Free Math Worksheets

- Each multiplication by \(10\) shifts every digit one place to the left. - Count how many factors of \(10\) appear.

1. \(10 \times 10 = 100\). 2. \(100 \times 10 = 1000\). 3. \(1000 \times 10 = 10{,}000\). 4. \(10{,}000 \times 10 = 100{,}000\). 5. \(100{,}000 \times 10 = 1{,}000{,}000\).

\(1{,}000{,}000\)

- Use place value to divide by \(10\). - Explain how each digit shifts one place to the right. - Check with multiplication by \(10\).

1. a) \(800 \div 10 = 80\). 2. b) \(360 \div 10 = 36\). 3. c) \(10 \div 10 = 1\). 4. d) \(990 \div 10 = 99\). 5. e) \(1000 \div 10 = 100\).

a) \(80\) b) \(36\) c) \(1\) d) \(99\) e) \(100\)

- Calculate each product. - Compare the four values. - Use place value when multiplying by a multiple of \(10\).

1. Calculate each product using place value: a) \(7 \times 80 = 560\) b) \(32 \times 20 = 640\) c) \(8 \times 70 = 560\) d) \(56 \times 10 = 560\) 2. Three expressions have a value of \(560\), while choice b) has a value of \(640\).

b) \(32 \times 20 = 640\)

- Begin with the basic multiplication fact hidden in the larger numbers. - How does the product change when one factor is multiplied by \(10\)? - What happens to a quotient when both the dividend and divisor are divided by \(10\)? - Use multiplication to check each division result.

1. Start with the basic fact: \(3 \times 7 = 21\). 2. Multiplying one factor by \(10\) multiplies the product by \(10\), so \(30 \times 7 = 210\) and \(3 \times 70 = 210\). 3. Use related multiplication facts for division: \(210 \div 3 = 70\) and \(210 \div 7 = 30\). 4. Divide both the dividend and divisor by \(10\): \(210 \div 30 = 21 \div 3 = 7\) and \(210 \div 70 = 21 \div 7 = 3\).

a) \(21\) b) \(210\) c) \(210\) d) \(70\) e) \(7\) f) \(30\) g) \(3\)

- Start with the basic division fact in each row. - How does the quotient change when the dividend has one more factor of \(10\)? - Compare the three equations in each row.

1. Find the basic quotients: \(24 \div 3 = 8\), \(24 \div 4 = 6\), and \(24 \div 6 = 4\). 2. When the dividend is multiplied by \(10\) and the divisor stays the same, the quotient is also multiplied by \(10\). 3. For a), the quotients are \(8\), \(80\), and \(800\). 4. For b), the quotients are \(6\), \(60\), and \(600\). 5. For c), the quotients are \(4\), \(40\), and \(400\).

a) \(8\), \(80\), \(800\) b) \(6\), \(60\), \(600\) c) \(4\), \(40\), \(400\) Pattern: When the dividend is multiplied by \(10\) and the divisor stays the same, the quotient is multiplied by \(10\).

- Find the quotients in the first row first. - What happens to the quotient when the divisor doubles and the dividend stays the same? - How are the dividends in each column related from one row to the next?

1. For the first row, \(400 \div 2 = 200\), \(400 \div 4 = 100\), and \(400 \div 8 = 50\). 2. Each dividend in the second row is \(10\) times the corresponding dividend in the first row, so the quotients are \(2000\), \(1000\), and \(500\). 3. Each dividend in the third row is \(10\) times the corresponding dividend in the second row, so the quotients are \(20{,}000\), \(10{,}000\), and \(5000\).

First row: \(200\), \(100\), \(50\) Second row: \(2000\), \(1000\), \(500\) Third row: \(20{,}000\), \(10{,}000\), \(5000\)

- Compare how the dividends change from one equation to the next. - How does the quotient change each time? - Look for a basic multiplication or division fact inside the larger numbers.

1. In each part, the dividend is multiplied by \(10\) while the divisor stays the same, so the quotient is also multiplied by \(10\). 2. For a), \(6300 \div 9 = 700\) and \(63{,}000 \div 9 = 7000\). 3. For b), \(6300 \div 7 = 900\) and \(63{,}000 \div 7 = 9000\). 4. For c), \(63{,}000 \div 10 = 6300\) and \(630{,}000 \div 10 = 63{,}000\).

a) \(700\), \(7000\) b) \(900\), \(9000\) c) \(6300\), \(63{,}000\)

- Start with the basic multiplication fact in each expression. - Then account for the factors of \(10\). - Compare your list of products and look for equal values.

1. Find each product: \(6 \times 70 = 420\), \(30 \times 9 = 270\), \(4 \times 900 = 3600\), \(42 \times 10 = 420\), \(270 \times 1 = 270\), and \(60 \times 60 = 3600\). 2. Match equal products: expressions 1 and 4 both equal \(420\); expressions 2 and 5 both equal \(270\); expressions 3 and 6 both equal \(3600\).

1 matches 4: both equal \(420\). 2 matches 5: both equal \(270\). 3 matches 6: both equal \(3600\).

- Use place-value relationships to simplify expressions with trailing zeros. - Dividing both the dividend and divisor by \(10\) preserves the quotient. - Check with multiplication.

1. \(180\div 3=60\) and \(420\div 7=60\). 2. \(360\div 60=6\) and \(480\div 80=6\). 3. \(210\div 30=7\) and \(49\div 7=7\). 4. Each pair has an equal quotient.

\(180\div 3\) and \(420\div 7\) both equal \(60\). \(360\div 60\) and \(480\div 80\) both equal \(6\). \(210\div 30\) and \(49\div 7\) both equal \(7\).

- Identify the basic multiplication fact inside each equation. - Use the inverse relationship between multiplication and division. - Track the factors of \(10\) in the numbers.

1. For a), divide the product by the known factor: \(640 \div 80 = 8\). 2. For b), use the related basic fact: \(420 \div 60 = 42 \div 6 = 7\). 3. For c), divide the product by the known factor: \(720 \div 9 = 80\). 4. For d), divide the dividend by the quotient to find the divisor: \(360 \div 4 = 90\).

a) \(8\) b) \(7\) c) \(80\) d) \(90\)

- “Times” tells you to multiply. - “One eighth of” tells you to divide by \(8\). - A product is the result of multiplication, and a quotient is the result of division.

1. For a), \(7 \times 60 = 420\). 2. For b), \(480 \div 8 = 60\). 3. For c), \(30 \times 50 = 1500\). 4. For d), \(400 \div 80 = 5\).

a) \(420\) b) \(60\) c) \(1500\) d) \(5\)

- Calculate each product separately. - Track how multiplying by \(10\) or \(100\) changes place value. - Decide whether the paired products are equal or differ by a factor of \(10\).

1. a) \(5 \times 10 = 50\) and \(5 \times 100 = 500\). The second product is ten times the first. 2. b) \(8 \times 100 = 800\) and \(80 \times 10 = 800\). The products are equal. 3. c) \(30 \times 10 = 300\) and \(3 \times 10 = 30\). The first product is ten times the second. 4. d) \(10 \times 10 = 100\) and \(1 \times 100 = 100\). The products are equal.

a) \(50\) and \(500\); the second is ten times the first. b) \(800\) and \(800\); they are equal. c) \(300\) and \(30\); the first is ten times the second. d) \(100\) and \(100\); they are equal.

- Calculate all six products first. - Mark expressions with equal products. - Then compare the hundreds, tens, and ones in the products.

1. The products are \(3 \times 1 = 3\), \(30 \times 1 = 30\), \(3 \times 10 = 30\), \(3 \times 100 = 300\), \(30 \times 10 = 300\), and \(300 \times 1 = 300\). 2. In order: \(3\), then \(30\), then \(300\), with equal-product expressions grouped together.

1. \(3 \times 1\), product \(3\) 2. \(30 \times 1\) and \(3 \times 10\), product \(30\) 3. \(3 \times 100\), \(30 \times 10\), and \(300 \times 1\), product \(300\)

- Determine the multiplier for each pair. - Compare how place value shifts from the input to the output. - Group pairs with the same multiplier.

