Free Math Worksheets

- Think about what happens when you subtract \(1\) from a number ending in zeros. - Then add \(1\) to \(600{,}000\).

1. Subtract \(1\) to find the number immediately before: \(600{,}000 - 1 = 599{,}999\). 2. Add \(1\) to find the next number: \(600{,}000 + 1 = 600{,}001\).

a) \(599{,}999\) b) \(600{,}001\)

- What happens to the place values when you subtract exactly \(1\)? - First find the next counting number, and then move forward one more number. - Think about the place-value change between \(999{,}999\) and \(1{,}000{,}000\).

1. a) Subtract \(1\): \(1{,}000{,}000-1=999{,}999\). 2. b) Add \(1\) twice: \(1{,}000{,}000+1=1{,}000{,}001\), then \(1{,}000{,}001+1=1{,}000{,}002\).

a) \(999{,}999\) b) \(1{,}000{,}002\)

- Write \(0\) for an empty column. - Read the columns from greatest place value to least.

1. Read the chip counts by place value: \(4\) hundred thousands, \(0\) ten thousands, \(7\) thousands, \(3\) hundreds, \(2\) tens, and \(8\) ones. 2. The number is \(407{,}328\).

The number is \(407{,}328\).

- The hundreds digit cannot be \(0\). - Choose the hundreds digit first, then arrange the remaining cards. - Compare the completed numbers from left to right.

1. A three-digit number cannot begin with \(0\), so the hundreds digit must be \(5\) or \(9\). 2. Beginning with \(5\) gives \(509\) and \(590\). Beginning with \(9\) gives \(905\) and \(950\). 3. There are \(4\) numbers. The least is \(509\), and the greatest is \(950\).

a) \(509, 590, 905, 950\); \(4\) numbers b) Greatest: \(950\); least: \(509\)

- Compare the last three digits in all three starting numbers. - First find how much must be added to \(417\) to reach \(1000\). - Decide whether the digits to the left change the needed amount.

1. For a), \(1000 - 417 = 583\). 2. In all three problems, the starting number ends in \(417\), and the target is the next thousand. 3. Therefore, the same amount is needed each time: \(26{,}000 - 25{,}417 = 583\) and \(804{,}000 - 803{,}417 = 583\).

a) \(583\) b) \(583\) c) \(583\)

- Identify the least number with exactly six digits. - Identify the greatest number with exactly six digits. - Think about the boundaries between five-, six-, and seven-digit numbers.

1. The least six-digit number is \(100{,}000\). One less is \(100{,}000 - 1 = 99{,}999\). 2. The greatest six-digit number is \(999{,}999\). One more is \(999{,}999 + 1 = 1{,}000{,}000\).

a) \(99{,}999\) b) \(1{,}000{,}000\)

- A halfway number has the same distance from both endpoints. - Find the distance between the two numbers, then take half of it. - Add that half-distance to the lesser number.

1. For a), the total distance is \(20{,}000\). Half is \(10{,}000\), so the halfway number is \(130{,}000\). 2. For b), the total distance is \(10{,}000\). Half is \(5000\), so the halfway number is \(125{,}000\). 3. For c), the total distance is \(2000\). Half is \(1000\), so the halfway number is \(123{,}000\).

a) \(130{,}000\) b) \(125{,}000\) c) \(123{,}000\)

- What is the greatest digit? - How can you make each of the six places as large as possible? - Think about what happens when you add \(1\) to a number made entirely of nines.

1. The greatest digit is \(9\). Placing a \(9\) in each of the six places gives \(999{,}999\), the greatest six-digit whole number. 2. Adding \(1\) causes every digit to regroup: \(999{,}999 + 1 = 1{,}000{,}000\). 3. The number name for \(1{,}000{,}000\) is one million.

a) \(999{,}999\) b) \(1{,}000{,}000\), one million

- A digit has the greatest value when it is placed farthest to the left. - Arrange the largest available digits from greatest to least. - To find the number immediately after a number, add \(1\).

1. To make the greatest six-digit number, place the six greatest available digits in descending order: \(9, 8, 7, 6, 5\), and \(4\). This gives \(987{,}654\). 2. The number immediately after it is found by adding \(1\): \(987{,}654 + 1 = 987{,}655\).

a) \(987{,}654\) b) \(987{,}655\)

- Think about which number comes directly before and directly after each given number when you count. - Pay special attention to numbers that end in several 9s. - What happens to the place values when you subtract \(1\) from a number ending in 00?

1. a) Subtract and add \(1\): \(9999-1=9998\) and \(9999+1=10{,}000\). 2. b) Subtract and add \(1\): \(20{,}100-1=20{,}099\) and \(20{,}100+1=20{,}101\). 3. c) Subtract and add \(1\): \(149{,}999-1=149{,}998\) and \(149{,}999+1=150{,}000\). 4. d) Subtract and add \(1\): \(1{,}000{,}000-1=999{,}999\) and \(1{,}000{,}000+1=1{,}000{,}001\).

a) Before: \(9998\); after: \(10{,}000\) b) Before: \(20{,}099\); after: \(20{,}101\) c) Before: \(149{,}998\); after: \(150{,}000\) d) Before: \(999{,}999\); after: \(1{,}000{,}001\)

- How are the number before, the given number, and the number after related? - When you know the number after, how can you work backward to the given number? - Starting with the number before, how many times do you add \(1\) to reach the number after?

1. In the first row, the number is \(10{,}099\). One less is \(10{,}098\), and one more is \(10{,}100\). 2. In the second row, the number before is \(49{,}999\), so the number is \(50{,}000\). The number after is \(50{,}001\). 3. In the third row, the number after is \(300{,}000\), so the number is \(299{,}999\). The number before is \(299{,}998\).

Row 1: Number before \(10{,}098\); number after \(10{,}100\) Row 2: Number \(50{,}000\); number after \(50{,}001\) Row 3: Number before \(299{,}998\); number \(299{,}999\)

- When you know the number immediately before a number, how can you find the number itself? - How many ones must you add to the number before \(n\) to reach the number after \(n\)? - Pay attention to the place-value changes across the tens, hundreds, and thousands places.

1. Add \(1\) to the number before \(n\): \(789{,}999+1=790{,}000\), so \(n=790{,}000\). 2. Add \(1\) to \(n\): \(790{,}000+1=790{,}001\).

The number \(n\) is \(790{,}000\), and the number immediately after it is \(790{,}001\).

- Use \(0\) for any place value that is not named. - A place-value chart can help you align the digits. - Identify the greatest place value in each description first.

1. a) Place the named digits in their positions and use zeros elsewhere: \(570{,}209\). 2. b) Place \(8\) in the hundred-thousands place, \(4\) in the thousands place, and \(3\) in the tens place: \(804{,}030\). 3. c) Place \(6\) in the ten-thousands place, \(5\) in the hundreds place, and \(1\) in the tens place: \(60{,}510\).

a) \(570{,}209\) b) \(804{,}030\) c) \(60{,}510\)

- Match each named place value to its digit position. - Use zeros for place values that are not named. - Write each description as a numeral before matching.

1. A gives \(705{,}020\), so A matches 2. 2. B gives \(70{,}502\), so B matches 3. 3. C gives \(750{,}200\), so C matches 1.

A–2, B–3, C–1

- Test possible ones digits in order. - Double each ones digit to get the tens digit. - A digit must be from \(0\) through \(9\), and a two-digit number cannot begin with \(0\).

1. The ones digit can be \(1, 2, 3\), or \(4\). If it were \(5\) or greater, twice the digit would not be a single digit. 2. Doubling these ones digits gives tens digits \(2, 4, 6\), and \(8\). 3. The numbers are \(21, 42, 63\), and \(84\).

\(21, 42, 63, 84\)

- List the odd digits greater than \(6\). - Subtract \(2\) from each possible tens digit. - Check both conditions for each number.

1. The odd digits greater than \(6\) are \(7\) and \(9\). 2. If the tens digit is \(7\), the ones digit is \(7-2=5\), giving \(75\). 3. If the tens digit is \(9\), the ones digit is \(9-2=7\), giving \(97\).

\(75\) and \(97\)

- Place the number in a place-value chart. - Count the places from right to left: ones, tens, hundreds, thousands, and so on. - Read the thousands period and the ones period separately.

1. The digits in \(305{,}704\), from left to right, are in the hundred-thousands, ten-thousands, thousands, hundreds, tens, and ones places. 2. The digit in the hundred-thousands place is \(3\). 3. The digit in the ten-thousands place is \(0\). 4. The number in words is three hundred five thousand seven hundred four.

a) \(3\) b) \(0\) c) three hundred five thousand seven hundred four

- The leftmost places have the greatest effect on a number’s value. - Pay attention to whether digits may repeat. - Use the greatest allowed digit in the greatest place when maximizing.

1. a) Use the five greatest digits in decreasing order: \(98{,}765\). 2. b) The first digit must be at least \(1\), and \(0\) is not allowed, so use \(1\) in every place: \(11{,}111\). 3. c) Repetition is allowed, so use the greatest allowed digit, \(7\), in every place: \(77{,}777\).

a) \(98{,}765\) b) \(11{,}111\) c) \(77{,}777\)

- Count the digits in each number. - Compare the digit in each place across all five numbers. - Add the digits of each number.

