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Lakeview Elementary Addition Strategies Definitions: Sum: The answer to an addition problem. Addend: Any one of the numbers that is added in an addition problem. Decompose: Separating a number into smaller numbers whose sum is equal to the original number. EX: The number 57 can be decomposed into 50+7, or 10+10+10+10+10+7 Problem: Jamal and Katie were counting the students on the playground. Jamal counted 36 students on the soccer field and Katie counted 45 students playing on the equipment. How many students did they count altogether? 36 + 45 = ? Strategy 1 (Counting Up by Ones on Open Number Line): This strategy can become lengthy, as the numbers get larger. Step 1: Begin with the largest addend (45) and count up by ones the number of times as the second addend (36). The number you end with is the sum of the addends. Strategy 2 (Counting Up by Tens and then Ones on Open Number Line): This strategy is a little shorter than counting up by ones. Step 1: Decompose the smallest addend (36) into groups of tens (3) and then ones (6). 36 = 10 + 10 + 10 + 6 Step 2: Starting at the largest addend (45), count up by ten as many times as there are tens (3) in the smallest addend (36). Then, count up by ones from the number you reached in step 2 (75) as many times as there are ones (6) in the smallest addend (36). The number you stop at is the sum. Strategy 3 (Landmark Numbers on Open Number Line): Step 1: Move to the nearest landmark number. (A number that is easy to count by and work with. Usually a multiple of 10.) Step 2: Hope forward the remaining groups of tens and ones. Strategy 4 (Left to Right Addition): This strategy helps the student to come up with a quick estimate that they should realize will be slightly smaller than the correct sum. Step 1: Set up the problem in a vertical arrangement with place values aligned under each other. 36 + 45 Step 2: Add the digits in the tens place first. Students should realize the 3 in 36 represents 30 and the 4 in 45 represents 40. 36 + 45 (30 + 40) = 70 (The sum of the entire problem should be slightly higher than 70) Step 3: Add the digits in the ones place (6) and (5) and record the sum below the sum (70) of the digits in the tens place. 36 + 45 70 11 = (6 + 5) Step 4: Add the two partial sums (70) and (11) to find the total sum. No regrouping is needed. 70 + 11 81 Strategy 5 (Decomposing into groups of tens and ones): Number sense is emphasized in this strategy. Students realize that numbers are made up of combinations of smaller numbers. Students continue to explore this as they solve multiplication problems in higher grades by using clusters of simpler problems. Step 1: Decompose both addends (36) and (45) into groups of tens and groups of ones. Addend 36: 30 + 6 Addend 45: 40 + 5 Step 2: Find the partial sums of the tens and the ones. 30 + 6 40 + 5 70 + 11 Step 3: Add the two partial sums (70) and (11) to find the final sum. 70 + 11 = 81 Strategy 6 (Compensation): This strategy is different than the compensation strategy for subtraction because it is based on maintaining the value of the whole and not the fact that subtraction is the distance between two numbers on the number line. With addition, as long the whole is maintained the sum will be the same. Therefore, what ever is done to one addend, the opposite must be done to the other addend. Step 1: Find the amount that can be added or subtracted from the addend (36) so that the digit in the ones place becomes a 0. 36 + 4 = 40 or 36 – 6 = 30 Step 2: Once you determine the number you are adding or subtracting from the addend (36), the opposite must be done to the addend (45). 45 - 4 = 41 or 45 + 6 = 51 Step 3: Find the sum of the two new numbers you created. 40 or 50 + 41 + 31 81 81 Strategy 7 (Regrouping): Although this is the traditional algorithm for solving addition and is most likely the strategy of choice for most adults, it does very little to teach students number sense. Most students remember this strategy as a series of steps with little understanding of why they are performing them. It is, therefore, not taught as the only strategy for addition in second grade. The term carrying is not used because carrying has many other meanings in the English language. Step 1: Add the digits in the ones place (6) and (5) of the two addends. 