater or smaller.

1. The differences are \(400 - 100 = 300\), \(450 - 100 = 350\), and \(500 - 100 = 400\). 2. Each time the first number increases by \(50\), the difference also increases by \(50\). 3. Increasing the number being subtracted in the original equation gives \(400 - 150 = 250\). 4. Compared with \(300\), the difference decreases by \(50\).

a) \(300\), \(350\), \(400\) b) The difference increases by \(50\). c) The difference decreases by \(50\), from \(300\) to \(250\).

- Compare the first numbers in the two equations. - What happens to a difference when the number being subtracted increases? - Find the difference between \(180\) and \(140\).

1. Both equations have the same first number, \(500\). 2. Ben subtracts \(180\), which is \(40\) more than \(140\). 3. Subtracting more produces a smaller difference, so Anya's difference is greater. 4. The two differences differ by \(180 - 140 = 40\).

Anya gets the greater difference. Her result is \(40\) greater than Ben's.

- Think about what happens when you start with more but subtract the same amount. - Think about what happens when you subtract more from the same amount. - Compare each new result with \(350\).

1. Increasing the first number gives \(620 - 230 = 390\), so the difference increases by \(40\). 2. Increasing the number being subtracted gives \(580 - 270 = 310\), so the difference decreases by \(40\).

a) The difference increases by \(40\), to \(390\). b) The difference decreases by \(40\), to \(310\).

- Think about what happens when you subtract less. - Test the pattern with \(10 - 5 = 5\). - Decide whether the difference changes in the same or opposite direction as the number being subtracted.

1. Decreasing the number being subtracted means less is subtracted. 2. Therefore, the difference increases by \(40\). 3. The new difference is \(380 + 40 = 420\).

The new difference is \(420\).

- Use neighboring pairs in the bottom row to build the middle row. - Find the top only after both middle-row values are known. - Compare the top with the special sum requested in part b).

1. The middle row is \(15 + 22 = 37\) and \(22 + 13 = 35\). 2. The top is \(37 + 35 = 72\). 3. The bottom-row sum is \(15 + 22 + 13 = 50\). 4. Adding the middle number again gives \(50 + 22 = 72\), which equals the top brick.

a) The middle row is \(37\), \(35\), and the top is \(72\). b) \(15 + 22 + 13 + 22 = 72\). The result equals the number in the top brick.

- Halve the number of rows first. - Halve the product of the large array to check.

1. The large array represents \(6 \times 8 = 48\). 2. Half of \(6\) rows is \(3\) rows. 3. The small array therefore represents \(3 \times 8 = 24\), which is half of \(48\).

The small array represents \(3 \times 8 = 24\). The number of rows is halved from \(6\) to \(3\), so the product is halved from \(48\) to \(24\).

- Compare each bar’s value with the value before it. - Add \(4\) to predict each next value. - Check whether the final bar fits after correcting the error.

1. The pattern begins \(8, 12, 16\), increasing by \(4\) each time. 2. The fourth value should be \(16+4=20\), but the chart shows \(16\). 3. The fifth value, \(24\), fits the intended pattern after the corrected fourth value.

Bar \(4\) is incorrect. Its value should be \(20\).

- Work upward from the bottom, recording each new number. - Try making the changed wall on a separate sheet so you can compare them. - If the two top answers differ, look for where the results first began to change.

1. The original middle row is \(2+3=5\) and \(3+5=8\), so the top is \(5+8=13\). 2. With \(4\) in the middle bottom brick, the middle row becomes \(2+4=6\) and \(4+5=9\). The new top is \(6+9=15\). 3. Changing the shared bottom brick increases both middle bricks by \(1\). The top therefore increases by \(2\).

Original: middle row \(5,8\); top \(13\). Changed: middle row \(6,9\); top \(15\). The middle bottom number affects both bricks above it, so the top increases by \(2\).

- Try marking the change between each pair of known numbers with an arrow. - Test a possible starting number by running the pattern forward.

1. The number before \(12\) must be four less. 2. \(12-4=8\).

\(8\), because \(8+4=12\).

- Compare the first two pictures. What is different? - Does the same change happen in the next pair?

1. The pictures have \(1\), \(2\), and \(3\) rows of \(2\) dots. 2. Each new picture adds one row of two dots. 3. The next picture has \(4\) rows and \(4\times2=8\) dots.

\(8\) dots. Each picture adds one row of \(2\) dots.

- Check the difference between neighboring numbers. - At what point does the change stop matching the rule?

1. Each of the first four numbers decreases by \(5\). 2. The next number must be \(35-5=30\), not \(20\).

The first wrong number is \(20\), in fifth place. It should be \(30\), because \(35-5=30\).

- Try each possible operation with the first finished wall. Which ones, if any, fit? - Test the possibilities again using the other finished wall. - Check that the rule you pick works for every example.

1. In wall a), both \(2+2=4\) and \(2\times2=4\) fit, so one example does not settle the rule. 2. In wall b), \(3+4=7\), but \(3\times4=12\); only addition fits both walls. 3. The missing top brick in wall c) is \(4+6=10\).

The rule is addition. Wall a) could use addition or multiplication, but wall b) fits only addition: \(3+4=7\). Wall c) has \(4+6=10\) at the top.

- Compare the numbers being subtracted in the three expressions. - Use the first difference to predict the next one. - Think about what happens when the number being subtracted increases by \(100\).

1. For a), \(500 - 142 = 358\). 2. In b), the number being subtracted is \(100\) greater than in a), so the difference is \(100\) less: \(358 - 100 = 258\). 3. In c), the number being subtracted increases by another \(100\), so the difference decreases by another \(100\): \(258 - 100 = 158\).

a) \(358\) b) \(258\) c) \(158\) The differences decrease by \(100\) each time.

- Decide which digits are in the tens places and which are in the ones places. - After choosing the tens digits, place the remaining two digits in the ones places in both possible orders. - Compare consecutive sums to find the change.

1. For a), the ones digits are \(6\) and \(7\): \(46 + 57 = 103\) and \(47 + 56 = 103\). 2. For b), the ones digits are \(5\) and \(7\): \(45 + 67 = 112\) and \(47 + 65 = 112\). 3. For c), the ones digits are \(5\) and \(6\): \(45 + 76 = 121\) and \(46 + 75 = 121\). 4. The sums increase by \(9\): \(112 - 103 = 9\) and \(121 - 112 = 9\).

a) \(46 + 57 = 103\) and \(47 + 56 = 103\) b) \(45 + 67 = 112\) and \(47 + 65 = 112\) c) \(45 + 76 = 121\) and \(46 + 75 = 121\) d) The sums are \(103\), \(112\), and \(121\), increasing by \(9\) each time.

- Calculate all four sums first. - Compare the tens digits and ones digits in equations with equal sums. - Consider how a digit's value changes when it moves from the ones place to the tens place.