1. Every number has four digits. 2. Every number has \(4\) in the hundreds place. 3. The digit sums are \(2+4+3+3=12\), \(5+4+1+2=12\), \(1+4+6+1=12\), \(8+4+0+0=12\), and \(3+4+4+1=12\). Therefore, every number has a digit sum of \(12\).

1. Every number has four digits. 2. Every number has a hundreds digit of \(4\). 3. Every number has a digit sum of \(12\).

- First determine how many groups of \(10{,}000\) the number contains. - A place-value chart can help you account for every place. - Consider what would happen to the other digits if a zero in the middle were removed.

1. Forty thousand is \(4\) ten thousands and \(0\) additional thousands. Eight hundred is \(8\) hundreds, \(0\) tens, and \(0\) ones. 2. The place-value amounts are \(4\) ten thousands, \(0\) thousands, \(8\) hundreds, \(0\) tens, and \(0\) ones. 3. These digits form \(40{,}800\). 4. The zero in the thousands place shows that there are no additional thousands, and the zero in the tens place shows that there are no tens. The zeros hold the other digits in their correct places.

a) \(4\) ten thousands, \(0\) thousands, \(8\) hundreds, \(0\) tens, and \(0\) ones b) \(40{,}800\) c) Those place values contain no units, so zeros serve as placeholders and keep the other digits in the correct positions.

- Missing place values need zeros in standard form. - A place-value chart can help you position each digit. - More than \(9\) of one unit may need to be regrouped. - Read each place-value name carefully before writing the digit.

1. For a), \(6 \times 1000 + 2 \times 100 + 5 = 6205\). 2. For b), \(4 \times 10{,}000 + 8 \times 1000 + 3 \times 10 = 48{,}030\). 3. For c), \(7 \times 100{,}000 + 1 \times 10{,}000 + 9 \times 100 = 710{,}900\). 4. For d), \(5 \times 1000 + 4 \times 10 = 5040\). 5. For e), \(23 \times 100 = 2300\).

a) \(6205\) b) \(48{,}030\) c) \(710{,}900\) d) \(5040\) e) \(2300\)

- Translate each place-value amount into its numerical value. - Missing places become zeros in standard form. - Regroup amounts greater than \(9\) units of one place when helpful. - Add the place-value amounts to check your number.

1. For a), \(60{,}000 + 2000 + 500 + 80 + 1 = 62{,}581\). 2. For b), \(300{,}000 + 40{,}000 + 700 + 5 = 340{,}705\). 3. For c), \(25 \times 1000 + 12 \times 10 = 25{,}120\). 4. For d), \(9 \times 10{,}000 + 40 \times 100 = 94{,}000\).

a) \(62{,}581\) b) \(340{,}705\) c) \(25{,}120\) d) \(94{,}000\)

- Picture each number in a place-value chart. - Compare the value of a \(5\) in the thousands place with a \(5\) in the tens place. - Think of each zero as holding an empty place open.

1. Fifty thousand eight hundred is \(50{,}800\), and five thousand eight is \(5008\). 2. In \(50{,}800\), zeros are in the thousands, tens, and ones places. 3. In \(5008\), the digit \(5\) is in the thousands place and has a value of \(5000\). 4. In \(58\), the digit \(5\) is in the tens place and has a value of \(50\). Removing the placeholder zeros changes the digit’s place and greatly reduces its value.

a) \(50{,}800\) and \(5008\) b) The thousands, tens, and ones places c) The \(5\) would move from the thousands place to the tens place. Its value would change from \(5000\) to \(50\).

- A place value that is not named has a digit of \(0\). - Use a place-value chart to align the digits. - Begin with the greatest place value in each description.

1. a) All places from thousands through ones are named, giving \(7253\). 2. b) The hundreds digit is \(0\), giving \(4081\). 3. c) The tens digit is \(0\), giving \(906\). 4. d) The thousands and ones digits are \(0\), giving \(30{,}520\).

a) \(7253\) b) \(4081\) c) \(906\) d) \(30{,}520\)

- Count the chips in each column carefully. - Write a zero for any column that has no chips. - Record the digits from the greatest place value to the ones place.

1. Chart a) shows \(2\) hundred-thousands, \(0\) ten-thousands, \(6\) thousands, \(3\) hundreds, \(9\) tens, and \(5\) ones. The number is \(206{,}395\). 2. Chart b) shows \(5\) hundred-thousands, \(4\) ten-thousands, \(1\) thousand, \(0\) hundreds, \(2\) tens, and \(8\) ones. The number is \(541{,}028\).

a) \(206{,}395\) b) \(541{,}028\)

- Move from the greatest place value to the ones place. - Record the number of chips in each column as one digit.

1. Count the chips from left to right: \(3\) ten-thousands, \(8\) thousands, \(2\) hundreds, \(5\) tens, and \(9\) ones. 2. These digits form \(38{,}259\).

The number is \(38{,}259\).

- Decrease the thousands digit by \(1\) and increase the ten-thousands digit by \(1\). - Find the net change in value caused by the move.

1. Remove one chip from the thousands column and add one chip to the ten-thousands column. 2. The net increase is \(10{,}000-1000=9000\). 3. Add the increase: \(241{,}000+9000=250{,}000\).

The new number is \(250{,}000\).

- Compare the thousands period in the numeral with the incorrect label. - Identify the value of the digit that was omitted.

1. The thousands period is \(321\), not \(301\). 2. The digit \(2\) in the ten-thousands place represents \(20{,}000\). 3. The correct number name is three hundred twenty-one thousand twenty-one.

The label omits the \(2\) in the ten-thousands place, which represents \(20{,}000\). The correct number name is three hundred twenty-one thousand twenty-one.

- Compare the thousands period in the numeral with the incorrect label. - Identify the value of the missing digit.

1. The thousands period is \(999\), not \(909\). 2. The omitted \(9\) is in the ten-thousands place and represents \(90{,}000\). 3. The correct number name is nine hundred ninety-nine thousand ninety-nine.

The label omits the \(9\) in the ten-thousands place, which represents \(90{,}000\). The correct number name is nine hundred ninety-nine thousand ninety-nine.

- Convert the incorrect number name to standard form. - Compare its thousands period with \(440\).

1. Four hundred four thousand forty represents \(404{,}040\), not \(440{,}040\). 2. In \(440{,}040\), the \(4\) in the ten-thousands place represents \(40{,}000\). 3. The correct word form is four hundred forty thousand forty.

The incorrect name omits the \(4\) in the ten-thousands place, which represents \(40{,}000\). The correct word form is four hundred forty thousand forty.

- The first digit of a four-digit number cannot be \(0\). - To make a number small, keep the digits on the left as small as possible. - To make a number large, place the greatest possible digits on the left.

1. To make the least number, use the smallest possible thousands digit, \(1\). Put as much of the remaining digit sum as possible in the ones place: \(1+0+2+9=12\). This gives \(1029\). 2. To make the greatest number, use the largest possible thousands digit, \(9\). Put the remaining sum of \(3\) in the hundreds place: \(9+3+0+0=12\). This gives \(9300\).

Least: \(1029\); greatest: \(9300\)

- Count how many nonzero digits the number must have. - For the greatest number, place large digits as far left as possible. - For the least number, use the smallest nonzero leading digit and place zeros as far left as possible after it.

1. A six-digit number with exactly four zeros has two nonzero digits. 2. To make the greatest number, place \(9\) in the two greatest places and zeros elsewhere: \(990{,}000\). 3. To make the least number, place the smallest nonzero digit, \(1\), in the hundred-thousands place. Put the four zeros immediately after it and the other \(1\) in the ones place: \(100{,}001\).

Greatest: \(990{,}000\); least: \(100{,}001\)

- Decide how many digits are not \(3\). - For the greatest number, use large digits in the leftmost available places. - For the least number, use the smallest nonzero leading digit and place a zero as far left as possible.

1. Four of the six places must contain \(3\), leaving two places for other digits. 2. To make the greatest number, use \(9\) in the two greatest places and put \(3\) in the remaining places: \(993{,}333\). 3. To make the least number, use the smallest possible leading digit, \(1\), then place \(0\) in the next place and \(3\) in the remaining four places: \(103{,}333\).

Greatest: \(993{,}333\); least: \(103{,}333\)

- Count how many digits must be nonzero. - To maximize the number, put large nonzero digits in the greatest places. - To minimize the number, use the smallest possible leading digit and place zeros immediately after it.

1. A six-digit number with exactly three zeros has three nonzero digits. 2. To make the greatest number, place \(9\) in the three greatest places and zeros elsewhere: \(999{,}000\). 3. To make the least number, use the smallest possible leading digit, \(1\), place the three zeros immediately after it, and put \(1\) in each remaining place: \(100{,}011\).

Greatest: \(999{,}000\); least: \(100{,}011\)

- Choose which two places contain \(3\). - Fill the other two places with \(7\). - Put the results in order to check for missing or repeated numbers.

