- 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 \(1\) from one million? - For part b, move backward one counting number at a time. - Think about the boundary between a six-digit number and \(1{,}000{,}000\).

1. a) Subtract \(1\): \(1{,}000{,}000-1=999{,}999\). 2. b) Count backward twice: \(1{,}000{,}000\), \(999{,}999\), \(999{,}998\).

a) \(999{,}999\) b) \(999{,}998\)

- 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 \(407{,}328\).

- Read the columns from the greatest place value to the ones. - Each empty column needs a zero when you write the number. - A chip's value depends on its column, not just its count.

1. The chart has \(2\) hundred-thousands, \(4\) thousands, \(3\) tens, and \(1\) one. 2. There are no chips in the ten-thousands or hundreds columns, so zeros go in those places. 3. The number is \(204{,}031\).

\(204{,}031\); the ten-thousands and hundreds columns are empty.

- Start at the hundred-thousands place and work right. - Think about how many places a six-digit number has. - Which places were not named in the question?

1. Put \(4\) in the hundred-thousands place, \(3\) in the ten-thousands place, and \(5\) in the tens place. 2. Use \(0\) in the thousands, hundreds, and ones places to get \(430{,}050\).

\(430{,}050\).

- 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 of each starting number and target before calculating. - If two numbers have the same final three digits, what happens to their distance from the next thousand?

1. Compare only the final three digits: each starting number ends in \(417\), while each target ends in \(000\) at the next thousand. 2. Therefore, all three equations have the same distance to the next thousand. 3. Since \(417+583=1000\), the common missing addend is \(583\).

Each starting number ends in \(417\), and each target is the next thousand, so the distance to the target is the same in all three equations. The common missing addend is \(583\).

- 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

- For a), the leftmost digits have the greatest place values. - Check that no card is used twice in a). - For b), think about the next whole number, not another digit-card arrangement.

1. To make the greatest six-digit number, use the six largest available digits in descending order: \(9,8,7,6,5,4\). The number is \(987{,}654\). 2. The next whole number is \(987{,}654+1=987{,}655\). The card rule does not apply to this part.

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 \(9\)s. - What happens to the place values when you add or subtract \(1\) at a boundary?

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 \(1\): \(999{,}999-1=999{,}998\) and \(999{,}999+1=1{,}000{,}000\).

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

- 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 \(790{,}001\).

- 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

- 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 \(2\).

- Look across the same place in all five numbers. - Add the digits of more than one number to check whether the sums match. - Check the remaining numbers before deciding that your findings always hold.

1. Compare the digits in each place. Every number has \(4\) in the hundreds place. 2. Add the digits of each number: \(2+4+3+3=12\), \(5+4+1+2=12\), \(1+4+6+1=12\), \(8+4+0+0=12\), and \(3+4+4+1=12\). 3. The shared digit is \(4\) in the hundreds place, and every digit sum is \(12\).

Shared digit: \(4\) in the hundreds place. Shared digit sum: \(12\).

- Add the three digits of each number. - Look for the one total that differs from the stated target of \(12\).

1. The digit sums of \(156\), \(453\), \(750\), and \(255\) are all \(12\). 2. The digit sum of \(355\) is \(3+5+5=13\). 3. Therefore, \(355\) is the unique outlier under the stated criterion.

The outlier is \(355\). Its digit sum is \(3+5+5=13\).

- 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 \(50\).

- 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 \(38{,}259\).

- First read the starting number from the chart. - Compare the value of a chip in the thousands column with a chip in the ten-thousands column. - Apply that change to the starting number.

1. The chart shows \(241{,}000\). 2. Moving one chip from the thousands column to the ten-thousands column increases the value by \(10{,}000-1000=9000\). 3. The new number is \(241{,}000+9000=250{,}000\).

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.

- Write the incorrect number name as a numeral. - Compare the thousands groups in the correct and incorrect numerals. - Decide which place the second \(4\) is supposed to occupy.

1. The incorrect name means \(404{,}040\), not \(440{,}040\). 2. In the given numeral, the thousands group is \(440\), so the \(4\) after the first \(4\) represents \(40{,}000\), not \(4000\). 3. The correct name is “four hundred forty thousand forty.”

four hundred forty thousand forty. The incorrect name puts the second \(4\) in the thousands place instead of the ten-thousands place.

- Read each chip in the column where it appears. - Look for a column with fewer chips in the second row. - How can you exchange chips between neighboring columns without changing the total?