36 + 45 11 Step 2: If the sum of the digits is greater than 10, the partial sum (11) should be regrouped into groups of tens and ones. The ones value (1) is placed under the ones and the tens value (1) is placed above the tens value of the addends. 1( the value of the tens place in the partial sum of (6 + 5) 36 +4 5 1(the value of the ones place in the partial sum of (6 + 5) Step 3: Add the values in the tens place (1 + 3 + 4) and record the partial sum (8) below the tens place of the addends. This should include the value that was regrouped in step 2. 1 36 +4 5 81 Strategy guide provided by Chet’s Creek Elementary Jacksonville, Florida Lakeview Elementary Subtraction Strategies Definitions: Difference: The answer to a subtraction problem. Minuend: The number from which another number is being subtracted. (The top number) Subtrahend: The number that is being subtracted from another number. (The bottom number) Landmark Numbers: Numbers which are familiar or easy to work with. These numbers are usually multiples of 5 or 10. EX: 10, 25, 50, 100, etc. Decompose: Separating a number into smaller numbers whose sum is equal to the original number. EX: The number 57 can be decomposed into 50+7, or 10+10+10+10+10+7 Problem: Sarah wanted to use red and blue construction paper for her art project. She took 57 sheets of paper. If 28 of the sheets were red and the rest were blue, how many sheets of blue paper did she take? 57 – 28 = ? Strategy 1 (Counting Down by Ones): This strategy can become lengthy, as the numbers get larger. Step 1: Begin at the minuend (57) and count down by ones until you reach the subtrahend (28). The difference (29) is the number of times you counted. Strategy 2 (Counting Down by Tens and then Ones): This strategy is a little shorter than counting down by ones. Step 1: Begin at the minuend (57) and count down by tens until you reach a number that if counted down ten more the result would be less than the subtrahend (28). Step 2: Begin at the number you ended with in step 1 (37) and count down by ones until you reach the subtrahend (28). Step 3: Add the two results (20) and (9) to find the difference (29). Strategy 3 (Landmark Numbers): Counting by landmark numbers is typically an easier task for most students. Counting by fives and tens is usually mastered before any addition or subtraction strategies. Students use this knowledge by creating situations where they can solve a majority of the problem by counting by tens. They then have two simpler problems to solve. Step 1: Begin with the minuend (57) and record the next lowest landmark number (50) to the left. Step 2: Next record the landmark number (30) that is just higher than the subtrahend (28) to the right and then record the subtrahend (28). Step 3: Find the difference between each pair of numbers. Step 4: The difference (29) is the sum of the differences (7), (20), and (2) of the easier problems calculated in step 3. Strategy 4 (Decomposing into groups of tens and ones): Number sense is emphasized in this strategy. Students realize that numbers are made up of combinations of smaller numbers. Students continue to explore this as they solve multiplication problems in higher grades by using clusters of simpler problems. Step 1: Decompose the minuend (57) and subtrahend (28) into groups of tens and groups of ones. Minuend (57) = 10 + 10 + 10 + 10 + 10 + 7 Subtrahend (28) = 10 + 10 + 8 Step 2: Find the difference of each of each problem. Fill in zeros as needed. Minuend (57) = 10 + 10 + 10 + 10 + 10 + 7 Subtrahend (28) = 10 + 10 + 8 + 0 + 0 + 0 0 0 2 10 10 7 Step 3: Find the sum of the differences found in step 2. 0 + 0 + 2 + 10 + 10 + 7 = 29 Strategy 5 (Compensation): This strategy is based on the fact that subtraction is the distance between two numbers on the number line. As long as the distance is maintained, you can adjust the numbers to make an easier problem. Step 1: Find the amount that can be added or subtracted from the minuend (57) so that the digit in the ones place becomes a 9. 57 + 2 = 59 or 57 – 8 = 49 Step 2: Once you determine the number you are adding or subtracting from the minuend (57), you must add or subtract the same number from the subtrahend (28). 28 + 2 = 30 or 28 – 8 = 20 Step 3: Find the difference between the two new numbers you created. 