1. Arrange two \(3\)s and two \(7\)s systematically. 2. Numbers beginning with \(3\) are \(3377, 3737\), and \(3773\). 3. Numbers beginning with \(7\) are \(7337, 7373\), and \(7733\). 4. There are \(6\) different numbers.

\(3377, 3737, 3773, 7337, 7373, 7733\)

- Which digits can appear in the ones place of an odd number? - For the greatest number, keep the largest digits as far left as possible. - For the least number, keep the smallest digits as far left as possible while preserving an odd ones digit.

1. An odd number must end in \(5\), \(7\), or \(9\). 2. To make the greatest number, place the smallest available odd digit, \(5\), in the ones place so the larger digits can occupy the higher places. Arrange the remaining digits in descending order to get \(987{,}225\). 3. To make the least number, place the largest available odd digit, \(9\), in the ones place so the smaller digits can occupy the higher places. Arrange the remaining digits in ascending order to get \(225{,}789\).

a) \(987{,}225\) b) \(225{,}789\)

- Locate the ten-thousands place in a six-digit number. - Choose the smallest possible hundred-thousands digit that still makes the number greater than \(500{,}000\). - Arrange the remaining digits from least to greatest.

1. The ten-thousands digit is fixed as \(1\), the second digit from the left. 2. To be greater than \(500{,}000\), the hundred-thousands digit must be \(5\), \(7\), or \(9\). Choose \(5\) to make the number as small as possible. 3. Arrange the remaining digits \(4, 5, 7\), and \(9\) in ascending order in the remaining places. This gives \(514{,}579\).

\(514{,}579\)

- For the greatest number, place the largest digits in the highest-value places. - Reverse that strategy for the least number. - For the closest number, compare the best possible number below the target with the best possible number above it.

1. Arrange the digits in descending order for the greatest number: \(876{,}432\). 2. Arrange the digits in ascending order for the least number: \(234{,}678\). 3. The closest number below \(500{,}000\) must begin with \(4\). Make the remaining digits as large as possible to get \(487{,}632\), which is \(500{,}000 - 487{,}632 = 12{,}368\) away. 4. The closest number above \(500{,}000\) must begin with \(6\). Make the remaining digits as small as possible to get \(623{,}478\), which is \(623{,}478 - 500{,}000 = 123{,}478\) away. 5. Since \(12{,}368 < 123{,}478\), the closest number is \(487{,}632\).

a) \(876{,}432\) b) \(234{,}678\) c) \(487{,}632\)

- For the number below the target, choose the greatest possible leading digit below \(6\). - Arrange the remaining digits to make that number as large as possible. - For the number above the target, begin with \(6\) and make the remaining places as small as possible. - Subtract to compare the two distances.

1. Number A must begin with \(5\). Arrange the remaining digits in descending order to make the greatest number below the target: \(597{,}610\). 2. Its distance is \(600{,}000 - 597{,}610 = 2390\). 3. Number B must begin with \(6\). Arrange the remaining digits in ascending order to make the least number above the target: \(601{,}579\). 4. Its distance is \(601{,}579 - 600{,}000 = 1579\). 5. Since \(1579 < 2390\), Number B is closer.

Number A: \(597{,}610\), distance \(2390\) Number B: \(601{,}579\), distance \(1579\) Number B is closer to \(600{,}000\).

- Find the greatest possible number that begins with \(24\). - Find the least possible number that begins with \(25\). - Calculate and compare both distances from \(250{,}000\).

1. The greatest possible number below \(250{,}000\) must begin with \(24\). Arrange the remaining digits in descending order to get \(246{,}531\). 2. Its distance is \(250{,}000 - 246{,}531 = 3469\). 3. The least possible number above \(250{,}000\) must begin with \(25\). Arrange the remaining digits in ascending order to get \(251{,}346\). 4. Its distance is \(251{,}346 - 250{,}000 = 1346\). 5. Since \(1346 < 3469\), \(251{,}346\) is closer to the target.

\(251{,}346\) is closest to \(250{,}000\). Its distance is \(1346\), compared with a distance of \(3469\) for \(246{,}531\).

- Use the stated interval to determine the first digit. - Find two remaining digits for which one is twice the other. - Find two remaining digits whose difference is \(3\). - Place the final unused digit in the only open place.

1. The hundred-thousands digit must be \(4\) because the number is between \(400{,}000\) and \(500{,}000\). 2. Among the remaining digits, the only pair in which one digit is twice the other is \(6\) and \(3\). Therefore, the thousands digit is \(6\), and the ones digit is \(3\). 3. Of the remaining digits \(5, 8\), and \(9\), the pair differing by \(3\) is \(8\) and \(5\). Therefore, the hundreds digit is \(8\), and the tens digit is \(5\). 4. The remaining digit, \(9\), goes in the ten-thousands place. The number is \(496{,}853\).

\(496{,}853\)

- Use the size condition to determine the first digit. - Use the even-number condition to determine the last digit. - Find two remaining digits whose sum is \(8\). - Remember that the repeated digit can occupy different places.

1. The hundred-thousands digit must be \(9\) because the number is greater than \(900{,}000\). 2. The ones digit must be \(2\), the only even digit available. 3. The remaining digits are \(1, 1, 5\), and \(7\). The only pair with a sum of \(8\) is \(1\) and \(7\), which can appear in either order in the ten-thousands and thousands places. 4. For each of those two orders, the remaining digits \(1\) and \(5\) can appear in either order in the hundreds and tens places. 5. The four numbers are \(917{,}152\), \(917{,}512\), \(971{,}152\), and \(971{,}512\).

\(917{,}152, 917{,}512, 971{,}152, 971{,}512\)

- Use the interval to determine the first two digits. - A whole number divisible by \(10\) must end in \(0\). - Among the remaining digits, find two whose difference is \(2\). - Place the final unused digit in the remaining position.

1. The interval fixes the hundred-thousands digit as \(2\) and the ten-thousands digit as \(4\). 2. Divisibility by \(10\) fixes the ones digit as \(0\). 3. The remaining digits are \(5, 6\), and \(8\). The only pair for which the thousands digit is \(2\) greater than the hundreds digit is \(8\) and \(6\). 4. The remaining digit, \(5\), goes in the tens place, giving \(248{,}650\).

\(248{,}650\)

- Digits in the thousands place have the greatest effect on the sum. - A four-digit number cannot begin with \(0\). - Distribute the available digits by place value before deciding which number receives each digit.

1. To maximize the sum, place the largest digits in the thousands places, the next largest in the hundreds places, and so on. One arrangement is \(7531 + 6420 = 13{,}951\). 2. To minimize the sum, use \(1\) and \(2\) in the thousands places, then place \(0\) and \(3\) in the hundreds places, \(4\) and \(5\) in the tens places, and \(6\) and \(7\) in the ones places. One arrangement is \(1046 + 2357 = 3403\).

a) \(13{,}951\) b) \(3403\)

- For part a), make one number as large as possible and the other as small as possible. - For a small difference, choose numbers with close leading digits. - Arrange the lower-place digits to bring the two numbers closer together. - Addition is the inverse operation of subtraction.

1. For the greatest difference, make the minuend as large as possible and the subtrahend as small as possible: \(98{,}765 - 10{,}234 = 88{,}531\). 2. One arrangement with a difference less than \(300\) is \(50{,}123 - 49{,}876 = 247\). 3. The inverse addition equation is \(247 + 49{,}876 = 50{,}123\).

a) \(98{,}765 - 10{,}234 = 88{,}531\) b) One possible answer is \(50{,}123 - 49{,}876 = 247\). c) \(247 + 49{,}876 = 50{,}123\)

- Arrange the digits from greatest to least for the greatest number and reverse the order for the least number. - “Difference” indicates subtraction. - For part c, write a missing-number equation such as \(97{,}531 - \Box = 80{,}000\).

1. The greatest number is \(97{,}531\), and the least number is \(13{,}579\). Their difference is \(97{,}531 - 13{,}579 = 83{,}952\). 2. The difference from \(50{,}000\) is \(50{,}000 - 13{,}579 = 36{,}421\). 3. Solve \(97{,}531 - x = 80{,}000\). The missing number is \(97{,}531 - 80{,}000 = 17{,}531\).

a) \(97{,}531 - 13{,}579 = 83{,}952\) b) \(36{,}421\) c) \(17{,}531\)

- Double the interval for the quotient to find the interval for the original number. - Use that interval to identify the hundreds digit. - Use the relationship between the hundreds and ones digits. - Place the given tens digit in the remaining position.

1. If half the number is between \(220\) and \(230\), the whole number is between \(440\) and \(460\). 2. Therefore, the hundreds digit is \(4\). 3. Since the hundreds digit is twice the ones digit, the ones digit is \(4 \div 2 = 2\). 4. With \(4\) in the tens place, the number is \(442\). 5. Check: \(442 \div 2 = 221\), which is between \(220\) and \(230\).