1. The sums are \(13 + 46 = 59\), \(16 + 43 = 59\), \(31 + 64 = 95\), and \(34 + 61 = 95\). 2. In the first two equations, the same digits occupy the tens places and the same digits occupy the ones places. Only the ones digits switch addends, so the total does not change. 3. In the last two equations, the larger digits occupy the tens places. A digit in the tens place has ten times its value in the ones place, making the total greater.

a) \(59\), \(59\), \(95\), \(95\) b) The same two digits are in the tens places and the same two digits are in the ones places; only their order between addends changes. c) The larger digits are in the tens places in the last two equations.

- Observe how both digits change from one line to the next. - Compare the difference between the two digits in each original number. - For part c), look for other two-digit numbers whose tens digit is \(2\) greater than the ones digit.

1. Increasing both digits by \(1\) gives \(74 - 47 = 27\) and \(85 - 58 = 27\). 2. The difference stays \(27\) because the tens digit is always \(3\) greater than the ones digit. 3. \(42 - 24 = 18\). A digit difference of \(2\) gives the same result, so examples include \(53 - 35 = 18\) and \(64 - 46 = 18\).

a) \(74 - 47 = 27\) and \(85 - 58 = 27\) b) Every difference is \(27\). c) The new difference is \(18\). Examples: \(53 - 35 = 18\) and \(64 - 46 = 18\).

- Compare each digit difference with the matching multiple of \(9\). - Notice how the result changes when the digit difference increases by \(1\). - For part c), find two digits that differ by \(5\).

1. The subtraction result is \(9\) times the difference between the digits. 2. For \(61 - 16\), the digits differ by \(5\), so \(5 \times 9 = 45\). 3. For \(81 - 18\), the digits differ by \(7\), so the prediction is \(7 \times 9 = 63\). Direct subtraction confirms \(81 - 18 = 63\). 4. To obtain \(45\), the digits must differ by \(5\). One example is \(72 - 27 = 45\).

a) \(45\) b) \(63\); indeed, \(81 - 18 = 63\). c) One example is \(72 - 27 = 45\).

- Start by finding the multiplication facts. - Compare \(5\times5\) with \(4\times6\) by finding both products. - Do the same for \(4\times4\) and \(3\times5\).

1. \(4\times4=16\), \(5\times5=25\), and \(6\times6=36\). 2. \(5\times4=20\), \(5\times6=30\), \(4\times6=24\), and \(6\times4=24\). 3. \(5\times5=25\) is \(1\) greater than \(4\times6=24\). 4. The same relationship happens because \(4\times4=16\) is \(1\) greater than \(3\times5=15\).

a) \(16,25,36\) b) \(5\times4=20\), \(5\times6=30\), \(4\times6=24\), \(6\times4=24\) c) In both comparisons, the product that uses the same number twice is \(1\) greater.

- Find the left side first. - Then ask which number multiplied by \(2\) gives that product. - Compare each first factor on the left with its missing factor.

1. The left-side products are \(4, 8, 12, 16\), and \(20\). 2. The matching 2s facts are \(2 \times 2 = 4\), \(4 \times 2 = 8\), \(6 \times 2 = 12\), \(8 \times 2 = 16\), and \(10 \times 2 = 20\). 3. The missing factor is always twice the factor multiplied by \(4\) on the left.

a) \(2\) b) \(4\) c) \(6\) d) \(8\) e) \(10\) Pattern: Each missing factor is twice the first factor on the left.

- Find the first two facts and compare their products. - How are \(5\) and \(10\) related? - Test the relationship with another factor.

1. \(2 \times 5 = 10\) and \(2 \times 10 = 20\). 2. The product \(20\) is twice the product \(10\). 3. Because \(10\) is twice \(5\), multiplying the same number by \(10\) gives twice the product of multiplying it by \(5\). 4. For example, \(3 \times 5 = 15\) and \(3 \times 10 = 30\); also, \(4 \times 5 = 20\) and \(4 \times 10 = 40\).

The product doubles. Possible examples: \(3 \times 5 = 15\) and \(3 \times 10 = 30\) \(4 \times 5 = 20\) and \(4 \times 10 = 40\)

- Evaluate the left side of each equation. - Which 8s fact has the same product? - Compare the first factors on the two sides.

1. The left-side products are \(16, 24, 32\), and \(40\). 2. Dividing each product by \(8\) gives the missing factors \(2, 3, 4\), and \(5\). 3. The factor paired with \(8\) is half the factor paired with \(4\). Equivalently, when one factor doubles from \(4\) to \(8\), the other factor must be halved to keep the product equal.

a) \(4 \times 4 = 2 \times 8\) b) \(6 \times 4 = 3 \times 8\) c) \(8 \times 4 = 4 \times 8\) d) \(10 \times 4 = 5 \times 8\) Rule: Doubling one factor and halving the other keeps the product the same.

- Evaluate each pair of facts. - Compare the product in the 3s facts with the product in the 6s facts. - How are \(3\) and \(6\) related?

1. For factor \(4\), \(4 \times 3 = 12\). Doubling \(12\) gives \(24\), and \(4 \times 6 = 24\). 2. For factor \(7\), \(7 \times 3 = 21\). Doubling \(21\) gives \(42\), and \(7 \times 6 = 42\). 3. Lucas is correct because \(6\) is twice \(3\). With the other factor unchanged, doubling one factor doubles the product.

Yes. \(4 \times 3 = 12\) and \(4 \times 6 = 24\); \(7 \times 3 = 21\) and \(7 \times 6 = 42\). Because \(6\) is twice \(3\), the matching products double.

- Compare the numbers you divide by in each pair. - Then compare the two answers. - How are doubling one number and halving the answer connected here? - Use multiplication to check your answers.

1. a) \(16\div2=8\) and \(16\div4=4\); b) \(20\div2=10\) and \(20\div4=5\); c) \(40\div4=10\) and \(40\div8=5\). 2. In each pair, the number being divided stays the same while the number you divide by doubles. 3. The answer is cut in half each time.

a) \(8\) and \(4\) b) \(10\) and \(5\) c) \(10\) and \(5\) The answer is cut in half when the number you divide by doubles in these exact division facts.

- What happens to the number being divided in each pair? - Find all six answers before comparing them. - What happens to the answer when the number being divided doubles?

1. a) \(8\div2=4\) and \(16\div2=8\); b) \(12\div3=4\) and \(24\div3=8\); c) \(15\div5=3\) and \(30\div5=6\). 2. In each pair, the number being divided doubles while the number you divide by stays the same. 3. The answer also doubles.

a) \(4\) and \(8\) b) \(4\) and \(8\) c) \(3\) and \(6\) The answer doubles.

- Find both answers in every pair. - Compare how both numbers in each division fact change. - What happens to the answer when both numbers are doubled?

