Free Math Worksheets

- Add each number on the cards exactly once. - Think about whether changing the order of addends changes a sum.

1. a) Add the numbers on the cards: \(0+1+2+3+4+5=15\). 2. b) Every arrangement uses the same six addends. Changing their order does not change the sum because addition is commutative.

a) \(15\) b) Every arrangement uses the same addends, so changing their order does not change the sum.

- Which multiplication facts have each product? - Remember that switching the order of unequal factors creates the related commutative fact. - Is either product also made by a square fact?

1. For product \(16\), the factor pairs within \(1\) through \(10\) are \(2\) and \(8\), and \(4\) and \(4\). Including both orders gives \(2 \times 8\), \(8 \times 2\), and \(4 \times 4\). 2. For product \(36\), the factor pairs within \(1\) through \(10\) are \(4\) and \(9\), and \(6\) and \(6\). Including both orders gives \(4 \times 9\), \(9 \times 4\), and \(6 \times 6\).

Product \(16\): \(2 \times 8\), \(8 \times 2\), \(4 \times 4\) Product \(36\): \(4 \times 9\), \(9 \times 4\), \(6 \times 6\)

- Find one multiplication fact for each product, then switch the factors. - Which multiplication facts have the given product? - Which two factors make the given product?

1. For \(14\), use factors \(2\) and \(7\): \(2 \times 7\) and \(7 \times 2\). 2. For \(45\), use factors \(5\) and \(9\): \(5 \times 9\) and \(9 \times 5\). 3. For \(28\), use factors \(4\) and \(7\): \(4 \times 7\) and \(7 \times 4\).

a) \(2 \times 7\) and \(7 \times 2\) b) \(5 \times 9\) and \(9 \times 5\) c) \(4 \times 7\) and \(7 \times 4\)

- Switch the positions of the two factors. - Then evaluate the product using either order. - Which order is easier for you to calculate?

1. Switch the order of the factors, then evaluate each product. 2. The completed equations are \(10 \times 4 = 4 \times 10 = 40\), \(2 \times 6 = 6 \times 2 = 12\), \(5 \times 7 = 7 \times 5 = 35\), and \(3 \times 8 = 8 \times 3 = 24\).

a) \(10 \times 4 = 4 \times 10 = 40\) b) \(2 \times 6 = 6 \times 2 = 12\) c) \(5 \times 7 = 7 \times 5 = 35\) d) \(3 \times 8 = 8 \times 3 = 24\)

- Do you need to evaluate both expressions in part a? - Does switching the order of factors change the product? - Which factor pairs have a product of \(12\)?

1. In part a, the factors are switched, so the products are equal: \(7 \times 4 = 4 \times 7\). 2. In part b, switching the factors does not change the product, so \(3 \times 9 = 27\). 3. One possible pair for part c is \(3 \times 4 = 12\) and \(4 \times 3 = 12\).

a) \(=\) b) \(3 \times 9 = 27\) c) For example, \(3 \times 4 = 12\) and \(4 \times 3 = 12\).

- Look for two addends that combine to make a multiple of \(100\). - The commutative property lets you change the order of addends. - The associative property lets you change how addends are grouped.

1. For a), use the commutative property to reorder the addends: \((245 + 355) + 178 = 600 + 178 = 778\). 2. For b), use the associative property to regroup: \((64 + 136) + 482 = 200 + 482 = 682\). 3. For c), use the commutative and associative properties: \((312 + 288) + 399 = 600 + 399 = 999\).

a) \(778\); commutative property b) \(682\); associative property c) \(999\); commutative and associative properties

- Look for two addends that make a multiple of \(100\). - The commutative property lets you change the order of addends. - The associative property lets you change how the addends are grouped.

1. Use the commutative property to place \(327\) and \(473\) together: \(327 + 473 + 158\). 2. Use the associative property to regroup: \((327 + 473) + 158\). 3. Calculate: \(800 + 158 = 958\).