59 or 49 - 30 - 20 29 29 Strategy 6 (Regrouping): Although this is the traditional algorithm for solving subtraction and is most likely the strategy of choice for most adults, it does very little to teach students number sense. Most students remember this strategy as a series of steps with little understanding of why they are performing them. It is, therefore, not taught as the only strategy for subtraction in second grade. The term borrowing is not used because borrowing implies that something will be returned. Step 1: Look at the digit (7) in the ones place of the minuend (57). Determine if it is greater than or less than the digit (8) in the ones place of the subtrahend (28). If is less than, as it is in this case, regroup a group of tens and add it to the digit (7). Be sure to decrease the digit in the tens place by 1 group of tens. 4 (50 – 40) Put a 4 in the tens place 5 17 (10 + 7) -2 8 Step 2: Find the difference of the two values now in the ones place (17 – 9) and record your result in the ones place of the your answer. 4 5 17 -2 8 9 Step 3: Find the difference of the two values in the tens place (40 – 20) and record the digit in the tens place of your result in the tens place of your answer. 4 5 17 -2 8 2 9 (40 – 20 = 20, record a two in the tens place) Lakeview Elementary Multiplication Strategies Definitions: Product: The answer to a multiplication problem. Factor: One of the two numbers used in multiplication to find a product. (factor x factor = product) A number that can be skip counted by and land on a given number. EX: 3 is a factor of 12, because I can skip count by 3 and land on 12. (3, 6, 9, 12) Multiple: The numbers you land on when skip counting by a specific number starting at zero. EX: The Multiples of five are 5, 10, 15, 20, 25, etc. Problem: Kevin was playing in a baseball tournament. The tournament had 7 teams and each team had 9 players. How many players were in the tournament? 7x9=? Strategy 1 (Tally Marks): This strategy can become lengthy, as the numbers get larger. Step 1: Draw a tally mark for each player. lllllllll lllllllll lllllllll lllllllll lllllllll lllllllll lllllllll Step 2: Count the number of tally marks. lllllllll lllllllll lllllllll lllllllll lllllllll lllllllll lllllllll 1...9 10 . . . 18 19 . . . 27 28 . . . 36 37 . . . 45 46 . . . 54 55 . . . 63 There are 63 players. Strategy 2 (Repeated addition): This strategy is a little shorter than tally marks. Step 1: Add the number of teams (7) as many times as there are players on each team (9). 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 63 There are 63 players. 1 2 3 4 5 6 7 8 9 Note: Due to the commutative property of multiplication, this problem can also be completed by adding the number of players (9) as many times as the number of teams (7). 9 + 9 + 9 + 9 + 9 + 9 + 9 = 63 There are 63 players. 1 2 3 4 5 6 7 Strategy 3 (Skip Counting using an Array): This strategy provides a model for the student to use while skip counting. Step 1: Draw an array with dimensions that match the two factors (7 by 9). 9 7 Step 2: Count the total number of square units within the array by skip counting by the number of rows (7) times the number of columns (9). Again, because of the commutative property, you can skip count by the number of columns (9) times the number of rows (7). 9 9 18 27 7 36 45 54 There are 63 players. 7 14 21 28 35 42 49 56 63 Strategy 4 (Decomposing into multiplication clusters): This strategy builds on the decomposing work done with addition and subtraction. The objective is to create a series of simpler multiplication problems to solve. Step 1: Decompose one of the factors (7) into groups of fives (5) and ones (2). Factor (7) = 5 + 2 Step 2: Find the partial product by multiplying the other factor (12) by the decomposed parts of the factor (16). 5 x 9 = 45 2 x 9 = 18 Step 3: Add the two partial products (45) and (18) to find the product. 45 + 18 = 63 There are 63 players. Note: Either factor can be decomposed. Strategy 5 (Array or Area Model): This strategy gives a visual representation of the problem and encourages students to understand that multiplication can be represented as the area of an array. Step 1: Draw an array with dimensions (7 x 9) or near dimensions (7 x 10) of the multiplication problem. 9 10 7 7 Step 2: Decompose the arrays into familiar dimensions with known products. 