\(442\)

- For a), first find the unknown number itself, and then find the number before it. - For b), the unknown number is halfway between the number before it and the number after it. - Try the same relationship with \(5\): what is the sum of the numbers before and after \(5\)?

1. a) If the number after is \(8000\), the unknown number is \(8000-1=7999\). The number before it is \(7999-1=7998\). 2. b) The numbers immediately before and after a number are equally far from it, so their sum is twice the number. Divide by \(2\): \(20{,}000\div 2=10{,}000\). 3. c) One less than \(99{,}909\) is \(99{,}908\), and one more is \(99{,}910\).

a) \(7998\) b) \(10{,}000\) c) Before: \(99{,}908\); after: \(99{,}910\)

- The leftmost digits have the greatest effect on the value. - Put smaller digits first for the least number and larger digits first for the greatest. - Subtract the least number from the greatest.

1. a) To make the least number, place the smaller digits first: \(44{,}477\). 2. b) To make the greatest number, place the larger digits first: \(77{,}444\). 3. c) Subtract: \(77{,}444-44{,}477=32{,}967\).

a) \(44{,}477\) b) \(77{,}444\) c) \(32{,}967\)

- Work through the clues one place value at a time. - A place with no value has the digit \(0\). - Translate “half” into a division calculation.

1. The ten-thousands digit is \(4\). 2. The thousands digit is \(4+3=7\). 3. The hundreds and tens digits are \(0\). 4. The ones digit is \(4\div 2=2\). 5. The number is \(47{,}002\), or \(4\) ten thousands, \(7\) thousands, and \(2\) ones.

Standard form: \(47{,}002\) Place-value description: \(4\) ten thousands, \(7\) thousands, and \(2\) ones

- Write each place-value amount in expanded form. - Remember that \(12\) tens is the same as \(120\). - Use zeros as placeholders for place values that are not named.

1. a) Add the place-value amounts: \(700{,}000+4000+20+5=704{,}025\). 2. b) Since \(12\) tens equals \(120\), \(125{,}000+800+120=125{,}920\). 3. c) Add the place-value amounts: \(300{,}000+300+30+3=300{,}333\).

a) \(704{,}025\) b) \(125{,}920\) c) \(300{,}333\)

- List tens digits from \(1\) through \(9\) and find the needed ones digit. - A number is odd when its ones digit is odd. - Count the odd numbers in your completed list.

1. a) The digit pairs that sum to \(10\) are \((1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2)\), and \((9,1)\). They form \(19, 28, 37, 46, 55, 64, 73, 82\), and \(91\). 2. b) The odd numbers are those with an odd ones digit: \(19, 37, 55, 73\), and \(91\). There are \(5\).

a) \(19, 28, 37, 46, 55, 64, 73, 82, 91\) b) \(5\)

- List the odd and even digits. - A four-digit number cannot begin with \(0\). - In part c, test tens digits whose product with \(4\) is still one digit.

1. a) The greatest odd digit is \(9\), so the greatest number is \(9999\). 2. b) The least possible nonzero even thousands digit is \(2\). Use the least remaining even digits in order: \(0, 4, 6\). This gives \(2046\). 3. c) The number has the form \(T0Z0\), with \(T=4\times Z\). The possible digit pairs are \((4,1)\) and \((8,2)\). The greater number is \(8020\).

a) \(9999\) b) \(2046\) c) \(8020\)

- Write each named amount in expanded form. - Combine the amounts carefully, using zeros as placeholders where needed. - For part b), add each digit of the completed numeral exactly once.

1. a) Add the place-value amounts: \(800{,}000+15{,}000+400+72=815{,}472\). 2. b) Add the digits: \(8+1+5+4+7+2=27\).

a) \(815{,}472\) b) The sum of the digits is \(27\).

- Write both numbers as numerals first. - Count all digits from the first nonzero digit through the ones place. - Subtract the smaller digit count from the larger one.

1. Number 1 is \(207{,}980\), which has \(6\) digits. 2. Number 2 is \(9800\), which has \(4\) digits. 3. The difference is \(6-4=2\) digits.

Number 1 has \(6\) digits. Number 2 has \(4\) digits. The digit counts differ by \(2\).

- Compare the number of digits. - Compare the digit in each place. - Find the digit sum of every number.

1. The first four numbers are three-digit numbers, have \(5\) in the tens place, and have a digit sum of \(12\). 2. The number \(355\) is also three-digit and has \(5\) in the tens place. 3. However, its digit sum is \(3+5+5=13\), not \(12\). Therefore, \(355\) does not fit.

\(355\), because its digit sum is \(13\) while the other numbers have a digit sum of \(12\).

- Arrange the given digits from greatest to least to make the greatest number. - To make the least number, put zeros as far left as possible. - A six-digit number cannot begin with \(0\).

1. The available digits are \(3, 3, 3, 0, 0\), and \(0\). 2. For the greatest number, place the \(3\)s in the greatest places and the zeros in the remaining places. This gives \(333{,}000\). 3. For the least number, the first digit must be \(3\). Place the three zeros immediately after it, then place the remaining two \(3\)s in the last two places. This gives \(300{,}033\).

Greatest: \(333{,}000\); least: \(300{,}033\)

- How many digits are nonzero? - Find all pairs of nonzero digits with a sum of \(12\). - Place the digits and zeros to maximize or minimize their place values. - The least number cannot begin with \(0\).

1. Exactly two digits are nonzero, and those two digits must have a sum of \(12\). The possible unordered pairs are \((3, 9), (4, 8), (5, 7)\), and \((6, 6)\). 2. For the greatest number, use \(9\) in the hundred-thousands place and \(3\) in the ten-thousands place, followed by four zeros. This gives \(930{,}000\). 3. For the least number, use the smallest possible leading digit, \(3\), place the four zeros next, and put \(9\) in the ones place. This gives \(300{,}009\).

Greatest: \(930{,}000\); least: \(300{,}009\)

- First add one copy of each required digit. - How many places remain after the four required digits are used? - Choose digits for the remaining places to maximize or minimize the sum.

1. Using each digit from \(0\) through \(3\) once accounts for four places and gives a digit sum of \(0+1+2+3=6\). 2. Two places remain. 3. To make the greatest digit sum, use \(3\) in both remaining places: \(6+3+3=12\). 4. To make the least digit sum, use \(0\) in both remaining places: \(6+0+0=6\). The digits can still be arranged so the number does not begin with \(0\).

Greatest possible digit sum: \(12\) Least possible digit sum: \(6\)

- Find each of the two numbers separately. - To make a number as small as possible, keep its leftmost digits small. - An odd number has an odd ones digit. - Add the two numbers after you identify them.

1. The least four-digit number with a digit sum of \(5\) begins with \(1\). Put zeros in the next two places and use \(4\) in the ones place, giving \(1004\). 2. The greatest three-digit number is \(999\), and it is odd. 3. Add the two numbers: \(1004+999=2003\).

\(2003\)

- Use the greatest possible digit in every unrestricted place of the first number. - A five-digit number cannot begin with \(0\). - Make sure no digit is repeated in the second number. - Find the difference by subtracting the lesser number from the greater number.

1. To make the greatest five-digit number with \(7\) in the hundreds place, use \(9\) in every other place. The number is \(99{,}799\). 2. To make the least five-digit number with no repeated digits, use \(1\) first, then arrange the smallest remaining digits in increasing order: \(0, 2, 3\), and \(4\). The number is \(10{,}234\). 3. Subtract: \(99{,}799-10{,}234=89{,}565\).

\(89{,}565\)

- A digit contributes more when it is placed in the hundreds place than in the ones place. - For the greatest sum, put the greatest digits in the greatest places. - For the least sum, put the least digits in the greatest places. - The way the two digits in the same place are divided between the numbers does not change the sum.

1. a) To maximize the sum, put the two greatest digits, \(8\) and \(7\), in the hundreds places; \(5\) and \(4\) in the tens places; and \(2\) and \(1\) in the ones places. 2. One possible pair is \(852\) and \(741\), and \(852+741=1593\). 3. b) To minimize the sum, put the two least digits, \(1\) and \(2\), in the hundreds places; \(4\) and \(5\) in the tens places; and \(7\) and \(8\) in the ones places. 4. One possible pair is \(147\) and \(258\), and \(147+258=405\).

a) \(1593\) b) \(405\)

- To make the number small, place smaller digits in higher-value places whenever the conditions allow. - A five-digit number must begin with a nonzero digit. - Compare the meanings of \(01234\) and \(1234\). - Fill the places from left to right using the least possible available digit.

1. The tens digit is fixed as \(9\). 2. The ten-thousands digit must be the least available nonzero digit, \(3\). 3. Place the two zeros in the thousands and hundreds places to keep the number as small as possible, and place \(5\) in the ones place. The number is \(30{,}095\). 4. Its digits by place are \(3\) ten-thousands, \(0\) thousands, \(0\) hundreds, \(9\) tens, and \(5\) ones. 5. A leading zero does not create a place value. For example, \(03095\) represents \(3095\), which has only four digits.

a) \(30{,}095\) b) Ten-thousands: \(3\); thousands: \(0\); hundreds: \(0\); tens: \(9\); ones: \(5\) c) A leading zero is not counted as a digit place, so the result would be a four-digit number.