\((327 + 473) + 158 = 800 + 158 = 958\)

- Look for two factors whose product is \(10\). - The commutative property lets you change the order of factors. - The associative property lets you change how the factors are grouped.

1. Use the commutative property to reorder the factors: \(2 \times 5 \times 7\). 2. Use the associative property to regroup: \((2 \times 5) \times 7\). 3. Calculate: \(10 \times 7 = 70\).

\((2 \times 5) \times 7 = 10 \times 7 = 70\); commutative and associative properties

- Look for pairs of addends that make \(100\). - You may reorder and regroup addends without changing the sum. - Add the two partial sums.

1. Reorder and regroup the addends: \((64 + 36) + (19 + 81)\). 2. Calculate the partial sums: \(100 + 100 = 200\).

\(200\)

- Look for addends that combine to make \(100\) or another multiple of \(100\). - Reorder and regroup the addends to place convenient pairs together. - Remember to include any addend that is not part of a pair.

1. Reorder and regroup the addends: \((235 + 165) + (88 + 12) + 50\). 2. Calculate the partial sums: \(400 + 100 + 50 = 550\).

\(550\)

- Look for pairs that make multiples of \(100\). - Reorder and regroup the addends to place those pairs together. - Add the partial sums.

1. Reorder and regroup the addends: \((125 + 75) + (430 + 170)\). 2. Calculate the partial sums: \(200 + 600 = 800\).

\(800\)

- Look for two addends that make a multiple of \(100\). - Think about which property lets you change how addends are grouped. - Regroup before calculating.

1. Use the associative property to regroup: \((357 + 143) + 89\). 2. Calculate: \(500 + 89 = 589\).

\(589\); associative property

- Look for pairs that make multiples of \(100\). - Reorder the addends to place convenient pairs together. - Regroup the addends before calculating.

1. Reorder and regroup the addends: \((225 + 375) + (187 + 113) + 100\). 2. Calculate the partial sums: \(600 + 300 + 100 = 1000\). 3. Reordering uses the commutative property, and regrouping uses the associative property.

\(1000\); commutative and associative properties

- Write one factor for the number of levels, rows per level, and spaces per row. - Look for two factors that make an easy product. - Reorder and regroup the factors before multiplying.

1. The total is \(4 \times 5 \times 10\). 2. Use the commutative and associative properties: \((4 \times 5) \times 10 = 20 \times 10 = 200\).

\(200\) parking spaces

- First identify whether the order of the addends changes. - Then identify whether the grouping changes. - Match each change to the name of a property.

1. The student changes \(15 + 38 + 85\) to \(15 + 85 + 38\) by using the commutative property to reorder addends. 2. The student then uses the associative property to group \(15\) and \(85\) so they are added first.

The commutative property changes the order of \(38\) and \(85\); then the associative property groups \(15\) and \(85\) to be added first.

- Count the rows and the dots in each row. - Then read the same array by columns. - Does the total number of dots change when the array is turned?

1. The array has \(3\) rows with \(7\) dots in each row, so \(3 \times 7 = 21\). 2. Reading the same array by columns gives \(7\) columns with \(3\) dots each, so \(7 \times 3 = 21\). 3. Switching rows and columns does not change the total number of dots, which demonstrates the commutative property.

a) \(3 \times 7 = 21\) b) \(7 \times 3 = 21\) c) Both equations describe the same array, so they have the same product.

- Count the rows in each colored half. - Express “two equal amounts” with both addition and multiplication.

1. The whole array has \(6\) rows of \(9\) dots, so \(6 \times 9 = 54\). 2. Each half covers \(3\) rows of \(9\) dots, so each half contains \(3 \times 9 = 27\) dots. 3. Two equations that show the relationship are \(6 \times 9 = 3 \times 9 + 3 \times 9\) and \(6 \times 9 = 2 \times (3 \times 9)\).

For example, \(6 \times 9 = 3 \times 9 + 3 \times 9 = 27 + 27 = 54\) and \(6 \times 9 = 2 \times 27 = 54\).