1. a) \(6\div3=2\) and \(12\div6=2\); b) \(10\div2=5\) and \(20\div4=5\); c) \(14\div2=7\) and \(28\div4=7\). 2. In each pair, both numbers in the division fact are doubled. 3. The answer stays the same.

a) \(2\) and \(2\) b) \(5\) and \(5\) c) \(7\) and \(7\) The answer stays the same when both numbers are doubled in these exact division facts.

- Decide whether each sequence increases or decreases. - Find the constant change between consecutive terms. - Continue by adding or subtracting that same amount.

1. For a), add \(25\) repeatedly after \(825\): \(850, 875, 900, 925, 950\). 2. For b), subtract \(5\) repeatedly after \(95\): \(90, 85, 80, 75, 70\). 3. For c), add \(20\) each time. The missing terms are \(380, 420,\) and \(460\).

a) \(850, 875, 900, 925, 950\) b) \(90, 85, 80, 75, 70\) c) \(380, 420, 460\)

- Use the ones digit to find even and odd numbers. - A multiple of \(10\) has \(0\) in the ones place. - After listing the numbers, describe the pattern you see in the ones digits.

1. Even numbers have \(0, 2, 4, 6,\) or \(8\) in the ones place. Counting by twos gives \(896\) through \(910\). 2. Odd numbers have \(1, 3, 5, 7,\) or \(9\) in the ones place. Counting by twos gives \(397\) through \(411\). 3. Multiples of \(10\) have \(0\) in the ones place. The requested numbers are \(680\) through \(720\) in steps of \(10\).

a) \(896, 898, 900, 902, 904, 906, 908, 910\); ones digits repeat \(0,2,4,6,8\). b) \(397, 399, 401, 403, 405, 407, 409, 411\); ones digits repeat \(1,3,5,7,9\). c) \(680, 690, 700, 710, 720\); every ones digit is \(0\).

- Track the change in the first number from line to line. - Track the change in the number being subtracted. - Combine those two changes to predict the next difference.

1. \(542 - 199 = 542 - 200 + 1 = 343\). 2. \(552 - 198 = 552 - 200 + 2 = 354\). 3. \(562 - 197 = 562 - 200 + 3 = 365\). 4. The first number increases by \(10\) while the number being subtracted decreases by \(1\), so each difference increases by \(11\). 5. The next expressions are \(572 - 196 = 376\) and \(582 - 195 = 387\).

\(542-199=343\) \(552-198=354\) \(562-197=365\) \(572-196=376\) \(582-195=387\) The first number increases by \(10\), and the number being subtracted decreases by \(1\). Both changes make the answer larger, so it increases by \(10+1=11\) each time.

- Compare the first numbers in consecutive expressions. - Compare the numbers being subtracted in the same way. - Think about the distance between two numbers when both increase equally.

1. \(735 - 399 = 735 - 400 + 1 = 336\). 2. \(736 - 400 = 336\). 3. \(737 - 401 = 336\). 4. From one expression to the next, both the first number and the number being subtracted increase by \(1\). Increasing both numbers by the same amount keeps their difference constant.

\(735 - 399 = 336\) \(736 - 400 = 336\) \(737 - 401 = 336\) The difference stays the same because both numbers increase by \(1\) each time.

- Compare one line with the line just before it. - In addition, if a number goes up, the sum goes up by the same amount. - In subtraction, changing the first number and changing the number being subtracted affect the answer in different directions.

1. a) Start with \(300+400=700\). The first number increases by \(20\), so the answer increases by \(20\) to \(720\). The second number increases by \(50\), so the answer increases by \(50\) to \(770\). The first number increases by \(6\), so the answer increases by \(6\) to \(776\). The second number increases by \(1\), so the answer increases by \(1\) to \(777\). 2. b) Start with \(800-500=300\). The first number increases by \(70\), so the answer increases by \(70\) to \(370\). The number being subtracted increases by \(30\), so the answer decreases by \(30\) to \(340\). The first number increases by \(9\), so the answer increases by \(9\) to \(349\). The number being subtracted increases by \(4\), so the answer decreases by \(4\) to \(345\).

a) Start: \(700\) The first number increases by \(20\), so the answer increases by \(20\) to \(720\). The second number increases by \(50\), so the answer increases by \(50\) to \(770\). The first number increases by \(6\), so the answer increases by \(6\) to \(776\). The second number increases by \(1\), so the answer increases by \(1\) to \(777\). b) Start: \(300\) The first number increases by \(70\), so the answer increases by \(70\) to \(370\). The number being subtracted increases by \(30\), so the answer decreases by \(30\) to \(340\). The first number increases by \(9\), so the answer increases by \(9\) to \(349\). The number being subtracted increases by \(4\), so the answer decreases by \(4\) to \(345\).

- Compare the two expressions in each pair and find what changed. - Decide whether that change makes the answer go up or down. - Say how the answer changes before giving the second answer.

1. a) \(450+300=750\). The second number goes up by \(20\), so the answer goes up by \(20\) to \(770\). 2. b) \(680-200=480\). The number being subtracted goes up by \(40\), so the answer goes down by \(40\) to \(440\). 3. c) \(512+100=612\). The second number goes up by \(7\), so the answer goes up by \(7\) to \(619\). 4. d) \(975-400=575\). The number being subtracted goes up by \(6\), so the answer goes down by \(6\) to \(569\).

a) The first answer is \(750\). The second number increases by \(20\), so the answer increases by \(20\) to \(770\). b) The first answer is \(480\). The number being subtracted increases by \(40\), so the answer decreases by \(40\) to \(440\). c) The first answer is \(612\). The second number increases by \(7\), so the answer increases by \(7\) to \(619\). d) The first answer is \(575\). The number being subtracted increases by \(6\), so the answer decreases by \(6\) to \(569\).

- Track how each digit changes from one line to the next. - Compare the three calculated differences. - Continue the same digit pattern for part c).

1. \(959 - 595 = 364\), \(848 - 484 = 364\), and \(737 - 373 = 364\). 2. Every difference is \(364\). 3. Each digit in both numbers decreases by \(1\) from one line to the next. 4. The next expressions are \(626 - 262 = 364\) and \(515 - 151 = 364\).

a) All three answers are \(364\). b) The answers are all the same. c) \(626 - 262 = 364\) and \(515 - 151 = 364\)

- Determine how the first number changes from line to line. - Notice that the number being subtracted remains fixed. - Look for a shared property of \(858\), \(747\), and \(636\).

1. The first number decreases by \(111\) each time, while \(121\) stays fixed. 2. Continue with \(646 - 121 = 525\), \(535 - 121 = 414\), and \(424 - 121 = 303\). 3. The results are also palindromes and decrease by \(111\) each time.

a) \(646 - 121 = 525\), \(535 - 121 = 414\), and \(424 - 121 = 303\) b) The results are palindromes that decrease by \(111\) each time.