1. The top row shows \(1\) hundred and \(10\) ones, worth \(100+10=110\). 2. Trade the \(10\) ones for \(1\) ten. The bottom row shows \(1\) hundred and \(1\) ten, also worth \(110\).

Trade \(10\) ones for \(1\) ten. Each row is worth \(110\).

- How many hundreds make one thousand? - After the trade, count the hundreds still left. - Keep the tens unchanged when you rewrite the number.

1. Ten hundreds make \(1\) thousand, leaving \(2\) hundreds. 2. Keep the \(3\) tens, so the number has \(1\) thousand, \(2\) hundreds, \(3\) tens, and \(0\) ones. 3. The standard form is \(1230\).

\(1230\).

- Find the thousands digit in the new number. - Work backward to undo the stated change. - Think about the value of four units in the thousands place.

1. The new thousands digit is \(5\). Before the change it was \(5+4=9\). 2. Restore that digit while keeping the other digits: the original number was \(329{,}014\). 3. Four thousands were removed, so the original was \(4000\) greater.

Original number: \(329{,}014\); it was \(4000\) greater.

- Locate the place of each occurrence of \(6\). - Write each digit's value before comparing them. - Think about how many factors of ten separate the two places.

1. The first \(6\) is in the ten-thousands place, so it is worth \(60{,}000\). 2. The second \(6\) is in the tens place, so it is worth \(60\). 3. \(60{,}000\div60=1000\). The first \(6\) is worth \(1000\) times as much.

First \(6\): \(60{,}000\); second \(6\): \(60\); factor: \(1000\).

- You may cross out a digit, but you cannot swap the remaining digits. - Compare the hundreds digits of the numbers you can make. - Check that you have considered every choice of one digit to cross out.

1. The four possible numbers are \(294\), \(794\), \(724\), and \(729\), depending on which digit is crossed out. 2. The greatest is \(794\), made by crossing out \(2\). 3. The least is \(294\), made by crossing out \(7\). Keeping the order of the digits rules out rearranging them.

a) Greatest: \(794\); cross out \(2\). b) Least: \(294\); cross out \(7\).

- 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 \(600{,}000\).

- First decide which unused-digit arrangement could be the closest number below the target. - Then find the closest possible arrangement above the target. - Compare the two 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 the relationship as a subtraction equation with an unknown amount and work backward from the result.

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. - If a place-value quantity is greater than \(9\), regroup it into the next place before combining the parts. - 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) \(5\)

- The leftmost digit affects the value the most. - Check which parts allow repeated digits. - In c), reserve a place for every required digit before maximizing the remaining places.

1. For a), use the five greatest digits in decreasing order: \(98{,}765\). 2. For b), use \(1\) in every place: \(11{,}111\). 3. For c), all three allowed digits must occur. Use three \(7\)s, then \(4\) and \(1\), in decreasing order to get \(77{,}741\).

a) \(98{,}765\). b) \(11{,}111\). c) \(77{,}741\).

- Decide which digits are allowed in each part. - The first digit of a four-digit number cannot be \(0\). - In c), the thousands digit must be one digit, even after multiplying the tens digit by \(4\).

1. In a), the greatest odd digit is \(9\), so using it in every place gives \(9999\). 2. In b), the smallest nonzero even thousands digit is \(2\). Put the remaining smallest even digits into the other places in order: \(0,4,6\), giving \(2046\). 3. In c), if the tens digit is \(1\), the thousands digit is \(4\) and the number is \(4010\). If the tens digit is \(2\), the thousands digit is \(8\) and the number is \(8020\). Larger tens digits would make the thousands digit exceed \(9\). 4. The greater possible 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 \(27\).

- 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) \(405\)

- The first digit of a five-digit number cannot be \(0\). - After choosing the first digit, place the remaining smallest available digits from left to right. - Check that no digit is repeated.

1. The first digit must be the smallest nonzero digit, \(1\). Use the remaining smallest digits in increasing order, including \(0\), to get \(10{,}234\). 2. Subtract: \(10{,}234 - 2489 = 7745\).

\(7745\)

- 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. - If a representation has more than \(9\) units of one place, regroup before comparing it with the target. - 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) \(90{,}000\)

- 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) \(72{,}000\)

- A place value that is not named contributes zero to the number. - For quantities such as \(14\) hundreds, regroup \(10\) hundreds as \(1\) thousand. - You may write the value of each place-value amount and add the values.