- Evaluate each representation separately. - Pay close attention to the difference between tens and groups of ten thousand. - Regroup \(60\) hundreds as \(6\) thousands. - Compare each resulting number with \(406{,}030\).

1. Representation 1 equals \(4 \times 100{,}000 + 6 \times 1000 + 3 \times 10 = 406{,}030\), so it is correct. 2. Representation 2 equals \(40 \times 10{,}000 + 6 \times 100 + 3 \times 10 = 400{,}630\), so it is incorrect. 3. Representation 3 equals \(406 \times 1000 + 3 = 406{,}003\), so it is incorrect. 4. Representation 4 equals \(400{,}000 + 6000 + 30 = 406{,}030\), so it is correct. 5. Representation 5 equals \(4 \times 100{,}000 + 60 \times 100 + 3 \times 10 = 406{,}030\), so it is correct.

1) Correct 2) Incorrect: \(400{,}630\) 3) Incorrect: \(406{,}003\) 4) Correct 5) Correct

- For part a, place each amount in its named place value. - For part b, identify the value of the digit \(4\) in \(34{,}071\). - For part c, regroup \(10\) hundreds as \(1\) thousand. - Convert both descriptions to standard form before comparing.

1. For a), \(40{,}000 + 2000 + 800 + 30 + 5 = 42{,}835\). 2. For b), the digit \(4\) in \(34{,}071\) is in the thousands place, so the missing description is \(4\) thousands. 3. For c), \(A = 2000 + 1300 + 5 = 3305\). 4. Also, \(B = 3000 + 300 + 1 = 3301\). 5. Since \(3305 > 3301\), number \(A\) is greater.

a) \(42{,}835\) b) \(4\) thousands c) \(A\) is greater because \(A = 3305\) and \(B = 3301\).

- Convert each place-value amount into a number. - Regroup \(10\) units of one place as \(1\) unit of the next greater place. - Add all the place-value amounts carefully.

1. For a), \(14{,}000 + 1200 + 5 = 15{,}205\). 2. For b), \(30{,}000 + 2500 + 180 = 32{,}680\). 3. For c), \(600{,}000 + 4000 + 220 = 604{,}220\). 4. For d), \(80{,}000 + 9000 + 1000 = 90{,}000\).

a) \(15{,}205\) b) \(32{,}680\) c) \(604{,}220\) d) \(90{,}000\)

- Write each number name in standard form before comparing the numbers. - In part b), identify every place value that is named and use zeros for the places that are not named. - In part c), determine the value of the final digit in \(900{,}009\).

1. Seven hundred thousand seven is \(700{,}007\). Seven hundred seven thousand is \(707{,}000\). Since \(707{,}000 > 700{,}007\), seven hundred seven thousand is greater. 2. For b), the specified place values are \(500{,}000 + 20{,}000 + 800 + 1\). The thousands and tens places contain zeros, so the number is \(520{,}801\). 3. In \(900{,}009\), the final \(9\) is in the ones place, so the number is nine hundred thousand nine. In nine hundred thousand ninety, the \(9\) must be in the tens place. The correct numeral is \(900{,}090\).

a) Seven hundred seven thousand, or \(707{,}000\), is greater because \(707{,}000 > 700{,}007\). b) \(520{,}801\) c) The \(9\) was placed in the ones place instead of the tens place. The correct number is \(900{,}090\).

- Use a place-value chart to write each representation in standard form. - For Numbers A and C, use zeros for place values that are not shown. - Find the difference by subtracting the least number from the greatest number.

1. Number A has digits \(4\), \(8\), \(0\), \(3\), \(0\), and \(2\) in order from the hundred-thousands place to the ones place, so Number A is \(480{,}302\). 2. Number B is four hundred eighty thousand plus three hundred two, so Number B is \(480{,}302\). 3. Number C is \(400{,}000 + 8000 + 300 + 2 = 408{,}302\). 4. Numbers A and B are equal because both are \(480{,}302\). 5. The greatest number is \(480{,}302\), and the least is \(408{,}302\). Their difference is \(480{,}302 - 408{,}302 = 72{,}000\).

a) Number A: \(480{,}302\); Number B: \(480{,}302\); Number C: \(408{,}302\) b) Numbers A and B are equal. c) \(72{,}000\)

- Use zeros as placeholders when you write each number. - Find the value of a \(9\) in the tens place and the value of a \(9\) in the hundreds place. - Compare the place values of the two differing digits.

1. Two hundred five thousand ninety is \(205{,}090\), so the first weight is \(205{,}090\,\text{tons}\). 2. Two hundred five thousand nine hundred is \(205{,}900\), so the second weight is \(205{,}900\,\text{tons}\). 3. In \(205{,}900\), the \(9\) is worth \(9 \times 100 = 900\). In \(205{,}090\), the \(9\) is worth \(9 \times 10 = 90\). All other corresponding digits are equal, and \(900 > 90\), so the student is correct.

a) First ship: \(205{,}090\,\text{tons}\); second ship: \(205{,}900\,\text{tons}\) b) Yes. The \(9\) in the hundreds place has a value of \(900\), while the \(9\) in the tens place has a value of \(90\). Since the other digits match, \(205{,}900\) is greater.

- Ten units of one place value can be regrouped as one unit of the next greater place value. - For example, \(12\) tens is \(1\) hundred and \(2\) tens. - You can also write each place-value amount as a number and add.

1. a) Regroup \(12\) tens as \(1\) hundred and \(2\) tens. This gives \(5\) hundreds, \(2\) tens, and \(5\) ones, or \(525\). 2. b) Regroup \(15\) hundreds as \(1\) thousand and \(5\) hundreds. This gives \(3\) thousands, \(5\) hundreds, \(3\) tens, and \(0\) ones, or \(3530\). 3. c) Regroup \(24\) hundreds as \(2\) thousands and \(4\) hundreds. This gives \(6\) ten-thousands, \(2\) thousands, \(4\) hundreds, \(0\) tens, and \(7\) ones, or \(62{,}407\).

a) \(525\) b) \(3530\) c) \(62{,}407\)

- Work through the clues one place at a time. - “No tens” means the tens digit is \(0\). - Check that the completed number has four digits.

1. The thousands digit is \(7\). 2. The hundreds digit is \(7-3=4\). 3. The tens digit is \(0\). 4. The ones digit is \(4\times2=8\). 5. The number is \(7\) thousands, \(4\) hundreds, \(0\) tens, and \(8\) ones, or \(7408\).

Place-value description: \(7\) thousands, \(4\) hundreds, \(0\) tens, and \(8\) ones Standard form: \(7408\)

- A five-digit number cannot begin with \(0\). - Work from left to right, choosing the least unused digit for each unrestricted place. - Keep the tens digit fixed at \(8\). - Check that no digit repeats.

1. Use the smallest possible nonzero digit, \(1\), in the ten-thousands place. 2. Use the smallest remaining digit, \(0\), in the thousands place, and then \(2\) in the hundreds place. 3. The tens digit is fixed at \(8\). 4. The smallest unused digit for the ones place is \(3\). The number is \(10{,}283\).

\(10{,}283\)

- First find the target number. - Which two place values differ by \(9000\)? - Check that the starting column has a chip available to move.

1. The chart shows \(24{,}167\). 2. The target number is \(24{,}167+9000=33{,}167\). 3. Moving one chip from the thousands column to the ten-thousands column changes its value from \(1000\) to \(10{,}000\), an increase of \(10{,}000-1000=9000\). 4. The new number is \(33{,}167\).

Move one chip from the thousands column to the ten-thousands column. The new number is \(33{,}167\).

- Find the value removed from the hundreds column. - Find the value added to the thousands column. - Subtract the decrease from the increase.

1. Two chips in the hundreds column have a value of \(2\times100=200\). 2. In the thousands column, those two chips have a value of \(2\times1000=2000\). 3. The net increase is \(2000-200=1800\). 4. The new number is \(42{,}517+1800=44{,}317\).

The new number is \(44{,}317\), so the number increases by \(1800\).

- Write \(1010\) as a sum of place values. - Which chip has a value of \(1000\)? - Which chip has a value of \(10\)?

1. Decompose the required decrease: \(1010=1000+10\). 2. Remove one chip from the thousands column and one chip from the tens column. 3. The thousands digit changes from \(2\) to \(1\), and the tens digit changes from \(8\) to \(7\). The new number is \(51{,}174\). 4. Check: \(52{,}184-51{,}174=1010\).

Remove one chip from the thousands column and one chip from the tens column. The new number is \(51{,}174\).

- What happens to a chip’s value when it moves one column to the left? - Find two columns whose values differ by exactly \(9\).

1. The chart shows \(10{,}004\). 2. The target is \(10{,}004+9=10{,}013\). 3. Moving one chip from the ones column to the tens column changes its value from \(1\) to \(10\), an increase of \(10-1=9\).

Move one chip from the ones column to the tens column.