- Notice how the first factor changes. - Compare \(30\) with \(60\). - Find the difference between consecutive products in each sequence.

1. Sequence a) continues with \(5 \times 60 = 300\), \(4 \times 60 = 240\), and \(3 \times 60 = 180\). The products decrease by \(60\). 2. Sequence b) continues with \(5 \times 30 = 150\), \(4 \times 30 = 120\), and \(3 \times 30 = 90\). The products decrease by \(30\). 3. Each product in b) is half the matching product in a) because \(30\) is half of \(60\).

a) \(5 \times 60 = 300\), \(4 \times 60 = 240\), \(3 \times 60 = 180\) b) \(5 \times 30 = 150\), \(4 \times 30 = 120\), \(3 \times 30 = 90\) The products decrease by \(60\) in a) and by \(30\) in b). Each b) product is half the matching a) product.

- Find what stays the same in each pair and what changes. - Compare \(2\) with \(4\). - Calculate both products in each pair before describing the pattern.

1. Part a: \(4 \times 2 = 8\) and \(4 \times 4 = 16\). 2. Part b: \(6 \times 2 = 12\) and \(6 \times 4 = 24\). 3. Part c: \(9 \times 2 = 18\) and \(9 \times 4 = 36\). 4. In each pair, the second factor doubles from \(2\) to \(4\), and the product also doubles.

a) \(8\) and \(16\) b) \(12\) and \(24\) c) \(18\) and \(36\) When one factor is doubled and the other factor stays the same, the product doubles.

- How does the first factor change in each pattern? - Which set of multiplication facts is used in Pattern A? - Which set of multiplication facts is used in Pattern B? - Can you find the next product by adding or subtracting one equal group?

1. In Pattern A, \(5 \times 6 = 30\) and \(6 \times 6 = 36\). The first factor increases by \(1\), so the product increases by \(6\) each step. 2. In Pattern B, \(7 \times 8 = 56\) and \(6 \times 8 = 48\). The first factor decreases by \(1\), so the product decreases by \(8\) each step.

Pattern A: \(30, 36\); the product increases by \(6\) each step. Pattern B: \(56, 48\); the product decreases by \(8\) each step.

- Compare the two smaller factors in each completed line. - Track how each smaller factor changes as the second factor increases by \(1\). - Check that the two smaller factors add to the original second factor.

1. The second factor is split into two addends that are equal or differ by \(1\). 2. Since \(6 = 3 + 3\), \(7 \times 6 = 7 \times 3 + 7 \times 3\). 3. Since \(7 = 3 + 4\), \(7 \times 7 = 7 \times 3 + 7 \times 4\). 4. Since \(8 = 4 + 4\), \(7 \times 8 = 7 \times 4 + 7 \times 4\).

\(7 \times 6 = (7 \times 3) + (7 \times 3)\) \(7 \times 7 = (7 \times 3) + (7 \times 4)\) \(7 \times 8 = (7 \times 4) + (7 \times 4)\)

- Consider each change separately. - One change raises the sum and the other lowers it. - Combine the two changes to find the overall change.

1. Increasing the first number being added increases the sum by \(210\). 2. Decreasing the second number being added decreases the sum by \(130\). 3. The overall change is \(210 - 130 = 80\). 4. Therefore the sum increases by \(80\).

The sum increases by \(80\).

- The answer increased by \(500\) altogether. - The first number caused \(320\) of that increase. - What increase is left for the second number?

1. The answer increases by \(500\) altogether. 2. Increasing the first number by \(320\) makes the answer increase by \(320\). 3. The rest of the increase is \(500-320=180\), so the second number increased by \(180\).

The second number increased by \(180\).

- Choose two numbers that are easy to add. - Increase one by \(15\) and decrease the other by \(15\). - Compare the old answer with the new answer.

1. For example, start with \(100+100=200\). 2. Increase the first number by \(15\): \(100+15=115\). 3. Decrease the second number by \(15\): \(100-15=85\). 4. The new equation is \(115+85=200\). 5. The answer stays the same because one change adds \(15\) and the other subtracts \(15\). The two changes cancel each other.

The answer stays the same. For example, \(100+100=200\) and \(115+85=200\). Adding \(15\) to one number and subtracting \(15\) from the other cancel each other.

- Decide how the first change affects the answer. - Decide how the second change affects the answer. - Combine the two changes, then use the starting answer of \(450\).

1. Raising the first number by \(80\) raises the answer by \(80\). 2. Lowering the second number by \(30\) lowers the answer by \(30\). 3. Together, the answer changes by \(80-30=50\). 4. The new answer is \(450+50=500\).

a) The sum increases by \(80\). b) The sum decreases by \(30\). c) The new sum is \(500\) because the overall change is an increase of \(50\).

- Decide how changing one number being added affects the sum. - The same change is made to both numbers being added. How many times does it affect the sum? - Combine the two decreases before changing the original sum.

1. Decreasing the first number being added by \(40\) decreases the sum by \(40\). 2. Decreasing the second number being added by \(40\) decreases the sum by another \(40\). 3. The total decrease is \(40 + 40 = 80\). 4. The new sum is \(500 - 80 = 420\).

The sum decreases by \(80\), so the new sum is \(420\).

- Think about what happens when you start with more. - Think about what happens when you subtract less. - Test the changes with an equation such as \(200 - 100\).

1. Increasing the first number by \(50\) increases the difference by \(50\). 2. Decreasing the number being subtracted by \(50\) also increases the difference by \(50\), because less is subtracted. 3. The total increase is \(50 + 50 = 100\).

The difference increases by \(100\).

- Decide how increasing one addend affects the sum. - Then decide how decreasing the other addend by the same amount affects it. - You may test the pattern with smaller numbers.

1. Increasing one addend by \(30\) increases the sum by \(30\): \(480 + 30 = 510\). 2. Decreasing the other addend by \(30\) decreases the new sum by \(30\): \(510 - 30 = 480\). 3. The final sum equals the original sum because the two changes cancel each other.

a) The new sum is \(510\). b) The resulting sum is \(480\), the same as the original sum.

- The first change raises the answer by \(50\). - The second change lowers the answer by \(80\). - Combine those two changes before applying them to \(550\).

1. The first change increases the answer by \(50\). 2. The second change decreases the answer by \(80\). 3. The decrease is \(30\) larger than the increase, so the answer decreases by \(30\). 4. The new answer is \(550-30=520\).

The answer decreases by \(30\). New answer: \(550-30=520\)

- Think of the difference as the distance between two numbers on a number line. - If one number moves \(40\) units to the right, what must happen to the other number to keep the distance unchanged? - You can represent the new number being subtracted with \(n\).