1. For a), \(5 \times 1000 + 3 \times 100 + 8 \times 10 + 2 = 5382\). 2. For b), \(9 \times 10{,}000 + 4 \times 1000 + 7 = 94{,}007\). 3. For c), \(2 \times 100{,}000 + 6 \times 10{,}000 + 1 \times 100 + 5 \times 10 = 260{,}150\). 4. For d), \(14 \times 100 + 6 \times 10 + 3 = 1400 + 60 + 3 = 1463\). 5. For e), \(3 \times 1000 + 25 \times 10 = 3000 + 250 = 3250\).

a) \(5382\) b) \(94{,}007\) c) \(260{,}150\) d) \(1463\) e) \(3250\)

- When a place has more than \(9\) units, regroup \(10\) of them as \(1\) unit in the next greater place. - Write zero in any place value that is not represented.

1. For a), the value is \(6 \times 10{,}000 + 14 \times 1000 + 2 \times 100 + 5\). 2. Regroup \(14\) thousands as \(1\) ten thousand and \(4\) thousands. This makes \(7\) ten thousands, \(4\) thousands, \(2\) hundreds, \(0\) tens, and \(5\) ones. The number is \(74{,}205\). 3. For b), the value is \(3 \times 100{,}000 + 8 \times 1000 + 12 \times 10\). 4. Regroup \(12\) tens as \(1\) hundred and \(2\) tens. The digits are \(3\) hundred thousands, \(0\) ten thousands, \(8\) thousands, \(1\) hundred, \(2\) tens, and \(0\) ones. The number is \(308{,}120\).

a) Hundred thousands: \(0\); ten thousands: \(7\); thousands: \(4\); hundreds: \(2\); tens: \(0\); ones: \(5\). Number: \(74{,}205\). b) Hundred thousands: \(3\); ten thousands: \(0\); thousands: \(8\); hundreds: \(1\); tens: \(2\); ones: \(0\). Number: \(308{,}120\).

- Ten units of one place value can be regrouped as one unit of the next greater place value. - Start with any place that has more than \(9\) units and exchange groups of ten for the next place. - 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 \(33{,}167\).

- Read the original number from the chart before finding the change. - Identify which two columns could account for the required decrease. - Which two adjacent place values differ by \(9000\)?

1. The chart shows \(215{,}400\). 2. The number decreased by \(215{,}400-206{,}400=9000\). 3. Moving a chip from the ten-thousands column to the thousands column decreases its value by \(10{,}000-1000=9000\). 4. 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.

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

- First read the value shown by the chart. - Compare the value of two hundreds with the value of two thousands. - Apply the net change to the starting number.

1. The chart shows \(42{,}517\). 2. Two chips in the hundreds column have a value of \(2\times100=200\). 3. In the thousands column, those two chips have a value of \(2\times1000=2000\). 4. The net increase is \(2000-200=1800\). 5. The new number is \(42{,}517+1800=44{,}317\).

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

- Read the starting number from the chart before making any change. - Write \(1010\) as a sum of place values. - Match those place values to chips that are actually present in the chart.

1. The chart shows \(52{,}184\). 2. Decompose the required decrease: \(1010=1000+10\). 3. Remove one chip from the thousands column and one chip from the tens column. 4. The new number is \(51{,}174\). 5. 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.

- Read the starting value from the chart. - Compare the value of the chip before and after the move. - Apply the net change to the starting number.

1. The chart shows \(123{,}456\). 2. Removing one hundreds chip decreases the number by \(100\), while adding one thousands chip increases it by \(1000\). 3. The net increase is \(1000-100=900\). 4. The new number is \(123{,}456+900=124{,}356\).

The new number is \(124{,}356\), and the number increases by \(900\).

- Read the starting number from the chart before changing any chips. - Express \(990\) as the difference of two place values. - The chip must move more than one column.

1. The chart shows \(111{,}111\). 2. Write the required increase as \(990=1000-10\). 3. Move one chip from the tens column, where it is worth \(10\), to the thousands column, where it is worth \(1000\). 4. 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. The new number is \(112{,}101\).

- 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 decide which digit can go in the thousands place. - After choosing a digit, make sure enough digits remain to fill the empty places. - Check your remaining choices from left to right without changing their order.

1. For the greatest number, start with \(9\) and leave enough digits to fill all four places. Then take \(8\), \(7\), and \(4\), giving \(9874\). 2. For the least number, start with \(1\). Keep \(2\), \(7\), and \(4\) afterward, giving \(1274\). 3. No larger first digit than \(9\) or smaller usable first digit than \(1\) is available. After each first choice, the remaining digits were chosen as large or small as possible while keeping enough later digits.