- To decrease the value, a chip must move to the right. - Which two columns have values that differ by exactly \(90\)?

1. The chart shows \(50{,}120\). 2. The target is \(50{,}120-90=50{,}030\). 3. Moving one chip from the hundreds column to the tens column changes its value from \(100\) to \(10\), a decrease of \(100-10=90\).

Move one chip from the hundreds column to the tens column.

- Find two adjacent place values whose difference is \(900\). - Check that the starting column contains at least one chip.

1. The chart shows \(40{,}500\). 2. The target is \(40{,}500+900=41{,}400\). 3. Moving one chip from the hundreds column to the thousands column changes its value from \(100\) to \(1000\), an increase of \(1000-100=900\).

Move one chip from the hundreds column to the thousands column.

- Find the value lost from the hundreds column. - Find the value gained in the thousands column. - Combine the two changes.

1. Removing one chip from the hundreds column decreases the number by \(100\). 2. Adding that chip to the thousands column increases the number by \(1000\). 3. The net change is \(1000-100=900\), so the number increases by \(900\).

The number increases by \(900\).

- Express \(990\) as the difference of two place values. - The chip must move more than one column. - Check the change by subtracting the chip's old value from its new value.

1. Write the required increase as \(990=1000-10\). 2. Move one chip from the tens column, where it is worth \(10\), to the thousands column, where it is worth \(1000\). 3. The new number is \(111{,}111-10+1000=112{,}101\), an increase of \(990\).

Move one chip from the tens column to the thousands column.

- Choose a hundreds digit first. - Find all pairs of tens and ones digits that make the remaining sum. - Organize the possibilities so none are missed.

1. If the hundreds digit is \(1\), the tens and ones digits must sum to \(4\): \(104, 113, 122, 131, 140\). There are \(5\). 2. If the hundreds digit is \(2\), there are \(4\): \(203, 212, 221, 230\). 3. If the hundreds digit is \(3\), there are \(3\): \(302, 311, 320\). 4. If the hundreds digit is \(4\), there are \(2\): \(401, 410\). 5. If the hundreds digit is \(5\), there is \(1\): \(500\). 6. The total is \(5+4+3+2+1=15\).

\(15\) three-digit numbers

- Choose four different digits. - For the least sum, repeat the smallest chosen digit. - For the greatest sum, repeat the largest chosen digit. - Make sure the example does not begin with \(0\).

1. The number uses four different digits, with one of them repeated. 2. a) For the least sum, use \(0, 1, 2\), and \(3\), and repeat \(0\): \(0+0+1+2+3=6\). One valid example is \(10{,}023\). 3. b) For the greatest sum, use \(6, 7, 8\), and \(9\), and repeat \(9\): \(6+7+8+9+9=39\). One valid example is \(99{,}876\).

a) Least digit sum: \(6\); example: \(10{,}023\) b) Greatest digit sum: \(39\); example: \(99{,}876\)

- Start with the maximum digit sum for a three-digit number. - Find all digit groups that are \(3\) below that maximum. - Rearrange each digit group and remove duplicates.

1. The greatest possible digit sum is \(9+9+9=27\). A sum of \(24\) is \(3\) less than this maximum. 2. The possible digit groups are \(6,9,9\); \(7,8,9\); and \(8,8,8\). 3. Arrange each group to get all numbers: \(699, 969, 996\); \(789, 798, 879, 897, 978, 987\); and \(888\).

\(699, 789, 798, 879, 888, 897, 969, 978, 987, 996\)

- Determine which digits must occupy the first three places. - List every arrangement of the remaining three digits. - Compare each completed number with \(844{,}100\).

1. To be greater than \(844{,}100\), the number must begin with \(844\). Any smaller arrangement of the first three digits would be below the target. 2. The remaining digits are \(0\), \(1\), and \(2\). Their possible three-digit endings are \(012, 021, 102, 120, 201\), and \(210\). 3. The endings \(012\) and \(021\) give numbers below \(844{,}100\). 4. The valid numbers are \(844{,}102\), \(844{,}120\), \(844{,}201\), and \(844{,}210\).

\(844{,}102 < 844{,}120 < 844{,}201 < 844{,}210\)

- For the greatest difference, create one very large number and one very small number. - For the smallest difference, begin the numbers with consecutive thousands digits. - Arrange the remaining digits to pull the two numbers as close together as possible.

1. For the greatest difference, make one number as large as possible and the other as small as possible: \(8765 - 1234 = 7531\). 2. For the smallest positive difference, the thousands digits must be consecutive. For each consecutive pair, put the smallest available digits after the greater thousands digit and the largest available digits after the lesser thousands digit. 3. The best differences for the possible thousands-digit pairs are \(2345 - 1876 = 469\), \(3145 - 2876 = 269\), \(4125 - 3876 = 249\), \(5123 - 4876 = 247\), \(6123 - 5874 = 249\), \(7123 - 6854 = 269\), and \(8123 - 7654 = 469\). 4. The smallest of these differences is \(247\), from \(5123 - 4876\).

a) \(8765 - 1234 = 7531\) b) \(5123 - 4876 = 247\)

- Choose ten-thousands digits whose sum is near \(8\). - Use the lower places to make the total approach the next ten-thousand. - To test whether exactly \(80{,}000\) is possible, track the required carries in each place.

1. One valid pair is \(10{,}243\) and \(69{,}758\); together they use each digit from \(0\) through \(9\) exactly once. 2. Their sum is \(10{,}243 + 69{,}758 = 80{,}001\), which is a distance of \(1\) from \(80{,}000\). 3. A sum of exactly \(80{,}000\) is impossible. In the ones place, the two digits would need a sum of \(10\). With the resulting carries, the digit pairs in the tens, hundreds, and thousands places would each need a sum of \(9\), and the ten-thousands digits would need a sum of \(7\). 4. Those required digit-pair sums total \(10 + 9 + 9 + 9 + 7 = 44\), but the digits from \(0\) through \(9\) have a total of \(45\). Therefore, distance \(0\) is impossible, and the achieved distance of \(1\) is optimal.

One possible pair is \(10{,}243\) and \(69{,}758\). Their sum is \(80{,}001\), and the distance from \(80{,}000\) is \(1\).

- Remember that the number from Group A cannot begin with \(0\). - For the smallest difference, choose ten-thousands digits that differ by \(1\), then arrange the remaining digits to close the gap. - For the greatest difference, compare the largest possible number from one group with the smallest possible number from the other.

1. For the smallest difference, the ten-thousands digits must differ by \(1\). For each adjacent pair, arrange the remaining digits so the greater number is as small as possible and the lesser number is as large as possible. 2. The closest differences for the possible adjacent leading digits are \(20{,}468 - 19{,}753 = 715\), \(31{,}579 - 28{,}640 = 2939\), \(40{,}268 - 39{,}751 = 517\), \(51{,}379 - 48{,}620 = 2759\), \(60{,}248 - 59{,}731 = 517\), \(71{,}359 - 68{,}420 = 2939\), \(80{,}246 - 79{,}531 = 715\), and \(91{,}357 - 86{,}420 = 4937\). 3. Therefore, the smallest possible positive difference is \(517\). It occurs for \(40{,}268\) and \(39{,}751\), and also for \(60{,}248\) and \(59{,}731\). 4. For the greatest difference, compare the extreme possibilities in both directions: \(97{,}531 - 20{,}468 = 77{,}063\) and \(86{,}420 - 13{,}579 = 72{,}841\). 5. Since \(77{,}063 > 72{,}841\), the greatest possible difference is \(77{,}063\).

a) The smallest possible difference is \(517\), for example \(40{,}268 - 39{,}751 = 517\). b) The greatest possible difference is \(77{,}063\), from \(97{,}531 - 20{,}468 = 77{,}063\).

- Look for two numbers whose sum is just below or just above \(100{,}000\). - Think about the ten-thousands digits and the carries from the lower places. - To test whether an exact sum is possible, determine the digit-pair sum required in each place.

1. One valid equation is \(49{,}876 + 50{,}123 = 99{,}999\). Every digit from \(0\) through \(9\) is used exactly once. 2. The distance from \(100{,}000\) is \(100{,}000 - 99{,}999 = 1\). 3. A sum of exactly \(100{,}000\) is impossible. The ones digits would need a sum of \(10\). After the carry, the digit pairs in the tens, hundreds, thousands, and ten-thousands places would each need a sum of \(9\). 4. Those required digit-pair sums total \(10 + 9 + 9 + 9 + 9 = 46\), but the digits from \(0\) through \(9\) total only \(45\). Therefore, distance \(0\) is impossible, and distance \(1\) is optimal.

One possible equation is \(49{,}876 + 50{,}123 = 99{,}999\). The distance from \(100{,}000\) is \(1\).

- For part a), look for digit pairs that differ by \(1\). - Use every digit exactly once in each equation. - For part b), begin with hundreds digits whose difference is about \(5\).

1. For part a), one valid arrangement is \(753 - 642 = 111\). 2. For part b), one valid arrangement is \(763 - 254 = 509\), and \(500 < 509 < 510\).

a) One possible answer is \(753 - 642 = 111\). b) One possible answer is \(763 - 254 = 509\).