1. The new first number is \(600+40=640\). 2. Let \(n\) be the new number being subtracted. It must satisfy \(640-n=350\). 3. The new number being subtracted is \(640-350=290\), so \(n=290\). 4. Since \(290-250=40\), the number being subtracted must also increase by \(40\). 5. Adding the same amount to both numbers keeps their difference unchanged.

He must increase \(250\) by \(40\), making it \(290\). Then \(640-290=350\).

- Compare the increase of \(80\) with the target increase of \(50\). - Decide whether the second number being added must increase or decrease. - Test your idea with an equation such as \(100 + 100 = 200\).

1. The answer needs to increase by \(50\). 2. Increasing the first number by \(80\) would increase the answer by \(80\). 3. That is \(30\) too much. 4. Therefore, the second number must decrease by \(30\).

The second number must decrease by \(30\).

- Find each new difference and compare it with \(440\). - Think about how changing the first number differs from changing the number being subtracted. - On a number line, what happens to the distance if both numbers move the same amount?

1. Increasing the first number gives \(710 - 230 = 480\), which is \(40\) greater than \(440\). 2. Increasing the number being subtracted gives \(670 - 270 = 400\), which is \(40\) less than \(440\). 3. Increasing both numbers gives \(770 - 330 = 440\). The difference stays the same because the distance between the two numbers does not change.

a) The new difference is \(480\), which is \(40\) greater. b) The new difference is \(400\), which is \(40\) less. c) The difference remains \(440\).

- What happens to the difference when the amount you start with decreases? - What happens when the amount you subtract increases? - Try to find each change without doing the whole subtraction again.

1. The original difference is \(780 - 150 = 630\). 2. Decreasing the first number gives \(730 - 150 = 580\). The difference decreases by \(50\). 3. Increasing the number being subtracted gives \(780 - 170 = 610\). The difference decreases by \(20\).

a) \(630\) b) The difference decreases by \(50\), to \(580\). c) The difference decreases by \(20\), to \(610\).

- Decide how each change affects the sum. - Apply the two changes one at a time. - You may also combine them into one overall change.

1. Increasing the first number by \(50\) increases the sum to \(420+50=470\). 2. Decreasing the second number by \(30\) decreases the sum to \(470-30=440\).

The new sum is \(440\).

- Test the situation with an equation such as \(400 - 100 = 300\). - After decreasing the first number, decide whether you must subtract more or less. - Remember that changing both numbers by the same amount keeps their difference unchanged.

1. Decreasing the first number by \(50\) would decrease the difference from \(300\) to \(250\). 2. To restore the difference to \(300\), the difference must increase by \(50\). 3. Decreasing the number being subtracted by \(50\) increases the difference by \(50\). 4. Therefore, both numbers must decrease by the same amount.

The number being subtracted must also decrease by \(50\).

- Compare the first numbers from one line to the next. - Compare the numbers being subtracted from one line to the next. - Describe the rule when both numbers in a subtraction increase by the same amount. - Think of the difference as the distance between the two numbers.

1. Both numbers increase by \(10\), so the next equation is \(570 - 240 = 330\). 2. In each step, the first number and the number being subtracted both increase by \(10\). 3. Adding the same amount to both numbers does not change the distance between them, so the difference remains unchanged.

a) \(570 - 240 = 330\) b) Both numbers increase by \(10\). c) The difference stays the same because both numbers change by the same amount.

- Decide how increasing the first number affects the difference. - Decide how decreasing the number being subtracted affects the difference. - Apply the two changes one at a time.

1. Increasing the first number by \(30\) increases the difference to \(150 + 30 = 180\). 2. Decreasing the number being subtracted by \(10\) means less is subtracted, so the difference increases by another \(10\). 3. The new difference is \(180 + 10 = 190\).

The new difference is \(190\). Increasing the first number by \(30\) raises the difference to \(180\), and decreasing the number being subtracted by \(10\) raises it by another \(10\).

- Identify which number changes in each pair of expressions. - Decide whether that change makes the result greater or less. - Use the size of the change to compare the results. - Compare the expressions closely before calculating from scratch.

1. \(270 + 80 = 350\) and \(270 + 90 = 360\). 2. The second addend increases by \(10\), so the sum also increases by \(10\). 3. \(640 - 50 = 590\) and \(660 - 50 = 610\). 4. The first number in the second subtraction is \(20\) greater while the number being subtracted is unchanged, so the second difference is \(20\) greater.

a) The sums are \(350\) and \(360\). The second is \(10\) greater. b) The differences are \(590\) and \(610\). The second is \(20\) greater because its first number is \(20\) greater.

- Change each number by the stated amount. - Compare how much one number increases and the other decreases. - For part c), think about the effect of two increases.

1. The changed numbers are \(340+20=360\) and \(160-20=140\). 2. Their sum is \(360+140=500\), so the claim works. 3. The increase and decrease cancel, producing an overall change of \(0\). 4. If both numbers increase by \(20\), the sum increases by \(40\) to \(540\).

a) The new numbers are \(360\) and \(140\), and their sum is \(500\). b) The increase and decrease cancel each other. c) The sum increases by \(40\), to \(540\).

- In part a, compare the number removed with the number added. - In part b, find the difference between the number added and the number removed. - Use that change to find the new total.

1. For part a, the baker removes and adds the same number of rolls. These changes cancel, so the total remains \(480\). 2. For part b, compare the number added with the number removed: \(80 - 70 = 10\). The total increases by \(10\). 3. Find the new total: \(480 + 10 = 490\).

a) There are still \(480\) rolls altogether. b) The total increases by \(10\), so there would be \(490\) rolls.

- Decide how each change affects the sum. - Treat the increase and decrease as opposite changes. - Find the overall change.

1. Increasing the first number increases the sum by \(120\). 2. Decreasing the second number decreases the sum by \(150\). 3. The decrease is \(30\) larger than the increase, so the sum decreases by \(30\).

The sum decreases by \(30\).

- Think of subtraction as the first number minus the number being subtracted. - Decide how decreasing the first number affects the difference. - Decide how decreasing the number being subtracted affects the difference. - Test the two changes in order with a simple example such as \(100 - 50\).

1. Decreasing the first number by \(40\) decreases the difference by \(40\). 2. Decreasing the number being subtracted by \(15\) increases the difference by \(15\), because less is subtracted. 3. The increase of \(15\) offsets part of the decrease of \(40\). The difference therefore decreases by \(40-15=25\).

The difference decreases by \(25\).

- Picture the two numbers as points on a number line. - Decide how moving the greater number toward the lesser number changes the distance. - Then decide how moving the lesser number toward the greater number changes it. - In each step, decide whether the gap becomes larger or smaller.

1. Decreasing the greater number by \(20\) reduces the difference to \(150 - 20 = 130\). 2. Increasing the lesser number by \(10\) reduces the difference by another \(10\). 3. The new difference is \(130 - 10 = 120\).