Greatest: \(9874\); choose \(9\) first, then the largest later digits that still leave four digits in all. Least: \(1274\); choose \(1\) first, then the smallest later digits that still leave four digits in all.

- 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 sums of the digits each student uses. - In the complete set of arrangements, how often does a digit appear in the hundreds place? - Consider the tens and ones places too before predicting the totals.

1. Lucas's digits sum to \(1+4+6=11\), and Sophie's digits sum to \(2+3+6=11\). 2. In all six arrangements of three different digits, each digit appears twice in the hundreds, tens, and ones places. Each total is therefore \(2\times11\times(100+10+1)=2442\). 3. Checking by addition: Lucas has \(146+164+416+461+614+641=2442\). Sophie has \(236+263+326+362+623+632=2442\).

The sums are equal; both are \(2442\). Each digit appears twice in every place, and both sets of digits have the same digit sum.

- 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) \(61-58=3\)

- In a), the tens digits matter more than the ones digits. - For b), consider how close a sum can be to the next hundred. - What must happen in the ones column for the sum to end in two zeros? - Check possible ones-digit pairs before deciding that \(100\) can be made.

1. To minimize the sum, the tens digits must be the smallest two nonzero digits, \(2\) and \(4\). Put the smallest remaining digits, \(0\) and \(7\), in the ones places. One example is \(20+47=67\). 2. For b), \(20+79=99\), which is \(1\) away from \(100\). 3. To total \(100\), the two ones digits would have to add to \(0\) or \(10\). Two different available digits cannot total \(0\), and no pair from \(0,2,4,7,9\) totals \(10\). So exactly \(100\) is impossible. 4. Since \(99\) is only \(1\) away, no achievable sum can be closer.

a) \(67\); for example, \(20+47=67\). b) \(20+79=99\). Exactly \(100\) is impossible because no allowed pair of distinct ones digits can add to \(0\) or \(10\).

- 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 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, compare the highest-value digits first. - For the least difference, try several choices for the hundreds digits before deciding. - How much can the smaller place values change the gap between two numbers? - A convincing answer should rule out other digit choices, not just give one example.

1. To maximize the difference, make the larger number as great as possible and the smaller number as small as possible. This gives \(985-304=681\). 2. To make a positive difference less than \(100\), the hundreds digits must be consecutive. The possible pairs are \(3,4\), \(4,5\), and \(8,9\). 3. With hundreds digits \(3\) and \(4\), the greatest possible difference between the two-digit endings is \(98-05=93\), so the full difference cannot be less than \(100-93=7\). 4. With hundreds digits \(4\) and \(5\), the largest possible endings gap is \(98-03=95\), so the difference cannot be less than \(100-95=5\). This is possible: \(503-498=5\). 5. With hundreds digits \(8\) and \(9\), the largest endings gap is \(54-03=51\), giving a difference of at least \(49\). Thus \(5\) is the least positive difference.

a) Greatest positive difference: \(681\), from \(985-304\) b) Least positive difference: \(5\), from \(503-498\)

- 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. \(74{,}500\)

- Think about how far apart the hundreds digits should be. - For a difference near \(200\), how close can you make the two-digit endings? - Check other possible pairs of hundreds digits before claiming your result is the closest. - Make sure each card is used exactly once.

1. A difference within \(5\) of \(200\) cannot use hundreds digits \(4\) or more apart: even the greatest possible change from the last two places is less than \(100\). 2. Hundreds digits one apart are also too close for these cards. The closest such pair is \(4\) and \(5\): with the remaining digits \(1,2,7,8\), the greatest possible endings gap is \(87-12=75\), giving a difference at most \(175\). Hundreds digits three apart are too far apart: the largest possible endings gap for these cards is \(73\), giving a difference of at least \(227\). 3. So the hundreds digits must be \(2\) and \(4\), or \(5\) and \(7\). The last two places must then make two-digit numbers as close as possible. With the remaining digits \(1,5,7,8\), the closest two-digit endings are \(75\) and \(81\), which differ by \(6\). With \(1,2,4,8\), they are \(18\) and \(24\), also differing by \(6\). 4. For example, \(475-281=194\), which is \(6\) away from \(200\); or \(481-275=206\), also \(6\) away. No arrangement is closer.

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