- The thousands digit cannot be \(0\). - Count the arrangements for each allowed thousands digit. - List numbers in groups based on their first digit.

1. a) The thousands digit has \(3\) choices: \(4, 5\), or \(9\). The remaining digits can be arranged in \(3\times2\times1=6\) ways. Thus, \(3\times6=18\) numbers can be made. 2. b) The six numbers beginning with \(4\) occupy positions \(1\) through \(6\): \(4059, 4095, 4509, 4590, 4905, 4950\). 3. The numbers beginning with \(5\) start \(5049, 5094, 5409, 5490\). Therefore, \(5490\) is tenth.

a) \(18\) numbers b) \(5490\)

- Count the choices for each place. - To be greater than \(6000\), decide which digits can be in the thousands place. - For decreasing order, choose the greatest available digit at each place first.

1. a) There are \(4\times3\times2\times1=24\) arrangements. 2. b) A number greater than \(6000\) must begin with \(6\) or \(8\). Each choice leaves \(3\times2\times1=6\) arrangements, so there are \(12\) such numbers. 3. c) In decreasing order, the numbers are \(8642, 8624, 8462, 8426, 8264, 8246, 6842, 6824, 6482, 6428, 6284, 6248\).

a) \(24\) b) \(12\) c) \(8642, 8624, 8462, 8426, 8264, 8246, 6842, 6824, 6482, 6428, 6284, 6248\)

- Choose the two positions for the digit \(6\). - Fill the other positions with \(1\). - Compare numbers from the greatest place value to the least.

1. Arrange the digits \(6, 6, 1, 1, 1\) systematically. There are \(10\) different arrangements. 2. Compare the digits from left to right to order the numbers. 3. The decreasing order is \(66{,}111>61{,}611>61{,}161>61{,}116>16{,}611>16{,}161>16{,}116>11{,}661>11{,}616>11{,}166\).

a) \(66{,}111, 61{,}611, 61{,}161, 61{,}116, 16{,}611, 16{,}161, 16{,}116, 11{,}661, 11{,}616, 11{,}166\) b) \(66{,}111>61{,}611>61{,}161>61{,}116>16{,}611>16{,}161>16{,}116>11{,}661>11{,}616>11{,}166\)

- To minimize a number, keep the leftmost digits as small as possible. - For part b, find different digits that sum to \(8\), then maximize their order. - For part c, add the four smallest different digits.

1. a) Use the smallest possible thousands digit, \(1\), then keep the hundreds and tens digits at \(0\). The ones digit must be \(7\), giving \(1007\). 2. b) The greatest possible set of four different digits with sum \(8\) is \(5, 2, 1, 0\). Arrange them in decreasing order to get \(5210\). 3. c) The four smallest different digits are \(0, 1, 2\), and \(3\), whose sum is \(6\). Any other four different digits have an even greater sum, so a sum of \(5\) is impossible.

a) \(1007\) b) \(5210\) c) The smallest possible sum of four different digits is \(0+1+2+3=6\), so \(5\) is impossible.

- Use the first two clues to determine the first two digits. - Find the remaining digit sum. - Keep only choices with an even ones digit.

1. The thousands digit is \(2\), so the hundreds digit is \(2\times2=4\). The number begins with \(24\). 2. The first two digits sum to \(6\), so the tens and ones digits must sum to \(4\). 3. The possible ordered pairs are \((0,4), (1,3), (2,2), (3,1)\), and \((4,0)\). 4. An even number must have an even ones digit. The valid pairs are \((0,4), (2,2)\), and \((4,0)\), giving \(2404, 2422\), and \(2440\).

\(2404, 2422\), and \(2440\)

- Organize the possibilities by hundreds digit. - Find all tens-and-ones pairs for each remaining sum. - Compare the two counts.

1. Digit sum \(3\): \(102, 111, 120, 201, 210, 300\). There are \(6\). 2. Digit sum \(4\): \(103, 112, 121, 130, 202, 211, 220, 301, 310, 400\). There are \(10\). 3. Since \(10>6\), there are more three-digit numbers with a digit sum of \(4\).

There are more with digit sum \(4\): \(10\) numbers, compared with \(6\) numbers with digit sum \(3\).

- Find the sum needed from the tens and ones digits. - Eliminate choices with repeated digits or an odd ones digit. - Compare the tens and ones digits in the remaining choices.

1. The tens and ones digits must sum to \(10-4=6\). 2. The ordered pairs with sum \(6\) are \((0,6), (1,5), (2,4), (3,3), (4,2), (5,1)\), and \((6,0)\). 3. The digits must be different, and the number must end in an even digit. The remaining possibilities are \((0,6)\) and \((6,0)\). 4. The ones digit must be less than the tens digit, so use \((6,0)\). The number is \(460\).

\(460\)

- Add the digits from \(0\) through \(5\). - Think about which digit to remove and which different digit to repeat to make the sum as small as possible. - Reverse those choices to make the sum as large as possible.

1. The first number has digit sum \(0+1+2+3+4+5=15\). 2. For the second number, omitting a digit \(x\) and repeating a different digit \(y\) changes the digit sum to \(15-x+y\). 3. The least possible sum occurs when \(5\) is omitted and \(0\) is repeated: \(15-5+0=10\). The digits can still be arranged as a six-digit number, such as \(100{,}234\). 4. The greatest possible sum occurs when \(0\) is omitted and \(5\) is repeated: \(15-0+5=20\).

First number: \(15\) Second number: least \(10\); greatest \(20\)

- A six-digit number cannot begin with \(0\). - To maximize the number of zeros, use one digit to make the entire digit sum if possible. - For part b, add the smallest possible different digits. - To minimize the number, keep the greatest place values as small as possible.

1. a) A six-digit number cannot begin with \(0\). Put \(5\) in the first place and \(0\) in all five remaining places. The number \(500{,}000\) has five zeros, so the greatest possible number is \(5\). 2. b) Four different digits would have a sum of at least \(0+1+2+3=6\), which is too large. Three different digits are possible, such as \(0, 1\), and \(4\) in the number \(100{,}004\). Therefore, the greatest possible number of different digits is \(3\). 3. c) Use the smallest possible leading digit, \(1\). Put zeros in the next four places and put the remaining digit-sum value, \(4\), in the ones place. The least number is \(100{,}004\).

a) \(5\) zeros; for example, \(500{,}000\) b) \(3\) different digits; for example, \(0, 1\), and \(4\) c) \(100{,}004\)

- A multi-digit number cannot begin with \(0\). - To make the least number, place small digits in the greatest places. - A whole number is divisible by \(2\) when its ones digit is even. - For part b, try the greatest possible hundreds digit first.

1. a) A five-digit number cannot begin with \(0\), so use the smallest available nonzero digit, \(3\), first. Place zeros in the next three places and put \(7\) in the ones place so every required digit appears. The least number is \(30{,}007\). 2. b) To make the greatest number, begin with \(9\). The tens and ones digits must then have a sum of \(15-9=6\). A number divisible by \(2\) must have an even ones digit. The greatest possible tens digit is \(6\), leaving \(0\) in the ones place. The number is \(960\).

a) \(30{,}007\) b) \(960\)

- Organize the numbers by their hundreds digit. - Write the numbers in columns and examine each place. - Look for how often each digit appears in each place. - Use place value to combine the repeated contributions.

1. a) The numbers are \(123, 132, 213, 231, 312\), and \(321\). 2. b) Their sum is \(123+132+213+231+312+321=1332\). 3. c) Each digit appears twice in each place. Since \(1+2+3=6\), the total value contributed in each place is twice that sum. 4. The hundreds contribute \(2\times6\times100=1200\), the tens contribute \(2\times6\times10=120\), and the ones contribute \(2\times6=12\). Therefore, \(1200+120+12=1332\).

a) \(123, 132, 213, 231, 312, 321\) b) \(1332\) c) Each digit appears twice in each place. Thus, \(2\times(1+2+3)\times100+2\times(1+2+3)\times10+2\times(1+2+3)=1332\).

- Compare the sum of Lucas’s three digits with the sum of Sophie’s three digits. - Determine how often each digit appears in each place when all arrangements are made. - Decide whether the total depends on the individual digits or only on their sum.

1. The two digit sums are equal: \(1+4+6=11\) and \(2+3+6=11\). 2. For three different digits, each digit appears twice in each place among all six arrangements. Therefore, the total depends only on the sum of the three digits, so the two totals must be equal. 3. Lucas’s total is \(2\times11\times(100+10+1)=2442\). 4. Sophie’s total is also \(2\times11\times(100+10+1)=2442\).

They get equal sums. Each total is \(2442\).

- Can an arrangement such as \(047\) count as a three-digit number? - List the valid numbers systematically. - Compare the number of valid arrangements in the two cases. - Focus on which digits can appear in the hundreds place.