1. The pairs \(6 \rightarrow 60\), \(8 \rightarrow 80\), and \(40 \rightarrow 400\) each multiply the input by \(10\). 2. The pairs \(2 \rightarrow 200\), \(5 \rightarrow 500\), and \(7 \rightarrow 700\) each multiply the input by \(100\).

Group 1: \(6 \rightarrow 60\), \(8 \rightarrow 80\), \(40 \rightarrow 400\); rule: multiply by \(10\). Group 2: \(2 \rightarrow 200\), \(5 \rightarrow 500\), \(7 \rightarrow 700\); rule: multiply by \(100\).

- Find the multiplier used in each pair. - Three pairs in each row share one multiplier. - Pay close attention to the existing zero in \(90\).

1. In a), the first three pairs multiply by \(10\). The pair \(5 \rightarrow 500\) multiplies by \(100\), so it is the outlier. 2. In b), \(2 \rightarrow 200\), \(6 \rightarrow 600\), and \(1 \rightarrow 100\) multiply by \(100\). The pair \(90 \rightarrow 900\) multiplies by \(10\), so it is the outlier.

a) \(5 \rightarrow 500\) b) \(90 \rightarrow 900\)

- Use place value to multiply by \(10\). - When the result is known, use division by \(10\). - Check each row against the example.

1. \(5 \times 10 = 50\). 2. \(30 \div 10 = 3\). 3. \(10 \times 10 = 100\). 4. \(100 \div 10 = 10\).

The missing numbers from top to bottom are \(50\), \(3\), \(100\), and \(10\).

- Compare the place value of each input with its output. - Apply division by \(100\) to \(500\). - Use multiplication by \(100\) to work backward from \(4\).

1. The completed pairs show division by \(100\): \(600 \div 100 = 6\), \(200 \div 100 = 2\), and \(800 \div 100 = 8\). 2. Apply the rule: \(500 \div 100 = 5\). 3. Use the inverse operation for the missing input: \(4 \times 100 = 400\).

Rule: \(\div 100\) Missing values: \(5\) and \(400\)

- Compare how the place value of each digit changes. - Apply division by \(10\) to \(90\). - Use multiplication by \(10\) to work backward from \(15\).

1. The completed pairs show division by \(10\): \(40 \div 10 = 4\), \(120 \div 10 = 12\), and \(300 \div 10 = 30\). 2. Apply the rule: \(90 \div 10 = 9\). 3. Use the inverse operation for the missing input: \(15 \times 10 = 150\).

Rule: \(\div 10\) Missing values: \(9\) and \(150\)

- Identify the basic multiplication or division fact in each expression. - Account for the factor of \(10\) after using the basic fact. - Evaluate every expression before sorting it.

1. The multiplication expressions have value \(360\): \(90 \times 4 = 360\), \(60 \times 6 = 360\), and \(12 \times 30 = 360\). Therefore, A, C, and E belong in the first column. 2. The division expressions have value \(60\): \(420 \div 7 = 60\), \(480 \div 8 = 60\), and \(360 \div 6 = 60\). Therefore, B, D, and F belong in the second column.

Value is \(360\): A, C, E Value is \(60\): B, D, F

- Use the inverse relationship between multiplication and division. - Check a division equation by multiplying the divisor and quotient. - Use a related basic fact for the equation with \(6300\).

1. For a), use the inverse operation: \(90 \times 4 = 360\), so the missing dividend is \(360\). 2. For b), divide the dividend by the quotient: \(480 \div 6 = 80\), so the missing divisor is \(80\). 3. For c), use \(63 \div 7 = 9\) and place value: \(6300 \div 7 = 900\).

a) \(360\) b) \(80\) c) \(900\)

- Use place value to divide by \(10\) or \(100\). - Reverse each division with multiplication. - Check that the multiplication returns the dividend.

1. a) \(300 \div 10 = 30\); check: \(30 \times 10 = 300\). 2. b) \(800 \div 100 = 8\); check: \(8 \times 100 = 800\). 3. c) \(600 \div 10 = 60\); check: \(60 \times 10 = 600\). 4. d) \(400 \div 100 = 4\); check: \(4 \times 100 = 400\).

a) \(30\); \(30 \times 10 = 300\) b) \(8\); \(8 \times 100 = 800\) c) \(60\); \(60 \times 10 = 600\) d) \(4\); \(4 \times 100 = 400\)

- Use multiplication as the inverse of division. - Multiply the quotient by \(10\) or \(100\). - Check by dividing your result by the stated divisor.

1. Multiply each quotient by its divisor. 2. In a), \(70 \times 10 = 700\). 3. In b), \(5 \times 100 = 500\). 4. In c), \(20 \times 10 = 200\). 5. In d), \(10 \times 100 = 1000\).

a) \(700\) b) \(500\) c) \(200\) d) \(1000\)

- Compare the first pair in each row before looking at the larger numbers. - How is each large number related to the corresponding smaller number? - Think about what happens when three zeros are added to both numbers.

1. For a), \(302 < 320\) and \(302{,}000 < 320{,}000\). 2. For b), \(740 > 470\) and \(740{,}000 > 470{,}000\). 3. For c), \(855 = 855\) and \(855{,}000 = 855{,}000\). 4. The symbols match in every row. Each large number is the corresponding small number multiplied by \(1000\). Multiplying both numbers by the same positive number preserves their order.

a) \(302 < 320\) and \(302{,}000 < 320{,}000\) b) \(740 > 470\) and \(740{,}000 > 470{,}000\) c) \(855 = 855\) and \(855{,}000 = 855{,}000\) The symbols are the same in each row because multiplying both numbers by \(1000\) does not change their order.

- A three-digit number cannot begin with \(0\). - To make the greatest number, place the greatest available digit in the hundreds place. - Multiplying by \(1000\) shifts every digit three places to the left.

1. The least three-digit number is \(305\), and the greatest is \(530\). Therefore, \(305 < 530\). 2. Multiplying by \(1000\) gives \(305 \times 1000 = 305{,}000\) and \(530 \times 1000 = 530{,}000\). 3. The new comparison is \(305{,}000 < 530{,}000\). 4. The order does not change because both numbers are multiplied by the same positive factor.

a) \(305 < 530\) b) \(305{,}000 < 530{,}000\) c) No. The order stays the same because both numbers are multiplied by \(1000\).

- Think about how many times the smaller place-value unit fits into one million. - A place-value chart can help you compare the values.

1. For thousands, calculate \(1{,}000{,}000 \div 1000 = 1000\). One million contains \(1000\) thousands. 2. For ten thousands, calculate \(1{,}000{,}000 \div 10{,}000 = 100\). One million contains \(100\) ten thousands. 3. For hundred thousands, calculate \(1{,}000{,}000 \div 100{,}000 = 10\). One million contains \(10\) hundred thousands.

a) \(1000\) thousands b) \(100\) ten thousands c) \(10\) hundred thousands

- Ten units of one place value make one unit of the next greater place value. - For part c), consider applying that relationship more than once.

1. \(10 \times 100{,}000 = 1{,}000{,}000\), so \(10\) hundred thousands make \(1\) million. 2. \(10 \times 10{,}000 = 100{,}000\), so \(10\) ten thousands make \(1\) hundred thousand. 3. \(100 \times 10{,}000 = 1{,}000{,}000\), so \(100\) ten thousands make \(1\) million.

a) \(1\) million b) hundred thousand c) \(1\) million

- Think of equal jumps on a number line and identify the size of each jump. - You can use the results of parts a) and b) to reason about part c).

1. For part a), \(1{,}000{,}000 \div 100{,}000 = 10\), so the machine makes \(10\) jumps. 2. For part b), \(100{,}000 \div 10{,}000 = 10\), so the machine makes \(10\) jumps. 3. For part c), \(1{,}000{,}000 \div 10{,}000 = 100\), so the machine makes \(100\) jumps.

a) \(10\) jumps b) \(10\) jumps c) \(100\) jumps

- Begin with the related basic division fact. - Use place value to account for the thousands. - Look for the relationship between the two facts in each pair.