The new difference is \(120\).

- Add adjacent bricks to fill the middle row. - Add the two middle-row bricks to find the top. - Compare the bottom middle numbers and note how much they increase. - Compare the top numbers and look for a relationship. - Find how many times the bottom middle number is used in the top.

1. In a), the middle row is \(10 + 15 = 25\) and \(15 + 12 = 27\), so the top brick is \(25 + 27 = 52\). 2. In b), the middle row is \(10 + 20 = 30\) and \(20 + 12 = 32\), so the top brick is \(30 + 32 = 62\). 3. In c), the middle row is \(10 + 25 = 35\) and \(25 + 12 = 37\), so the top brick is \(35 + 37 = 72\). 4. The top bricks increase by \(10\). The bottom middle number is used in both middle-row sums, so increasing it by \(5\) increases the top by \(2 \times 5 = 10\).

1. a) Middle row: \(25\), \(27\); top: \(52\) b) Middle row: \(30\), \(32\); top: \(62\) c) Middle row: \(35\), \(37\); top: \(72\) 2. Each increase of \(5\) in the bottom middle number increases the top by \(10\), because that number is used in both bricks in the middle row.

- Start where an upper brick and one brick below it are known. - Use subtraction to find the other lower brick. - Use addition when both lower bricks are known. - After completing both walls, compare the same positions in a) and b).

1. In a), the bottom-left brick is \(32 - 15 = 17\), the right brick in the middle row is \(15 + 24 = 39\), and the top is \(32 + 39 = 71\). 2. In b), the bottom-left brick is \(35 - 15 = 20\), the right brick in the middle row is \(15 + 24 = 39\), and the top is \(35 + 39 = 74\). 3. The bottom-left brick increases by \(3\), from \(17\) to \(20\), and the top also increases by \(3\), from \(71\) to \(74\).

a) Bottom row: \(17\), \(15\), \(24\); middle row: \(32\), \(39\); top: \(71\) b) Bottom row: \(20\), \(15\), \(24\); middle row: \(35\), \(39\); top: \(74\) The bottom-left brick and the top brick both increase by \(3\).

- Complete the original wall before changing any bottom values. - Compare how much each bottom brick changes with how much the top changes. - Track how often each bottom brick contributes to the top.

1. The middle row is \(6 + 8 = 14\) and \(8 + 12 = 20\), so the top is \(14 + 20 = 34\). 2. The new bottom row is \(8\), \(10\), \(14\). The middle row is \(18\), \(24\), and the top is \(18 + 24 = 42\). It is \(42 - 34 = 8\) greater. 3. If every bottom number increases by the same amount, the top increases by four times that amount: the outer numbers contribute once each, and the middle number contributes twice. An increase of \(5\) in each bottom number increases the top by \(4 \times 5 = 20\).

a) Middle row: \(14\), \(20\); top: \(34\) b) New bottom row: \(8\), \(10\), \(14\); middle row: \(18\), \(24\); top: \(42\). The top is \(8\) greater. c) The top would be \(20\) greater.

- Finish each wall separately before comparing them. - Track which two bricks in the next row use the changed middle value. - Compare the two top values after both walls are complete.

1. In Wall 1, the middle row is \(8 + 10 = 18\) and \(10 + 12 = 22\), so the top is \(18 + 22 = 40\). 2. In Wall 2, the middle row is \(8 + 11 = 19\) and \(11 + 12 = 23\), so the top is \(19 + 23 = 42\). 3. The bottom middle brick increases by \(1\), while the top increases by \(42 - 40 = 2\).

a) In Wall 1, the middle row is \(18\), \(22\), and the top is \(40\). In Wall 2, the middle row is \(19\), \(23\), and the top is \(42\). b) Increasing the bottom middle brick by \(1\) increases the top by \(2\).

- When an upper brick and one lower brick are known, subtract to find the other lower brick. - After completing the bottom row, compare the sums of each adjacent pair. - Look for repeated values in the rows above and connect them to the bottom-row pattern.

1. Find the first and third bricks in the bottom row: \(150-50=100\) in each case. 2. The middle brick in the second row is \(50+100=150\), so all three bricks in that row are \(150\). 3. Each brick in the third row is \(150+150=300\). 4. The top brick is \(300+300=600\). 5. The bottom row alternates \(100,50,100,50\). Because every adjacent pair sums to \(150\), the entire second row repeats \(150\); that repetition then creates two equal \(300\) bricks above.

Bottom row: \(100\), \(50\), \(100\), \(50\) Second row: \(150\), \(150\), \(150\) Third row: \(300\), \(300\) Top: \(600\) Pattern: the bottom row alternates \(100\) and \(50\), so every adjacent pair adds to \(150\). That makes every number in the second row the same.

- Start with the known \(250\) and the \(100\) directly below it. - Use the top value to work backward after the left side is known. - Compare the four completed bottom values only after the wall is finished.

1. The second bottom brick is \(250 - 100 = 150\). 2. The left brick in the second row is \(100 + 150 = 250\), so the left brick in the third row is \(250 + 250 = 500\). 3. The right brick in the third row is \(1000 - 500 = 500\). 4. The right brick in the second row is \(500 - 250 = 250\). 5. The fourth bottom brick is \(250 - 100 = 150\). 6. The bottom row alternates between \(100\) and \(150\).

Bottom row: \(100\), \(150\), \(100\), \(150\) Second row: \(250\), \(250\), \(250\) Third row: \(500\), \(500\) Top: \(1000\) The bottom-row pattern is alternating \(100, 150, 100, 150\).

- Track where each bottom brick is used in the row above. - A change in the middle bottom brick affects both bricks in the next row. - Compare one change that is used twice with two changes that are each used once.

1. The original middle row is \(15,15\), so the original top is \(30\). 2. The middle bottom brick is used in both middle-row bricks. Increasing it by \(5\) therefore increases the top by \(5+5=10\). 3. Each outer bottom brick is used on one side only. Increasing both outer bricks by \(5\) increases the top by \(5+5=10\). 4. Both changes increase the top by \(10\), though they change different bottom bricks.

a) Middle row: \(15,15\); top: \(30\) b) Top change: \(+10\) c) Top change: \(+10\) d) The middle brick is counted twice in the top, while the two outer bricks are each counted once, so both changes add \(10\) altogether.

- Finish one complete row before starting the next. - Compare the values in the second row before adding upward again. - Ask what happens when every neighboring pair in one row has the same sum.

1. The second row is \(50+60=110\), \(60+40=100\), \(40+70=110\), and \(70+30=100\). 2. The third row is \(110+100=210\), \(100+110=210\), and \(110+100=210\). 3. The fourth row is \(210+210=420\) and \(210+210=420\). 4. The top brick is \(420+420=840\). 5. The second row alternates \(110,100,110,100\). Every adjacent pair there sums to \(210\), so the whole third row repeats \(210\); that repetition then produces two equal \(420\) bricks.