- Write one equation for the rows and one for the columns. - Describe what happens to the dots when the page is turned.

1. Reading by rows gives \(4 \times 7 = 28\). 2. Reading by columns gives \(7 \times 4 = 28\). 3. Turning the page changes only the viewing direction; no dots are added or removed. The array shows the commutative property.

\(4 \times 7 = 7 \times 4 = 28\). Turning the page changes only how the array is viewed, not the number of dots.

- Find the number of dots in one half. - The whole array consists of two identical parts.

1. Each half has \(3\) rows of \(8\) dots, so each half contains \(3 \times 8 = 24\) dots. 2. The whole array contains two equal halves, so \(2 \times 24 = 48\). 3. Therefore, \(6 \times 8 = 2 \times (3 \times 8) = 48\).

\(6 \times 8 = 2 \times (3 \times 8) = 48\).

- Multiply the number of rows by the sum of each row. - Think about whether regrouping the same addends can change their total. - Each number appears once when you add by rows and once when you add by columns.

1. There are three row sums of \(15\), so the total is \(3 \times 15 = 45\). 2. Regrouping the same nine addends by columns does not change their sum. Therefore, the three column sums also have a total of \(45\).

a) \(45\) b) \(45\). The same nine numbers are being added, only grouped differently.

- Look for addends that combine to make a multiple of \(100\) or \(1000\). - In a product, look for factors that make \(10\). - The commutative and associative properties let you reorder and regroup addends or factors.

1. For a), regroup the addends: \((389 + 111) + (250 + 750) = 500 + 1000 = 1500\). 2. For b), reorder and regroup the addends: \((123 + 77) + 456 = 200 + 456 = 656\). 3. For c), reorder and regroup the factors: \((5 \times 2) \times 7 = 10 \times 7 = 70\).

a) \((389 + 111) + (250 + 750) = 1500\) b) \((123 + 77) + 456 = 656\) c) \((5 \times 2) \times 7 = 70\)

- Find pairs of addends that total \(100\). - Reorder the addends so each pair is together. - Add the three partial sums.

1. Pair addends that make \(100\): \(13 + 87 = 100\), \(26 + 74 = 100\), and \(39 + 61 = 100\). 2. Reorder and regroup the addends: \((13 + 87) + (26 + 74) + (39 + 61)\). 3. Add the partial sums: \(100 + 100 + 100 = 300\).

\(300\)

- Calculate the total using each grouping. - Think about whether addition changes when addends are reordered. - Think about whether addition changes when addends are regrouped.

1. Adding in the given order gives \(456 + 23 + 102 + 304 = 885\). 2. Lisa groups the two largest numbers and the two smallest numbers: \(456 + 304 = 760\) and \(23 + 102 = 125\). Then \(760 + 125 = 885\). 3. Both methods give the same result because the commutative and associative properties allow addends to be reordered and regrouped without changing the sum.

The sum is \(885\). Both methods give the same result because changing the order and grouping of addends does not change their sum.

- Write “three times a number” as multiplication by \(3\). - Complete the two multiplication steps before comparing with multiplication by \(6\). - Consider how \(3\), \(2\), and \(6\) are related.

1. For \(4\): \((4 \times 3) \times 2 = 12 \times 2 = 24\), and \(4 \times 6 = 24\). 2. For \(7\): \((7 \times 3) \times 2 = 21 \times 2 = 42\), and \(7 \times 6 = 42\). 3. For \(10\): \((10 \times 3) \times 2 = 30 \times 2 = 60\), and \(10 \times 6 = 60\). 4. Because \(3 \times 2 = 6\), multiplying first by \(3\) and then by \(2\) gives the same result as multiplying by \(6\). This is an application of the associative property.

The results are \(24\), \(42\), and \(60\). For each number, multiplying by \(3\) and then by \(2\) gives the same result as multiplying by \(6\).