1. For a), use \(45 \div 5 = 9\) and \(45 \div 9 = 5\). The quotients are \(9000\) and \(5000\). 2. For b), use \(54 \div 6 = 9\) and \(54 \div 9 = 6\). The quotients are \(9000\) and \(6000\). 3. For c), use \(56 \div 7 = 8\) and \(56 \div 8 = 7\). The quotients are \(8000\) and \(7000\).

a) \(9000\) and \(5000\) b) \(9000\) and \(6000\) c) \(8000\) and \(7000\)

- Use the related basic division facts. - Keep track of the thousands in each quotient. - When the dividend stays the same, how does a larger divisor affect the quotient?

1. For a), \(20{,}000 \div 4 = 5000\) and \(20{,}000 \div 5 = 4000\). Therefore, \(5000 > 4000\). 2. For b), \(36{,}000 \div 6 = 6000\) and \(36{,}000 \div 4 = 9000\). Therefore, \(6000 < 9000\).

a) \(5000 > 4000\) b) \(6000 < 9000\)

- Compare the value of one counter in the ones column with its value in the hundred thousands column. - Think about what \(10\) hundred thousands regroup to.

1. A counter has the greatest value in the hundred thousands column, the greatest available place. 2. Ten counters in that column represent \(10 \times 100{,}000 = 1{,}000{,}000\). 3. The number is \(1{,}000{,}000\), or one million.

a) The hundred thousands column b) \(1{,}000{,}000\); one million

- Begin with the related one-digit multiplication fact. - Interpret the result as a number of hundreds or thousands. - Use place value rather than treating the zeros as separate digits to append.

1. For a), \(4 \times 8 = 32\), so \(4\) hundreds multiplied by \(8\) is \(32\) hundreds, or \(3200\). 2. For b), \(5 \times 6 = 30\), so \(5\) hundreds multiplied by \(6\) is \(30\) hundreds, or \(3000\). 3. For c), \(3 \times 7 = 21\), so \(3\) thousands multiplied by \(7\) is \(21\) thousands, or \(21{,}000\). 4. For d), \(8 \times 4 = 32\), so \(8\) thousands multiplied by \(4\) is \(32\) thousands, or \(32{,}000\).

a) \(3200\) b) \(3000\) c) \(21{,}000\) d) \(32{,}000\)

- Compare the changing factor within each pair. - Determine how many times as large the second factor is. - Compare the place values in the two products.

1. Part a gives \(16\times 4=64\) and \(16\times 40=640\). 2. Part b gives \(35\times 3=105\) and \(35\times 30=1050\). 3. Part c gives \(125\times 6=750\) and \(125\times 60=7500\). 4. In each pair, one factor is multiplied by \(10\), so the product is also multiplied by \(10\). Every digit in the product shifts one place to the left.

a) \(64\) and \(640\) b) \(105\) and \(1050\) c) \(750\) and \(7500\) Multiplying one factor by \(10\) multiplies the product by \(10\).

- Compare each new factor with the corresponding factor in \(14\times 5\). - Track whether a factor is multiplied by \(10\) or \(100\). - Apply the same scale factor to the product.

1. In part a, \(5\) is multiplied by \(10\), so \(70\) is multiplied by \(10\): \(14\times 50=700\). 2. In part b, \(14\) is multiplied by \(10\), so the product is multiplied by \(10\): \(140\times 5=700\). 3. In part c, both factors are multiplied by \(10\), so the product is multiplied by \(100\): \(140\times 50=7000\). 4. In part d, \(5\) is multiplied by \(100\), so the product is multiplied by \(100\): \(14\times 500=7000\).

a) \(700\) b) \(700\) c) \(7000\) d) \(7000\)

- Compare each pair of factors with the factors in part a). - Determine whether a factor was multiplied by \(10\), \(100\), or \(1000\). - Apply the same scale factor to the product.

1. \(64 \times 8=512\). 2. In b), one factor is \(10\) times as great, so the product is \(512 \times 10=5120\). 3. In c), one factor is \(100\) times as great, so the product is \(512 \times 100=51{,}200\). 4. In d), the first factor is \(100\) times as great and the second factor is \(10\) times as great. The product is \(1000\) times as great: \(512 \times 1000=512{,}000\).

a) \(512\) b) \(5120\) c) \(51{,}200\) d) \(512{,}000\)

- Compare the place value of the factor in the basic fact with the factor in the related problem. - Decide whether it is ten or one hundred times as great. - Scale the product by the same factor.

1. a) Since \(20\) is ten times \(2\), \(20 \times 4 = 80\). 2. b) Since \(60\) is ten times \(6\), \(3 \times 60 = 180\). 3. c) Since \(400\) is one hundred times \(4\), \(400 \times 2 = 800\). 4. d) Since \(50\) is ten times \(5\), \(50 \times 9 = 450\). 5. e) Since \(300\) is one hundred times \(3\), \(2 \times 300 = 600\).

a) \(80\) b) \(180\) c) \(800\) d) \(450\) e) \(600\)

- Use related basic multiplication and division facts. - For multiplication by a multiple of \(10\), solve the basic fact first and then multiply the product by \(10\). - For division, think of the dividend as a number of tens and divide those tens. - Check division answers with multiplication.

1. \(60 \times 7 = 420\). 2. \(8 \times 40 = 320\). 3. \(540 \div 9 = 60\). 4. \(280 \div 4 = 70\). 5. \(9 \times 30 = 270\).

1. \(420\) 2. \(320\) 3. \(60\) 4. \(70\) 5. \(270\)

- Think of making groups of \(10\). - The ones digit tells the remainder when a whole number is divided by \(10\). - A number ending in zero can be divided into groups of \(10\) with no remainder.

1. For a), \(860 \div 10 = 86\), so \(86\) boxes are needed. 2. For b), \(2400 \div 10 = 240\), so \(240\) boxes are needed. 3. For c), \(347 \div 10 = 34\) remainder \(7\). Therefore, \(34\) boxes can be filled, and \(7\) pencils remain. 4. For d), \(1005 \div 10 = 100\) remainder \(5\), so \(5\) pencils remain.

a) \(86\) boxes b) \(240\) boxes c) \(34\) full boxes and \(7\) pencils left over d) \(5\) pencils left over

- Think about how place values shift when dividing by \(10\). - For division by \(100\), consider what the final two digits tell you about the remainder.

1. When dividing a whole number by \(10\), the ones digit is the remainder and the other digits form the whole-number quotient. 2. For a), \(5420 \div 10 = 542\) with remainder \(0\). 3. For b), \(5427 \div 10 = 542\) remainder \(7\). 4. When dividing by \(100\), the final two digits form the remainder and the preceding digits form the whole-number quotient. 5. For c), \(19{,}300 \div 100 = 193\) with remainder \(0\). 6. For d), \(19{,}306 \div 100 = 193\) remainder \(6\). 7. For e), \(19{,}340 \div 100 = 193\) remainder \(40\).

a) \(542\) b) \(542\) remainder \(7\) c) \(193\) d) \(193\) remainder \(6\) e) \(193\) remainder \(40\)

- Dividing by \(100\) shifts every digit two places to the right. - Work one division at a time from left to right.

1. \(1{,}000{,}000 \div 100 = 10{,}000\). 2. \(10{,}000 \div 100 = 100\). 3. \(100 \div 100 = 1\).

\(1\)

- Relate the number of zeros in \(10\), \(100\), or \(1000\) to the number of place-value shifts. - Decide whether multiplication shifts digits left or division shifts them right.

1. Dividing \(300{,}000\) by \(100\) shifts the digits two places to the right, giving \(3000\). 2. Dividing \(30{,}000\) by \(10\) shifts the digits one place to the right, giving \(3000\). 3. Dividing \(300{,}000\) by \(1000\) shifts the digits three places to the right, giving \(300\). 4. Multiplying \(3000\) by \(100\) shifts the digits two places to the left, giving \(300{,}000\).

a) \(3000\) b) \(3000\) c) \(300\) d) \(300{,}000\)

- Compare the place value of the leading nonzero digit before and after each operation. - A shift of one, two, or three places corresponds to a factor of \(10\), \(100\), or \(1000\).