9 10 7 7 7 x 5 = 35 (shaded) 7 x 4 = 28 (unshaded) 7 x 1 = 7 (shaded) 7 x 10 = 70 (the entire array) Step 3: Combine the two partial products (35 + 28) in the first array to find the total product (63). Subtract the “extra” or shaded portion (7) from the total (70) of the second array to find the product (63) of the “remaining” unshaded 7 x 9 arrray. Strategy guide provided by Chet’s Creek Elementary Jacksonville, Florida Lakeview Elementary Division Strategies Definitions: Quotient: The answer to a division problem. Divisor: The number that you are dividing by. EX: In the problem 12 divided by 3, 3 is the divisor. Dividend: The number that is divided. EX: In the problem 12 divided by 3, 12 is the dividend. Problem: Kevin was playing in a baseball tournament. The tournament had 16 teams and a total of 192 players. If each team had the same number of players, how many players were on each team? 192 16 = ? Strategy 1 (Tally Marks): This strategy can become lengthy, as the numbers get larger. Step 1: Draw a tally mark for each player. IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII II Step 2: Circle groups of tally marks with each group equaling the divisor (16). IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII II Step 3: Count the number of circles. IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII IIII II There are 12 players on each team. Strategy 2 (Successive Subtraction): This strategy is a little shorter than tally marks. Step 1: Subtract the divisor (16) from the quotient (192). Continue this process until you reach zero. 192 – 16 = 176, 176 – 16 = 160, 160 – 16 = 144, 144 – 16 = 128, 128 – 16 = 112, 112 – 16 = 96, 96 – 16 = 80, 80 – 16 = 64, 64 – 16 = 48, 48 – 16 = 32, 32 – 16 = 16, 16 – 16 = 0 Step 2: Count the amount of groups of 16 that were subtracted. 192 – 16 = 176, 176 – 16 = 160, 160 – 16 = 144, 144 – 16 = 128, 128 – 16 = 112, 112 – 16 = 96, 96 – 16 = 80, 1 2 3 4 5 6 7 80 – 16 = 64, 64 – 16 = 48, 48 – 16 = 32, 32 – 16 = 16, 16 – 16 = 0 8 9 10 11 12 There are 12 players on each team. Strategy 3 (Division Clusters/Landmark Numbers): This strategy uses familiar facts to subtract larger multiples of the divisor from the dividend. Step 1: Create a list of familiar division facts with the divisor (16) the same in the problem. This is done by using multiples of ten and doubling and halving. 16 ÷ 16 = 1 160 ÷ 16 = 10 32 ÷ 16 = 2 320 ÷ 16 = 20 80 ÷ 16 = 5 800 ÷ 16 = 50 Step 2: Using the list created in step 1, find a combination of dividends that will add up to the dividend in the problem. 160 ÷ 16 = 10 32 ÷ 16 = 2 160 + 32 = 192 Step 3: Add the quotients in the problems you selected in step 2 to find the total quotient. 160 ÷ 16 = 10 32 ÷ 16 = 2 10 + 2 = 12 There are 12 players on each team. Strategy 4 (Russian Peasant Method): Similar to clusters, but set up more like our traditional algorithm. Step 1: Set up the problem to look like a long division problem. 16 192 Step 2: Using a multiple of the divisor (16) which is known, place the factor other than the divisor on top and subtract the multiple from the dividend. 10 16 192 - 160 32 Step 3: Repeat the process used in step 2 with the remaining dividend (32). Place the new factor above the factor (10) that was used before. 1 10 16 192 - 160 32 - 16 16 Step 4: Repeat the process used in step 2 with the remaining dividend (16). Place the new factor above the factor (1) that was used before. 1 1 10 16 192 - 160 32 - 16 16 - 16 0 Step 5: Once the remaining dividend is 0 or less than your divisor, add all the factors placed above the long division symbol to find the quotient. 1 1 + 1 + 10 = 12 1 10 There are 12 players on each team. 16 192 - 160 32 - 16 16 - 16 0 Strategy 5 (Traditional Long Division Algorithm): Students tend to loose track of place value when using this algorithm. Step 1: Set up the problem using the long division symbol. 16 192 Step 2: Look at the digits in the hundreds and tens place of the dividend (19). Determine how many groups of the divisor (16) will go into (19). Place that amount in the tens place above the dividend. Then multiply the number of groups (1) times the divisor (16) and subtract that amount from the two digits (19) in the dividend. 1 16 192 -16 (1 x 16) 3 Step 3: Take the 2 ones from the original dividend (192) and add it to the 3 groups of 10. Determine how many groups of the divisor (16) will go into (32). Place the amount in the ones place above the dividend. Then multiply the number of groups (2) times the divisor (16) and subtract that amount from the two digits (32). This will leave a remainder of 0. 12 16 192 - 16 32 - 32 0 There are 12 players on the team. Strategy guide provided by Chet’s Creek Elementary Jacksonville, Florida