1. a) The valid three-digit numbers are \(407, 470, 704\), and \(740\). The arrangements \(047\) and \(074\) do not represent three-digit numbers. Therefore, there are \(4\) valid numbers. 2. b) Their sum is \(407+470+704+740=2321\). 3. c) The digits \(1, 4\), and \(6\) make six three-digit numbers because none begins with \(0\). With \(0, 4\), and \(7\), two of the six arrangements begin with \(0\), so only four numbers are included. That is why the sum is less.

a) \(4\) numbers: \(407, 470, 704, 740\) b) \(2321\) c) Two arrangements of \(0, 4\), and \(7\) begin with \(0\), so they are not three-digit numbers. Only four numbers are added instead of six.

- The first kept digit has the greatest effect on the number. - You must leave enough digits to complete a five-digit number. - The order of the kept digits cannot change. - Write both resulting numbers before subtracting.

1. For the greatest number, choose the greatest possible first digit while leaving four later digits available. This is \(9\). From the remaining sequence \(1, 5, 3, 8, 6\), cross out \(1\). The greatest number is \(95{,}386\). 2. For the least number, choose the least possible first digit while leaving four later digits available. This is \(1\). The four later digits must then be kept, giving \(15{,}386\). 3. Subtract: \(95{,}386-15{,}386=80{,}000\).

1. \(95{,}386\) 2. \(15{,}386\) 3. \(80{,}000\)

- The tens place has a greater effect than the ones place. - For the greatest difference, make the first number large and the second number small. - For the least positive difference, make the two numbers as close as possible. - Try nearby tens digits and then adjust the ones digits.

1. a) For the greatest difference, put the greatest available digits in the minuend and the least available digits in the subtrahend, giving \(86-13=73\). 2. b) For the least positive difference, make the two numbers as close as possible. The arrangement \(61-58=3\) uses four different given digits. Checking the possible arrangements shows that no positive difference of \(1\) or \(2\) can be formed.

a) \(86-13=73\) b) \(61-58=3\)

- A two-digit number cannot begin with \(0\). - For the least sum, put small nonzero digits in the tens places. - For a sum near \(100\), first consider the sum of the tens digits. - Then use the ones digits to get as close as possible.

1. a) To minimize the sum, use the two least possible nonzero digits, \(2\) and \(4\), in the tens places. Use \(0\) and the least remaining digit, \(7\), in the ones places. One equation is \(20+47=67\). 2. b) To get close to \(100\), use tens digits whose sum is \(9\), then choose the ones digits to approach the remaining amount. The equation \(20+79=99\) is only \(1\) away from \(100\). No valid arrangement has a sum of exactly \(100\).

a) \(67\); for example, \(20+47=67\) b) \(20+79=99\)

- Start with the five least different digits and find their sum. - Once the digits are determined, place the greatest digit in the greatest place. - Arrange the remaining digits from greatest to least.

1. Five different digits with the least possible sum are \(0, 1, 2, 3\), and \(4\). Their sum is \(0+1+2+3+4=10\), so these must be the digits. 2. To make the greatest number, arrange the digits from greatest to least: \(4, 3, 2, 1, 0\). 3. The greatest number is \(43{,}210\).

\(43{,}210\)

- First determine how much the number decreased. - Identify the columns whose chip counts changed. - Which two adjacent place values differ by \(9000\)?

1. Find the change: \(215{,}400-206{,}400=9000\), so the number decreased by \(9000\). 2. Moving a chip from the ten-thousands column to the thousands column decreases its value by \(10{,}000-1000=9000\). 3. The chart changes from \(1\) chip in the ten-thousands column and \(5\) chips in the thousands column to \(0\) chips and \(6\) chips, respectively, while all other columns remain unchanged.

The chip moved from the ten-thousands column to the thousands column.

- Consider each possible place for the \(8\) separately. - Fill every other place with zero, then read each resulting number.

1. If the \(8\) is in the ten-thousands place, the number is \(280{,}000\), read as two hundred eighty thousand. 2. If the \(8\) is in the thousands place, the number is \(208{,}000\), read as two hundred eight thousand. 3. The chart does not identify which of those two places contains the \(8\), so both numbers satisfy the information.

The chart is ambiguous because it does not specify whether the \(8\) is in the ten-thousands place or the thousands place. \(280{,}000\): two hundred eighty thousand \(208{,}000\): two hundred eight thousand

- First find sets of three different digits that sum to \(6\). - Rearrange each set to form three-digit numbers. - Do not allow \(0\) in the hundreds place.

1. The sets of three different digits that sum to \(6\) are \(\{0,1,5\}\), \(\{0,2,4\}\), and \(\{1,2,3\}\). 2. From \(\{0,1,5\}\), the valid three-digit numbers are \(105, 150, 501, 510\). A number cannot begin with \(0\). 3. From \(\{0,2,4\}\), the valid numbers are \(204, 240, 402, 420\). 4. From \(\{1,2,3\}\), all six arrangements work: \(123, 132, 213, 231, 312, 321\).

\(105, 123, 132, 150, 204, 213, 231, 240, 312, 321, 402, 420, 501, 510\)

- Find all groups of four digits that sum to \(4\). - Rearrange each group, but do not put \(0\) first. - Organize the list by the thousands digit.

1. Organize the possibilities by digit groups that sum to \(4\): \(4,0,0,0\); \(3,1,0,0\); \(2,2,0,0\); \(2,1,1,0\); and \(1,1,1,1\). 2. Arrange each group without placing \(0\) first. This gives \(4000\); \(3100, 3010, 3001, 1300, 1030, 1003\); \(2200, 2020, 2002\); \(2110, 2101, 2011, 1210, 1201, 1021, 1120, 1102, 1012\); and \(1111\). 3. The total is \(1+6+3+9+1=20\).

There are \(20\) numbers: \(1003, 1012, 1021, 1030, 1102, 1111, 1120, 1201, 1210, 1300, 2002, 2011, 2020, 2101, 2110, 2200, 3001, 3010, 3100, 4000\).

- Compare \(43\) with the maximum digit sum \(45\). - Distribute the missing total of \(2\) among the five digits. - List every different arrangement of each digit group.

1. The maximum digit sum is \(5\times9=45\). The target sum is \(2\) less. 2. One possibility is to reduce one \(9\) by \(2\), giving one \(7\) and four \(9\)s. This produces \(79{,}999, 97{,}999, 99{,}799, 99{,}979\), and \(99{,}997\). 3. The other possibility is to reduce two \(9\)s by \(1\), giving two \(8\)s and three \(9\)s. The ten arrangements are \(88{,}999, 89{,}899, 89{,}989, 89{,}998, 98{,}899, 98{,}989, 98{,}998, 99{,}889, 99{,}898\), and \(99{,}988\). 4. There are \(15\) numbers in all.

\(79{,}999, 88{,}999, 89{,}899, 89{,}989, 89{,}998, 97{,}999, 98{,}899, 98{,}989, 98{,}998, 99{,}799, 99{,}889, 99{,}898, 99{,}979, 99{,}988, 99{,}997\)

- For the greatest difference, make the greater number large and the lesser number small. - For the least positive difference, try to make the hundreds digits consecutive. - Then make the greater number's last two digits small and the lesser number's last two digits large. - Check that every digit is used exactly once.

1. a) To maximize the difference, make one number as great as possible and the other as small as possible. The numbers are \(985\) and \(304\), so the greatest difference is \(985-304=681\). 2. b) To minimize the positive difference, the two numbers must be as close as possible. Using consecutive hundreds digits allows the lesser number to have large tens and ones digits while the greater number has small tens and ones digits. 3. The numbers \(503\) and \(498\) use every digit exactly once and give \(503-498=5\). No positive difference less than \(5\) can be formed with the given digits, so the least positive difference is \(5\).

a) \(681\) b) \(5\)

- A five-digit number cannot begin with \(0\). - For the greatest number, move a large digit as far left as the crossing-out limit permits. - For the least number, keep small digits early while leaving enough later digits. - The order of the kept digits cannot change.

1. For the greatest number, choose the greatest possible first digit while leaving four later digits. This is \(8\). From \(4, 2, 7, 3, 9\), cross out \(2\), giving \(84{,}739\). 2. For the least number, keep \(1\) first and \(0\) second. From the remaining digits, choose the least possible three-digit subsequence, \(239\). The least number is \(10{,}239\). 3. Subtract: \(84{,}739-10{,}239=74{,}500\).

1. \(84{,}739\) 2. \(10{,}239\) 3. \(74{,}500\)

- Begin by choosing hundreds digits whose difference is near \(2\). - Use the remaining tens and ones digits to adjust the difference. - Compare how far each result is from \(200\). - Remember to use every digit exactly once.

1. A difference near \(200\) is likely when the hundreds digits differ by about \(2\). 2. Using \(4\) and \(2\) in the hundreds places, the remaining digits can be arranged as \(475\) and \(281\). Then \(475-281=194\), which is \(6\) away from \(200\). 3. Another optimal arrangement is \(481-275=206\), also \(6\) away from \(200\). 4. Checking all valid arrangements shows that no difference is closer than \(6\) to \(200\).

\(475-281=194\), or \(481-275=206\). Each result is \(6\) away from \(200\).