1. From \(50{,}000\) to \(500\), the digits shift two places to the right, so divide by \(100\). 2. From \(500\) to \(500{,}000\), the digits shift three places to the left, so multiply by \(1000\). 3. From \(500{,}000\) to \(50{,}000\), the digits shift one place to the right, so divide by \(10\). 4. From \(5000\) to \(50{,}000\), the digits shift one place to the left, so multiply by \(10\).

a) \(100\) b) \(1000\) c) \(10\) d) \(10\)

- Divide by the size of the group and count only complete groups. - You can cover the digits to the right of the place value named in each part.

1. \(528{,}314 \div 10 = 52{,}831\) remainder \(4\), so there are \(52{,}831\) complete groups of \(10\). 2. \(528{,}314 \div 100 = 5283\) remainder \(14\), so there are \(5283\) complete groups of \(100\). 3. \(528{,}314 \div 1000 = 528\) remainder \(314\), so there are \(528\) complete groups of \(1000\). 4. \(528{,}314 \div 10{,}000 = 52\) remainder \(8314\), so there are \(52\) complete groups of \(10{,}000\).

a) \(52{,}831\) complete groups of \(10\) b) \(5283\) complete groups of \(100\) c) \(528\) complete groups of \(1000\) d) \(52\) complete groups of \(10{,}000\)

- First determine how many letters are in one full crate. - Once you know the number per crate, find the number in six crates. - For part b), think about how dividing by \(10\) and \(100\) changes place value.

1. Find the number of letters in one crate: \(10 \times 10 = 100\). 2. Find the number of letters in six crates: \(6 \times 100 = 600\). 3. Find the number of bundles made from nine hundred letters: \(900 \div 10 = 90\). 4. Find the number of full crates: \(900 \div 100 = 9\).

a) Six crates contain \(600\) letters. b) The mail carrier makes \(90\) bundles and fills \(9\) crates completely.

- Reverse a division with remainder by multiplying the quotient by the divisor and then adding the remainder. - Pay attention to whether the divisor is \(100\) or \(1000\).

1. For a), \(34{,}500 \div 100 = 345\). 2. For b), \(6780 \div 100 = 67\) remainder \(80\). 3. For c), \(125{,}000 \div 1000 = 125\). 4. For d), reverse the division: \(8 \times 100 + 15 = 815\). 5. For e), \(89{,}432 \div 1000 = 89\) remainder \(432\).

a) \(345\) b) \(67\) remainder \(80\) c) \(125\) d) \(815\) e) \(89\) remainder \(432\)

- Use place value to divide by \(10\). - Check each quotient by multiplying by \(10\). - Compare your work with the example.

1. a) \(400 \div 10 = 40\); check: \(40 \times 10 = 400\). 2. b) \(730 \div 10 = 73\); check: \(73 \times 10 = 730\). 3. c) \(50 \div 10 = 5\); check: \(5 \times 10 = 50\). 4. d) \(1000 \div 10 = 100\); check: \(100 \times 10 = 1000\).

a) \(40\) b) \(73\) c) \(5\) d) \(100\)

- First find ten times the number. - Since \(5\) is half of \(10\), halve that product. - Carry out the strategy in two steps for each problem.

1. Part a: \(28 \times 10 = 280\), and \(280 \div 2 = 140\). Therefore, \(28 \times 5 = 140\). 2. Part b: \(46 \times 10 = 460\), and \(460 \div 2 = 230\). Therefore, \(46 \times 5 = 230\). 3. Part c: \(72 \times 10 = 720\), and \(720 \div 2 = 360\). Therefore, \(72 \times 5 = 360\). 4. Part d: \(84 \times 10 = 840\), and \(840 \div 2 = 420\). Therefore, \(84 \times 5 = 420\).

a) \(140\) b) \(230\) c) \(360\) d) \(420\)

- First try multiplying from left to right. - Then multiply the powers-of-ten factors together before multiplying by the first number. - Use place-value shifts to check both methods.

1. For a), step by step: \(52 \times 10 = 520\), \(520 \times 10 = 5200\), and \(5200 \times 10 = 52{,}000\). Combining factors: \(10 \times 10 \times 10 = 1000\), so \(52 \times 1000 = 52{,}000\). 2. For b), step by step: \(81 \times 100 = 8100\), then \(8100 \times 100 = 810{,}000\). Combining factors: \(100 \times 100 = 10{,}000\), so \(81 \times 10{,}000 = 810{,}000\). 3. For c), step by step: \(125 \times 10 = 1250\), then \(1250 \times 100 = 125{,}000\). Combining factors: \(10 \times 100 = 1000\), so \(125 \times 1000 = 125{,}000\).

a) \(52{,}000\) b) \(810{,}000\) c) \(125{,}000\)

- Find the products with \(10\) first. - Compare the factors \(5\) and \(10\). - Check whether one product is half of the other.

1. The first pair is \(32 \times 10 = 320\) and \(32 \times 5 = 160\). 2. The second pair is \(54 \times 10 = 540\) and \(54 \times 5 = 270\). 3. The third pair is \(86 \times 10 = 860\) and \(86 \times 5 = 430\). 4. The fourth pair is \(28 \times 10 = 280\) and \(28 \times 5 = 140\). 5. Since \(5\) is half of \(10\), each product with \(5\) is half the related product with \(10\).

1. \(320\) and \(160\) 2. \(540\) and \(270\) 3. \(860\) and \(430\) 4. \(280\) and \(140\) In each pair, the product with \(5\) is half the product with \(10\).

- Dividing by \(10\) shifts every digit one place to the right. - Look for the pattern created by dividing by \(10\) twice.

1. For the first row, \(45{,}000 \div 10 = 4500\), and \(4500 \div 10 = 450\). 2. For the second row, \(12{,}300 \div 10 = 1230\), and \(1230 \div 10 = 123\). 3. For the third row, \(800 \div 10 = 80\), and \(80 \div 10 = 8\). 4. For the fourth row, \(7000 \div 10 = 700\), and \(700 \div 10 = 70\).

<table><tr><th>Starting Number</th><th>\(\div 10\)</th><th>\(\div 10\) Again</th></tr><tr><td>\(45{,}000\)</td><td>\(4500\)</td><td>\(450\)</td></tr><tr><td>\(12{,}300\)</td><td>\(1230\)</td><td>\(123\)</td></tr><tr><td>\(800\)</td><td>\(80\)</td><td>\(8\)</td></tr><tr><td>\(7000\)</td><td>\(700\)</td><td>\(70\)</td></tr></table>

- Track how the digits in the ten-thousands and thousands places change. - Compare the value of the chip before and after the move.

1. The starting chart represents \(4\) ten-thousands, \(1\) thousand, \(2\) hundreds, \(5\) tens, and \(0\) ones, or \(41{,}250\). 2. After the move, the chart has \(3\) ten-thousands and \(2\) thousands, with all other places unchanged. The new number is \(32{,}250\). 3. The decrease is \(41{,}250 - 32{,}250 = 9000\). Equivalently, \(10{,}000 - 1000 = 9000\).

a) \(32{,}250\) b) The value decreased by \(9000\).

- Decrease the hundreds digit by \(1\). - Increase the tens digit by \(1\), then reread the chart.

1. The starting counts are \(1\) thousand, \(4\) hundreds, \(2\) tens, and \(6\) ones. 2. Removing one chip from hundreds changes \(4\) hundreds to \(3\) hundreds. 3. Adding that chip to tens changes \(2\) tens to \(3\) tens. 4. The new number is \(1336\).

The new number is \(1336\).

- The number must decrease, so move a chip to a less valuable place. - Which two neighboring place values differ by \(9\)?

1. The target number is \(4032 - 9 = 4023\). 2. Moving a chip from a greater place to a lesser place decreases the number by the difference between those place values. 3. A tens chip is worth \(10\), and an ones chip is worth \(1\). The difference is \(10 - 1 = 9\). 4. Therefore, move one chip from tens to ones.

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

- Focus only on the value of the chip before and after the move. - Subtract the old value from the new value.

1. A chip in the tens column is worth \(10\). 2. The same chip in the hundreds column is worth \(100\). 3. The number loses \(10\) and gains \(100\), so the net change is \(100 - 10 = 90\). 4. The number increases by \(90\).