The top value is \(840\). The second row alternates \(110\) and \(100\), so every adjacent pair sums to \(210\). Therefore, the third row is all \(210\), and the fourth row is \(420,420\).

- Use subtraction when an upper brick and one lower brick are known. - After completing each row, compare values that are the same distance from the left and right ends. - Think about why adding matching pairs on opposite sides makes the same mirror pattern in the row above.

1. The missing bottom values are \(7-4=3\) and \(7-4=3\), so the bottom row is \(2,3,4,3,2\). 2. The outside blanks in the second row are \(2+3=5\) and \(3+2=5\), giving \(5,7,7,5\). 3. The third row is \(5+7=12\), \(7+7=14\), and \(7+5=12\). 4. The fourth row is \(12+14=26\) and \(14+12=26\). 5. The top is \(26+26=52\). 6. Each row reads the same from left to right as from right to left. Because matching values on opposite sides are added in matching positions, the same mirror pattern appears in the row above.

Bottom row: \(2,3,4,3,2\) Second row: \(5,7,7,5\) Third row: \(12,14,12\) Fourth row: \(26,26\) Top: \(52\) Pattern: each row reads the same from left to right and right to left. Adding matching pairs on the two sides makes matching numbers in the row above.

- Complete both walls first and compare the top values. - Notice which bottom brick is used in both middle-row sums. - Compare the middle bottom values in the two walls while remembering that the bottom-row totals are equal.

1. In wall a), the second row is \(5+5=10\) and \(5+5=10\), so the top is \(10+10=20\). 2. In wall b), the second row is \(4+7=11\) and \(7+4=11\), so the top is \(11+11=22\). 3. Both bottom rows total \(15\), but the middle bottom brick is used to both second-row bricks. Wall b) has middle brick \(7\) instead of \(5\), so that greater middle value is counted twice in the top and makes wall b) larger.

Wall b) has the greater top value: \(22\), compared with \(20\) for wall a). The middle bottom brick is used in both sums above it, so wall b)’s larger middle value affects the top twice even though the two bottom rows have the same total.

- Complete both walls using the stated subtraction rule. - Compare the two top bricks. - Track how the bottom middle number affects both bricks in the middle row.

1. In a), the middle row is \(50 - 15 = 35\) and \(15 - 5 = 10\), so the top is \(35 - 10 = 25\). 2. In b), the middle row is \(50 - 16 = 34\) and \(16 - 5 = 11\), so the top is \(34 - 11 = 23\). 3. When the bottom middle number increases by \(1\), the top decreases from \(25\) to \(23\), a decrease of \(2\).

The top is \(25\) in a) and \(23\) in b). Increasing the bottom middle number by \(1\) decreases the top by \(2\).

- Read the row and column counts from each diagram before describing the change. - Extend both dimension patterns by one step. - Compare the total dots in the third and fourth arrays.

1. The diagrams show arrays of \(2 \times 3\), \(3 \times 4\), and \(4 \times 5\). Both the number of rows and the number of columns increase by \(1\) each time. 2. The next array has dimensions \(5 \times 6\), so it contains \(5 \times 6 = 30\) dots. 3. The third array contains \(4 \times 5 = 20\) dots, so the increase is \(30 - 20 = 10\) dots.

The rows and columns each increase by \(1\). The next array is \(5 \times 6\) with \(30\) dots, an increase of \(10\) dots.

- Read the height of each labeled bar from the vertical scale. - The halfway height must have the same difference from the lower and higher values. - Do not use left-to-right position; compare the numerical heights.

1. The bar heights are \(8\), \(24\), and \(16\) for A, B, and C. 2. Bar C has height \(16\). 3. \(16-8=8\) and \(24-16=8\). 4. Because the two differences are equal, C is exactly halfway in height between A and B.

a) Bar C b) \(16-8=8\) and \(24-16=8\)

- Compare the same bar in both graphs. - Find all three increases. - Decide whether the increases are equal.

1. A increased by \(15 - 10 = 5\). 2. B increased by \(25 - 20 = 5\). 3. C increased by \(35 - 30 = 5\). 4. Every bar increased by \(5\), so the statement is true.

A increased by \(5\), B increased by \(5\), and C increased by \(5\). The statement is true.

- The two rows in each row-pair have the same number of dots. - Match dots that are in the same column. - Ask whether any dot can be left without a partner when there are four equal rows.

1. Each pair of rows has the same number of dots in both rows. Pair each dot in one row with the dot directly below it in the other row, so no dot is left without a partner. 2. Any \(4\times n\) array has four equal rows. Pair the rows two at a time. Every dot belongs to a pair, so the total is even.

a) Each dot can be paired with the dot in the same column of the matching row, so no dot is left over. b) In every \(4\times n\) array, the four equal rows can be paired two at a time. Every dot has a partner, so every product in the \(4\)s facts is even.

- Can one group of \(6\) be split completely into pairs? - For the \(3\)s facts, look at what happens when one more group of \(3\) is added. - Does adding \(3\) switch an even answer to odd and an odd answer to even?

1. Every product in the \(6\)s facts is even. Each group of \(6\) can be split into \(3\) pairs, so any number of groups still has only pairs. 2. In the \(3\)s facts, the products alternate odd, even, odd, even as the second factor increases by \(1\). 3. Each new fact adds another group of \(3\), which is odd. Adding an odd number switches a total from odd to even or from even to odd.

a) All the \(6\)s products are even because each group of \(6\) can be split into pairs with nothing left over. b) The \(3\)s products alternate odd, even, odd, even. Each new fact adds \(3\), so the total switches between odd and even each time.

- Try a multiplication fact with one even factor and one odd factor. - One example is enough to show that Mason is wrong. - After finding an example, say what must be true about at least one factor to make an even product.

1. Mason’s claim is false. For example, \(4\times3=12\). The product \(12\) is even though \(3\) is odd. 2. One even factor is enough to make a product even. The even factor can be split into pairs, and repeating complete pairs keeps the total even.

Mason is not correct. For example, \(4 \times 3=12\). A correct pattern is: if at least one factor is even, the product is even.

- Think about the value of the two tens digits and the value of the two ones digits. - If the ones digits trade places, does their combined value change? - For the least and greatest sums, decide which two digits should be in the tens places.

1. Swapping the ones digits does not change the total value of the ones digits, and the tens digits stay where they are. So the sum does not change. 2. Once the two tens digits are chosen, the other two cards must be the ones digits. That fixes both the tens total and the ones total. 3. For the least sum, put \(2\) and \(4\) in the tens places. One arrangement is \(25+48=73\). 4. For the greatest sum, put \(5\) and \(8\) in the tens places. One arrangement is \(52+84=136\). 5. There are \(6\) possible pairs of tens digits, giving the sums \(73,82,100,109,127,136\).

a) Swapping the ones digits keeps their total the same, so the sum does not change. b) After the tens digits are chosen, the other two digits must be the ones digits, so the sum is fixed. c) Least sum: \(73\); greatest sum: \(136\); \(6\) different sums.