The number increases by \(90\).

- Compare the value of a chip in the hundreds column with its value in the thousands column. - Find the number that is \(900\) greater than \(1100\).

1. The target number is \(1100 + 900 = 2000\). 2. Moving one chip from hundreds to thousands increases its value from \(100\) to \(1000\). 3. The increase is \(1000 - 100 = 900\). 4. The chart then shows \(2\) thousands and no hundreds, so the new number is \(2000\).

The new number is \(2000\).

- The number must decrease, so move a chip to a less valuable place. - Which move loses \(1000\) but adds back \(100\)?

1. The target number is \(3000 - 900 = 2100\). 2. Move one chip from the thousands column to the hundreds column. 3. This changes the chip's value from \(1000\) to \(100\), a decrease of \(1000 - 100 = 900\). 4. The chart then shows \(2\) thousands and \(1\) hundred, so the new number is \(2100\).

The new number is \(2100\).

- Find the number that is \(9000\) greater than \(11{,}000\). - How many ten-thousands are in that number?

1. The target number is \(11{,}000 + 9000 = 20{,}000\). 2. Move one chip from the thousands column to the ten-thousands column. 3. This changes the chip's value from \(1000\) to \(10{,}000\), an increase of \(10{,}000 - 1000 = 9000\). 4. The ten-thousands column then contains \(2\) chips.

There will be \(2\) chips in the ten-thousands column.

- The number must decrease, so move a chip to a less valuable place. - Which two neighboring place values differ by \(9000\)?

1. The chart shows \(50{,}000\). 2. The target number is \(50{,}000 - 9000 = 41{,}000\). 3. Moving one chip from ten-thousands to thousands changes its value from \(10{,}000\) to \(1000\). 4. The decrease is \(10{,}000 - 1000 = 9000\), so take the chip from the ten-thousands column.

Take the chip from the ten-thousands column and move it to the thousands column.

- Read each column from left to right, including the column with no chips. - Adding two thousands changes only the thousands digit.

1. The chart shows \(2\) ten-thousands, \(0\) thousands, \(9\) hundreds, \(4\) tens, and \(3\) ones, which is \(20{,}943\). 2. Adding \(2\) chips to the thousands column changes the thousands digit from \(0\) to \(2\). 3. The new number is \(22{,}943\).

The original number is \(20{,}943\). The new number is \(22{,}943\).

- Read the chart one place at a time from left to right. - Change only the thousands digit and the tens digit.

1. The chart shows \(4\) ten-thousands, \(3\) thousands, \(2\) hundreds, \(1\) ten, and \(9\) ones, which is \(43{,}219\). 2. Adding one chip to the thousands column changes the thousands digit from \(3\) to \(4\). 3. Adding one chip to the tens column changes the tens digit from \(1\) to \(2\). 4. The other digits stay the same, so the new number is \(44{,}229\).

The original number is \(43{,}219\). The new number is \(44{,}229\).

- Solve all the equations first. - Compare matching answers, such as the first answer in a) and the first answer in b). - Look for a place-value pattern involving a zero.

1. For a), subtract each given addend from \(1000\): \(1000 - 720 = 280\), \(1000 - 340 = 660\), \(1000 - 810 = 190\), and \(1000 - 470 = 530\). 2. For b), subtract each given addend from \(100\): \(100 - 72 = 28\), \(100 - 34 = 66\), \(100 - 81 = 19\), and \(100 - 47 = 53\). 3. Each missing addend in a) is \(10\) times the matching missing addend in b).

a) \(280\), \(660\), \(190\), \(530\) b) \(28\), \(66\), \(19\), \(53\) c) Each answer in a) is \(10\) times the matching answer in b).

- Use the related basic facts and place value. - Multiplying a factor by \(10\) multiplies the product by \(10\). - When both dividend and divisor have a factor of \(10\), divide both by \(10\).

1. For a), \(9 \times 40=360\), so the equation is false. 2. For b), \(360 \div 9=40\), so the equation is correct. 3. For c), \(360 \div 4=90\), so the equation is correct. 4. For d), \(360 \div 40=9\), so the equation is false.

a) False; the correct value is \(360\). b) Correct c) Correct d) False; the correct value is \(9\).

- Begin with the basic fact involving \(8\), \(4\), and \(32\). - Use the first equation to build the equations with multiples of ten. - Rewrite a division equation as a related multiplication equation to find a missing value. - Check that every completed equation is true.

1. For a), \(8 \times 4 = 32\). 2. For b), multiplying \(8\) by \(10\) also multiplies the product by \(10\), so \(80 \times 4 = 320\). The missing factor is \(4\). 3. For c), use the related multiplication equation \(40 \times 8 = 320\). The missing dividend is \(320\). 4. For d), use \(8 \times 40 = 320\), so \(320 \div 40 = 8\).

a) \(32\) b) \(4\) c) \(320\) d) \(8\)

- Begin with the related basic multiplication fact. - Track the factors of \(10\) in the factors and product. - Use a related division equation to find a missing factor. - Check each completed equation by multiplying.

1. For a), \(42 \div 6 = 7\). Since \(4200\) is \(42\) hundreds, the missing factor is \(700\): \(6 \times 700 = 4200\). 2. For b), \(56 \div 8 = 7\). One factor already contains one factor of \(10\), so the missing factor is \(70\): \(70 \times 80 = 5600\). 3. For c), \(9 \times 4 = 36\), and the two factors contribute two factors of \(10\), so \(90 \times 40 = 3600\). 4. For d), \(15 \div 3 = 5\). Since \(300\) contributes two factors of \(10\) and the product has three, the missing factor is \(50\).

a) \(700\) b) \(70\) c) \(3600\) d) \(50\)

- Find the basic multiplication fact in each product first. - Then account for the factors of \(10\). - When the basic facts are the same, compare the total number of factors of \(10\). - Check whether each number is a multiple of \(10\), \(100\), or \(1000\).

1. For a), \(4 \times 6 = 24\), so \(40 \times 60 = 2400\). Also, \(5 \times 5 = 25\), so \(50 \times 50 = 2500\). Therefore, \(2400 < 2500\). 2. For b), \(8 \times 7 = 56\), so \(8 \times 700 = 5600\). Also, \(9 \times 6 = 54\), so \(90 \times 60 = 5400\). Therefore, \(5600 > 5400\). 3. For c), both products are based on \(3 \times 3 = 9\) and contain three factors of \(10\). Each product is \(9000\), so they are equal.

a) \(<\) b) \(>\) c) \(=\)

- Identify the related basic multiplication fact. - Track the factors of \(10\) on both sides of each equation. - Use division to find a missing factor when helpful.

1. For a), \(8 \times 4 = 32\), so \(8 \times 40 = 320\). The missing factor is \(8\). 2. For b), \(7 \times 8 = 56\). To make \(5600\), the missing factor is \(800\): \(7 \times 800 = 5600\). 3. For c), \(9 \times 3 = 27\), and the factors contribute two factors of \(10\), so \(90 \times 30 = 2700\). 4. For d), \(8 \times 5 = 40\), so \(8 \times 500 = 4000\). The missing factor is \(8\).

a) \(8\) b) \(800\) c) \(2700\) d) \(8\)

- First find the value of the side with no blank. - Then use division to find the missing factor on the other side. - Use the basic facts and place-value patterns to simplify the calculations.

1. For a), \(3 \times 80 = 240\). Then \(240 \div 4 = 60\), so the missing factor is \(60\). 2. For b), \(60 \times 5 = 300\). Then \(300 \div 30 = 10\), so the missing factor is \(10\). 3. For c), \(40 \times 9 = 360\). Then \(360 \div 6 = 60\), so the missing factor is \(60\).

a) \(60\) b) \(10\) c) \(60\)

- Use basic multiplication facts and place-value patterns. - Compare how the factors change from one expression to the other. - For division, use a related basic fact before accounting for the factor of \(10\).

1. For a), \(6 \times 80 = 480\) and \(7 \times 70 = 490\), so \(480 < 490\). 2. For b), \(540 \div 6 = 90\) and \(630 \div 7 = 90\), so the values are equal. 3. For c), \(4 \times 900 = 3600\) and \(5 \times 700 = 3500\), so \(3600 > 3500\).

a) \(<\) b) \(=\) c) \(>\)

- Evaluate the side with no missing number first. - Use place value to relate multiplication by \(10\) and multiplication by \(100\). - Check that both sides of each equation have the same value.

1. In a), \(4 \times 100 = 400\), and \(40 \times 10 = 400\), so the missing number is \(40\). 2. In b), \(60 \times 10 = 600\), and \(6 \times 100 = 600\), so the missing number is \(100\). 3. In c), \(9 \times 10 = 90\), and \(90 \times 1 = 90\), so the missing number is \(90\). 4. In d), \(1 \times 100 = 100\), and \(10 \times 10 = 100\), so the missing number is \(10\).

a) \(40\) b) \(100\) c) \(90\) d) \(10\)

- Determine the rule from the completed example in each column. - Apply the rule when the input is known. - Use the inverse operation when the output is known.

1. Column 1 uses multiplication by \(10\), because \(3 \times 10 = 30\). 2. Therefore, \(9 \times 10 = 90\), and \(70 \div 10 = 7\). 3. Column 2 uses multiplication by \(100\), because \(4 \times 100 = 400\). 4. Therefore, \(800 \div 100 = 8\), and \(6 \times 100 = 600\).

Column 1: \(9 \to 90\) and \(7 \to 70\). Column 2: \(8 \to 800\) and \(6 \to 600\).

- Use place value to compare division by \(10\) and division by \(100\). - Check each quotient with multiplication. - Explain how the place value of each digit changes.

1. a) \(500 \div 10 = 50\) and \(500 \div 100 = 5\), so \(50 > 5\). The related equations are \(50 \times 10 = 500\) and \(5 \times 100 = 500\). 2. b) \(900 \div 100 = 9\) and \(900 \div 10 = 90\), so \(9 < 90\). The related equations are \(9 \times 100 = 900\) and \(90 \times 10 = 900\). 3. c) \(1000 \div 10 = 100\) and \(1000 \div 100 = 10\), so \(100 > 10\). The related equations are \(100 \times 10 = 1000\) and \(10 \times 100 = 1000\).

a) \(50 > 5\); \(50 \times 10 = 500\), \(5 \times 100 = 500\) b) \(9 < 90\); \(9 \times 100 = 900\), \(90 \times 10 = 900\) c) \(100 > 10\); \(100 \times 10 = 1000\), \(10 \times 100 = 1000\)

- Read the rule at the top of each table. - Use place value to multiply by \(10\) or \(100\). - Use division when the output is given.

1. a) \(4 \times 100 = 400\), \(600 \div 100 = 6\), and \(9 \times 100 = 900\). 2. b) \(20 \times 10 = 200\), \(800 \div 10 = 80\), and \(40 \times 10 = 400\).

a) \(400\), \(6\), \(900\) b) \(200\), \(80\), \(400\)

- Compare each input with its output using place value. - Decide whether the rule is multiplication by \(10\) or \(100\). - Use the inverse operation when the output is given.

1. a) The rule is multiply by \(100\). Therefore, \(8 \times 100 = 800\), and \(500 \div 100 = 5\). 2. b) The rule is multiply by \(10\). Therefore, \(90 \times 10 = 900\), and \(200 \div 10 = 20\).

a) Rule: \(\times 100\); missing values: \(800\) and \(5\) b) Rule: \(\times 10\); missing values: \(900\) and \(20\)

- Write an equation with a missing number for each riddle. - Use multiplication as the inverse of division. - Use place-value relationships involving \(10\) and \(100\).

1. For a), use the inverse operation: \(7 \times 10 = 70\). 2. For b), solve \(500 \div x = 5\). Since \(5 \times 100 = 500\), the mystery number is \(100\). 3. For c), use the inverse operation: \(10 \times 100 = 1000\).

a) \(70\) b) \(100\) c) \(1000\)

- Compare the place value of each first number with its second number. - Test division by \(10\) and division by \(100\). - Group pairs that use the same operation.

1. The pairs \((800, 80)\), \((400, 40)\), and \((900, 90)\) use division by \(10\). 2. The pairs \((600, 6)\), \((300, 3)\), and \((100, 1)\) use division by \(100\).

Rule \(\div 10\): \((800, 80)\), \((400, 40)\), \((900, 90)\) Rule \(\div 100\): \((600, 6)\), \((300, 3)\), \((100, 1)\)

- Apply the stated rule from the first coordinate to the second. - Use the inverse operation when the second coordinate is given. - Use place value to multiply or divide by \(10\) or \(100\).

1. a) \(3 \times 10 = 30\), \(10 \times 10 = 100\), and \(600 \div 10 = 60\). 2. b) \(500 \div 100 = 5\), \(800 \div 100 = 8\), and \(2 \times 100 = 200\).

a) \((3, 30)\), \((10, 100)\), \((60, 600)\) b) \((500, 5)\), \((800, 8)\), \((200, 2)\)

- Track the change in the second factor. - Multiplying by \(10\) shifts place value one position. - Compare corresponding products across the two sequences.

1. Sequence 1 has products \(200\), \(300\), and \(400\), followed by \(10 \times 50 = 500\), \(10 \times 60 = 600\), and \(10 \times 70 = 700\). 2. Sequence 2 has products \(100\), \(150\), and \(200\), followed by \(5 \times 50 = 250\), \(5 \times 60 = 300\), and \(5 \times 70 = 350\). 3. In Sequence 1, the second factor increases by \(10\), so each product increases by \(10 \times 10 = 100\).

Sequence 1: \(10 \times 50 = 500\), \(10 \times 60 = 600\), \(10 \times 70 = 700\) Sequence 2: \(5 \times 50 = 250\), \(5 \times 60 = 300\), \(5 \times 70 = 350\) Sequence 1 increases by \(100\) each step.

- Begin with the basic division fact in each group. - Compare how the dividend and quotient change. - Use multiplication to check each missing divisor. - Use place-value scaling rather than tracking written zeros.

1. In a), \(48 \div 6 = 8\). Keeping the divisor and multiplying the dividend by \(10\) gives \(480 \div 6 = 80\). To keep the quotient at \(8\) when the dividend is multiplied by \(10\), the divisor must also be multiplied by \(10\): \(480 \div 60 = 8\). 2. In b), \(32 \div 8 = 4\). Keeping the divisor gives \(320 \div 8 = 40\). Multiplying both dividend and divisor by \(10\) gives \(320 \div 80 = 4\).

a) \(48 \div 6 = 8\), \(480 \div 6 = 80\), \(480 \div 60 = 8\) b) \(32 \div 8 = 4\), \(320 \div 8 = 40\), \(320 \div 80 = 4\)

- Start with a related basic division fact. - Compare the dividend and quotient in parts b) and c). - Decide when the divisor must be ten times as great and when it stays the same.

1. In a), \(72 \div 8 = 9\), so the divisor is \(8\). 2. In b), the dividend is ten times as great while the quotient stays the same, so the divisor is ten times as great: \(720 \div 80 = 9\). 3. In c), both the dividend and quotient are ten times those in a), so the divisor remains \(8\): \(720 \div 8 = 90\). 4. In d), the related fact \(24 \div 8 = 3\) gives \(240 \div 8 = 30\).

a) \(8\) b) \(80\) c) \(8\) d) \(8\)

- Compare place values after division by \(10\) and by \(100\). - Use multiplication to check each quotient. - Compare the two quotients within each pair.