- Subtract carefully by place value. - Compare the tens digits of the results. - Add the hundreds and ones digits in each result. - Think about why different starting numbers can produce the same difference.

1. \(421 - 124 = 297\). 2. \(623 - 326 = 297\). 3. \(825 - 528 = 297\). 4. Every result is \(297\). Its tens digit is \(9\), and its hundreds and ones digits add to \(2 + 7 = 9\).

a) \(297\) b) \(297\) c) \(297\) d) All three results are \(297\). The tens digit is \(9\), and the outer digits add to \(9\).

- Calculate all three expressions in Set A and compare the results. - Repeat for Set B. - Compare the two repeated results by finding their difference.

1. Set A gives \(654 - 456 = 198\), \(765 - 567 = 198\), and \(876 - 678 = 198\). 2. Set B gives \(642 - 246 = 396\), \(753 - 357 = 396\), and \(864 - 468 = 396\). 3. All three answers in each set are the same. Compare the two repeated answers: \(396 - 198 = 198\). 4. Therefore, every result in Set B is \(198\) greater than the result in the same position in Set A.

a) Set A: \(198\), \(198\), \(198\); Set B: \(396\), \(396\), \(396\) b) Each Set B result is \(198\) greater than the result in the same position in Set A.

- Rearrange the digits again after every subtraction. - Put the greatest digit in the hundreds place for the greatest number. - Stop as soon as a result has appeared before. - Check each subtraction carefully, especially when regrouping is needed.

1. Arrange \(8,4,1\): \(841 - 148 = 693\). 2. Arrange \(9,6,3\): \(963 - 369 = 594\). 3. Arrange \(9,5,4\): \(954 - 459 = 495\). 4. Repeating the rule with \(9,5,4\) gives \(954 - 459 = 495\) again. 5. The result repeats on the fourth subtraction step.

\(841 - 148 = 693\) \(963 - 369 = 594\) \(954 - 459 = 495\) \(954 - 459 = 495\) The first repeated result appears on step \(4\).

- Arrange the digits from greatest to least and then from least to greatest. - Keep using all three digit positions, including a zero when needed. - Look for a number that appears in both chains.

1. Maya: \(721-127=594\). Theo: \(830-38=792\). 2. Maya then finds \(954-459=495\). 3. Theo finds \(972-279=693\), then \(963-369=594\), then \(954-459=495\). 4. Both chains reach \(594\) and then \(495\), but Theo needs more steps.

a) Maya: \(721-127=594\); Theo: \(830-38=792\). b) Maya: \(594\rightarrow495\). Theo: \(792\rightarrow693\rightarrow594\rightarrow495\). c) Both chains reach \(594\) and then \(495\).

- Look at how many sides touch one corner. - Track how many times a corner is included in all three side sums. - How is the side number connected to the two corners beside it?

1. Each corner touches two sides, so its number is included in two side sums. 2. The three corners together are half of the combined side sums: \(42\div2=21\). 3. The side labeled \(17\) contains the two corners next to it. The opposite corner is \(21-17=4\).

a) Each corner belongs to two sides and is counted in two side sums. b) \(21\) c) \(4\), because the other two corner numbers together make \(17\), and \(21-17=4\).

- First find the difference after only the first number increases. - Compare that answer with \(320\). - To make a difference smaller, should the number being subtracted increase or decrease? - Test the same idea with a simple subtraction such as \(10-5=5\).

1. Increasing the first number by \(50\) would increase the difference to \(300 + 50 = 350\). 2. The target difference, \(320\), is \(350 - 320 = 30\) less than \(350\). 3. To decrease the difference by \(30\), increase the number being subtracted by \(30\).

The number being subtracted must increase by \(30\). Raising the first number by \(50\) would raise the difference from \(300\) to \(350\). To get \(320\), subtract \(30\) more.

- Trace which bricks in the middle row use each bottom-row value. - The middle bottom brick is used in both sums directly above the bottom row. - Predict the effect of the changes on the top before calculating any new wall bricks.

1. The middle row is \(1+4=5\) and \(4+2=6\), so the original top is \(11\). 2. In a three-brick bottom row, the middle bottom brick is used in both middle-row sums, while each outer bottom brick is used once. 3. Increasing the left brick by \(3\) adds \(3\) to the top. Increasing the middle brick by \(3\) adds \(6\). Increasing the right brick by \(3\) adds \(3\). 4. The total change is \(3+6+3=12\), so the new top would be \(23\).

a) Original top: \(11\) b) The top increases by \(12\), to \(23\). c) The middle bottom brick is used in both bricks above it, so its change is counted twice; each outer bottom brick is used only once.

- Look for given numbers that occur at equal distances to the left and right of the center. - After finding one unknown on the left, check whether the same relationship appears at the matching position on the right. - The explanation should name the mirror relationship, not only list the completed numbers.

1. The bottom outside bricks are each \(75-25=50\), giving bottom row \(50,25,100,25,50\). 2. The next row is \(75,125,125,75\); then \(200,250,200\); then \(450,450\); and the top is \(900\). 3. Mirror pairs are equal: bottom positions \(1\) and \(5\), bottom positions \(2\) and \(4\); second-row positions \(1\) and \(4\), positions \(2\) and \(3\); third-row positions \(1\) and \(3\); and the two fourth-row bricks. 4. Because the givens mirror each other around the center, once a value is found on one side, its matching position on the other side has the same value.

a) Bottom: \(50,25,100,25,50\) Second row: \(75,125,125,75\) Third row: \(200,250,200\) Fourth row: \(450,450\) Top: \(900\) b) Matching positions the same distance from the center have equal numbers. c) The starting numbers match on the two sides, so the same calculations repeat. Once you find a number on one side, the matching place on the other side has the same number.

- Complete each wall upward before comparing the tops. - Track how the \(+20\) change in the left outer brick reaches the top. - Track how the \(-10\) change in the middle brick affects both second-row bricks.

1. In wall 1, the second row is \(200+100=300\) and \(100+200=300\), so the top is \(600\). 2. In wall 2, the second row is \(220+90=310\) and \(90+200=290\), so the top is \(600\). 3. Increasing the left outer brick by \(20\) increases the top by \(20\), because an outer brick is used once. Decreasing the middle brick by \(10\) decreases both second-row bricks by \(10\), lowering the top by \(20\). The two changes cancel.

Both top values are \(600\). The \(+20\) change in an outer bottom brick adds \(20\) to the top, while the \(-10\) change in the middle bottom brick subtracts \(20\) from the top because the middle brick is used twice. The overall change is \(0\).