1. a) \(900 \div 10 = 90\), checked by \(90 \times 10 = 900\); \(900 \div 100 = 9\), checked by \(9 \times 100 = 900\). 2. b) \(200 \div 10 = 20\), checked by \(20 \times 10 = 200\); \(200 \div 100 = 2\), checked by \(2 \times 100 = 200\). 3. c) \(500 \div 10 = 50\), checked by \(50 \times 10 = 500\); \(500 \div 100 = 5\), checked by \(5 \times 100 = 500\). 4. In each pair, the quotient after division by \(100\) is one tenth of the quotient after division by \(10\).

a) \(900 \div 10 = 90\), \(90 \times 10 = 900\); \(900 \div 100 = 9\), \(9 \times 100 = 900\) b) \(200 \div 10 = 20\), \(20 \times 10 = 200\); \(200 \div 100 = 2\), \(2 \times 100 = 200\) c) \(500 \div 10 = 50\), \(50 \times 10 = 500\); \(500 \div 100 = 5\), \(5 \times 100 = 500\) Dividing by \(100\) gives a quotient one tenth as large as dividing the same number by \(10\).

- Identify which factor changed and by what factor. - A product changes by the same factor when only one factor changes. - In d), consider the combined effect of both factor changes.

1. a) The factor \(4\) becomes ten times as large, so the product becomes \(12 \times 10 = 120\). 2. b) The product becomes ten times as large while the factor \(2\) stays fixed, so \(5\) must become \(50\). 3. c) The factor \(10\) becomes \(100\), so the product becomes \(80 \times 10 = 800\). 4. d) One factor is divided by \(10\) while the other is multiplied by \(10\), so the product remains \(600\).

a) \(120\) b) \(50\) c) \(800\) d) \(600\)

- Use multiplication to find a missing divisor. - Compare how the dividend and quotient change from one equation to another. - Check each completed equation.

1. In a), \(45 \div 9 = 5\), so the missing divisor is \(9\). 2. In b), the dividend and quotient are each ten times those in a), so the divisor remains \(9\): \(450 \div 9 = 50\). 3. In c), the dividend is ten times as great while the quotient stays \(5\), so the divisor is ten times as great: \(450 \div 90 = 5\). 4. In d), \(450 \div 5 = 90\).

a) \(9\) b) \(9\) c) \(90\) d) \(90\)

- Begin with the basic fact \(72 \div 9 = 8\). - Compare the dividend and quotient in parts b) and c). - Use multiplication to find the missing dividend in part d).

1. In a), \(72 \div 9 = 8\). 2. In b), the dividend is ten times as great while the quotient stays the same, so the divisor is ten times as great: \(720 \div 90 = 8\). 3. In c), \(720 \div 9 = 80\). 4. In d), use the inverse operation: \(80 \times 9 = 720\).

a) \(9\) b) \(90\) c) \(80\) d) \(720\)

- How are \(7\) and \(70\) related? - How should the product change when one factor is multiplied by \(10\)? - What happens when the two factors switch places? - A related division equation begins with the multiplication product.

1. The factor \(70\) is \(10\) times \(7\), so the product is \(10\) times \(63\): \(9 \times 70 = 630\). 2. Switch the factors to write the commutative equation: \(70 \times 9 = 630\). 3. Use the product as the dividend to write the related division equations: \(630 \div 9 = 70\) and \(630 \div 70 = 9\).

Because \(70\) is \(10\) times \(7\), \(630\) is \(10\) times \(63\). Commutative equation: \(70 \times 9 = 630\) Related division equations: \(630 \div 9 = 70\) and \(630 \div 70 = 9\)

- How many groups of \(6\) are in \(540\)? - Use \(54 \div 6\) and then account for the factor of \(10\). - Once you have all three numbers, use them in two multiplication and two division equations. - The greatest number is the dividend in each division equation.

1. Use \(54 \div 6 = 9\) and the place-value relationship to find \(540 \div 6 = 90\). The missing number is \(90\). 2. The multiplication equations are \(90 \times 6 = 540\) and \(6 \times 90 = 540\). 3. The division equations are \(540 \div 6 = 90\) and \(540 \div 90 = 6\).

The missing number is \(90\). \(90 \times 6 = 540\) \(6 \times 90 = 540\) \(540 \div 6 = 90\) \(540 \div 90 = 6\)

- Start with the basic division fact. - Use multiplication as the inverse operation to fill a missing dividend. - Track how the factors of \(10\) change across the equations.

1. Use the basic fact: \(42 \div 6 = 7\). 2. To make a quotient of \(70\), multiply by the divisor: \(70 \times 6 = 420\). 3. The missing divisor is \(4200 \div 700 = 6\). 4. Using the place-value pattern, \(42{,}000 \div 6 = 7000\). 5. The missing dividend is \(70{,}000 \times 6 = 420{,}000\).

\(42 \div 6 = 7\) \(420 \div 6 = 70\) \(4200 \div 6 = 700\) \(42{,}000 \div 6 = 7000\) \(420{,}000 \div 6 = 70{,}000\)

- Use the related one-digit multiplication fact before accounting for the zeros. - Estimate which digit will place the product inside the interval. - Test nearby one-digit numbers. - Check both inequality signs for each product.

1. For a), use the basic fact and place value: \(9 \times 4000 = 36{,}000\), which lies between \(35{,}000\) and \(38{,}000\). The digit is \(9\). 2. For b), \(6 \times 80{,}000 = 480{,}000\), which lies between \(450{,}000\) and \(500{,}000\). The digit is \(6\). 3. For c), \(3 \times 50{,}000 = 150{,}000\), which lies between \(120{,}000\) and \(200{,}000\). The digit is \(3\).

a) \(9\) b) \(6\) c) \(3\)

- Begin with the inequality containing the smallest numbers. - Compare the number of zeros in corresponding values. - Test the first missing digit in the two larger inequalities. - Describe how each line is scaled from the previous line.

1. In the first inequality, \(5 \times 4 = 20\), and \(18 < 20 < 22\). 2. The same digit works in the second inequality because \(5 \times 400 = 2000\), and \(1800 < 2000 < 2200\). 3. It also works in the third inequality because \(5 \times 40{,}000 = 200{,}000\), and \(180{,}000 < 200{,}000 < 220{,}000\). 4. From one line to the next, the two bounds and the multiplication factor are each multiplied by \(100\), so the missing digit remains the same.

The missing digit is \(5\). From one line to the next, both bounds and the multiplication factor are multiplied by \(100\), so the same digit works each time.

- 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: \(308{,}120\).

- Compare how the factors are scaled by \(10\). - Parts a) and b) distribute a factor of \(10\) differently. - Use the product \(24 \times 13\) to reason about all three expressions.

1. \(240 \times 13=240 \times 10+240 \times 3=2400+720=3120\). 2. \(24 \times 130=(24 \times 13) \times 10=312 \times 10=3120\). 3. \(240 \times 130=(24 \times 13) \times 100=312 \times 100=31{,}200\).

a) \(3120\) b) \(3120\) c) \(31{,}200\)

- Rewrite each multiple of \(10\) as a one-digit factor times \(10\). - Compare the basic multiplication facts first. - Use the place value of the products to finish each comparison.

1. Part a: \(50 \times 4 = 200\) and \(30 \times 7 = 210\), so \(200 < 210\). 2. Part b: \(6 \times 80 = 480\) and \(8 \times 60 = 480\), so the values are equal. Each expression represents \(6 \times 8\) tens. 3. Part c: \(90 \times 3 = 270\) and \(40 \times 6 = 240\), so \(270 > 240\).

a) \(<\) b) \(=\) c) \(>\) Sample explanation for b): Both expressions equal \(6 \times 8\) tens, or \(48\) tens, so both equal \(480\).

- Work from left to right, one division at a time. - Use place-value shifts for division by \(10\) and \(100\). - For division by \(3\) or \(4\), divide the nonzero part of the number before accounting for the zeros.

1. For a), \(400{,}000 \div 10 = 40{,}000\), then \(40{,}000 \div 10 = 4000\), and \(4000 \div 4 = 1000\). 2. For b), \(90{,}000 \div 10 = 9000\), then \(9000 \div 3 = 3000\), and \(3000 \div 10 = 300\). 3. For c), \(1{,}000{,}000 \div 100 = 10{,}000\), then \(10{,}000 \div 10 = 1000\), and \(1000 \div 10 = 100\).

a) \(1000\) b) \(300\) c) \(100\)

- First evaluate the known part of each equation. - You may also work backward using multiplication. - Compare the place values in the intermediate result and the final result.

1. For a), \(280{,}000 \div 10 = 28{,}000\). Since \(28{,}000 \div 10 = 2800\), the missing divisor is \(10\). 2. For b), work backward: \(450 \times 10 = 4500\). Because \(450{,}000 \div 100 = 4500\), the missing divisor is \(100\). 3. For c), \(1{,}000{,}000 \div 10 \div 10 = 10{,}000\). Since \(10{,}000 \div 100 = 100\), the missing divisor is \(100\).

a) \(10\) b) \(100\) c) \(100\)

- For parts a) and b), divide by the size of each group and use only complete groups. - For part c), find the next multiple of \(100\) greater than \(73{,}412\).

1. For a), \(73{,}412 \div 10 = 7341\) remainder \(2\), so the number contains \(7341\) complete groups of \(10\). 2. For b), \(73{,}412 \div 1000 = 73\) remainder \(412\), so it contains \(73\) complete groups of \(1000\). 3. The next multiple of \(100\) after \(73{,}412\) is \(73{,}500\). 4. The amount needed is \(73{,}500 - 73{,}412 = 88\), so \(88\) ones must be added.

a) \(7341\) complete groups of \(10\) b) \(73\) complete groups of \(1000\) c) \(88\) ones

- First find the total difference between the two numbers. - Then determine how many groups of \(100\) are in that difference.

1. Find the amount needed: \(1{,}000{,}000 - 998{,}000 = 2000\). 2. Since \(2000 = 20 \times 100\), the difference contains \(20\) hundreds. 3. Therefore, \(20\) hundreds must be added.

\(20\) hundreds must be added.

- Find the smallest number that contains exactly that many complete tens. - Determine how many ones can be added without creating another complete ten.

1. The least number with exactly \(3057\) complete groups of \(10\) is \(3057 \times 10 = 30{,}570\). 2. The number of complete tens does not increase until the number reaches \(30{,}580\). 3. Therefore, the greatest possible number is one less than \(30{,}580\), which is \(30{,}579\).

The least possible number is \(30{,}570\), and the greatest possible number is \(30{,}579\).

- Multiplying by \(5\) can be done by multiplying by \(10\) and then dividing by \(2\). - In part b, identify the number whose product with \(10\) is \(820\). - Use the same two-step structure in every line.

1. Part a: Since multiplying by \(5\) is equivalent to multiplying by \(10\) and dividing by \(2\), \(46 \times 5 = 460 \div 2 = 230\). 2. Part b: The number multiplied by \(10\) to make \(820\) is \(82\). Then \(820 \div 2 = 410\), so \(82 \times 5 = 410\). 3. Part c: \(28 \times 10 = 280\), and \(280 \div 2 = 140\). 4. Part d: \(54 \times 10 = 540\), and \(540 \div 2 = 270\).

a) \(46 \times 5 = 460 \div 2 = 230\) b) \(82 \times 5 = 820 \div 2 = 410\) c) \(28 \times 5 = 280 \div 2 = 140\) d) \(54 \times 5 = 540 \div 2 = 270\)

- Evaluate the known factors first. - Compare the place value of the partial product with the final product. - Recall that factors may be reordered without changing the product.

1. For a), \(37 \times 10 \times 10 = 3700\). Since \(3700 \times 10 = 37{,}000\), \(\square = 10\). 2. For b), \(5 \times 100 = 500\). Since \(500 \times 100 = 50{,}000\), \(\square = 100\). 3. For c), \(10 \times 10 \times 10 \times 10 = 10{,}000\), so \(\square = 10{,}000\). 4. For d), both products equal \(22 \times 1000 = 22{,}000\). The order of factors does not change a product.

a) \(\square = 10\) b) \(\square = 100\) c) \(\square = 10{,}000\) d) Yes. Both products equal \(22{,}000\) because only the order of the factors changed.

- Since \(50\) is half of \(100\), multiply by \(100\) first. - Then divide that result by \(2\). - Use place value to multiply by \(100\).

1. Use \(n \times 50=(n \times 100) \div 2\). 2. \((16 \times 100) \div 2=1600 \div 2=800\). 3. \((38 \times 100) \div 2=3800 \div 2=1900\). 4. \((54 \times 100) \div 2=5400 \div 2=2700\). 5. \((72 \times 100) \div 2=7200 \div 2=3600\). 6. \((110 \times 100) \div 2=11{,}000 \div 2=5500\).

a) \(800\) b) \(1900\) c) \(2700\) d) \(3600\) e) \(5500\)

- Determine the original number first. - Compare the value of one chip in the hundred-thousands column with one chip in the ten-thousands column. - Remember that the same chip is removed from one column and added to another.

1. The chart shows \(3\) hundred-thousands, \(2\) ten-thousands, \(5\) thousands, \(4\) hundreds, \(1\) ten, and \(6\) ones, so the original number is \(325{,}416\). 2. The target number is \(325{,}416 - 90{,}000 = 235{,}416\). 3. Moving one chip from the hundred-thousands column to the ten-thousands column decreases the value by \(100{,}000 - 10{,}000 = 90{,}000\). 4. Therefore, that move creates the target number.

Move one chip from the hundred-thousands column to the ten-thousands column.

- Read the number by counting the chips in each column. - For each move, subtract the value of the original column and add the value of the new column. - For part c), find two place values whose difference is \(9900\).

1. The chart has \(4\) ten-thousands, \(2\) thousands, \(6\) hundreds, \(1\) ten, and \(5\) ones. The number is \(42{,}615\), or forty-two thousand six hundred fifteen. 2. Moving one chip from ten-thousands to thousands subtracts \(10{,}000\) and adds \(1000\), for a net change of \(-9000\). The new number is \(33{,}615\). 3. To increase the number by \(9900\), move a chip from the hundreds column to the ten-thousands column because \(10{,}000 - 100 = 9900\). 4. The new place-value counts are \(5\) ten-thousands, \(2\) thousands, \(5\) hundreds, \(1\) ten, and \(5\) ones, giving \(52{,}515\).

a) \(42{,}615\); forty-two thousand six hundred fifteen b) \(33{,}615\); the value decreases by \(9000\). c) Move one chip from the hundreds column to the ten-thousands column. The new number is \(52{,}515\).

- Read the starting number by counting the chips in each column. - For part b), compare the value of one chip in the two columns. - For part c), return to the original number before adding \(3\) hundreds.

1. The chart shows \(2\) ten-thousands, \(5\) thousands, \(1\) hundred, \(4\) tens, and \(3\) ones, so the starting number is \(25{,}143\). 2. Moving one chip from ten-thousands to thousands subtracts \(10{,}000\) and adds \(1000\). The net decrease is \(10{,}000 - 1000 = 9000\). 3. Starting from \(25{,}143\), adding three hundreds adds \(300\), giving \(25{,}443\).

a) \(25{,}143\) b) The number decreases by \(9000\). c) \(25{,}443\)

- Count the chips in every column, including the empty hundreds column. - To minimize the increase, add chips to the least valuable place. - To maximize the increase, add chips to the greatest place.

1. The chart shows \(1\) hundred-thousand, \(3\) ten-thousands, \(4\) thousands, \(0\) hundreds, \(2\) tens, and \(5\) ones. The starting number is \(134{,}025\). 2. To make the least possible number, place both new chips in the ones column. This adds \(2\), giving \(134{,}027\). 3. To make the greatest possible number, place both chips in the hundred-thousands column. This adds \(200{,}000\), giving \(334{,}025\). 4. Adding one hundred-thousand and one ten increases the number by \(100{,}010\), giving \(234{,}035\).

a) \(134{,}025\) b) Least: \(134{,}027\) Greatest: \(334{,}025\) One chip in tens and one in hundred-thousands: \(234{,}035\)

- Write the chart’s number in standard form first. - Find the difference between that number and one million. - Express the difference as groups of \(1000\).

1. The chart shows \(9\) hundred-thousands, \(8\) ten-thousands, \(4\) thousands, and zeros in the remaining places. The number is \(984{,}000\). 2. The difference to one million is \(1{,}000{,}000 - 984{,}000 = 16{,}000\). 3. Since \(16{,}000 = 16 \times 1000\), exactly \(16\) thousands must be added.

\(16\) thousands must be added.