Prerequisites The prerequisite for teaching this unit is that children are already familiar with the material covered in "Integers Up To Ten Thousand". Children will use the 5 by 4, 5 by 3, and 5 by 2 boards to carry out integer multiplication.
The boards shown above will be used by children to carry out integer multiplication.
Explanation of the Method That Will Be Used You are given two written positive integers, for example 5 and 12. To compute 5*12, you need to find the number that would be produced if you added 5 copies of 12 on a counting board. So in this example 5*12 = 12 + 12 + 12 + 12 + 12 = 60. Children should work several simple examples on their counting boards. Remark 1 The teacher should not say that multiplication is repeated addition, because this is true only when the multiplier is a positive integer. For example, when ^{1}⁄_{2} * ^{2}⁄_{3} = ^{1}⁄_{3}, the result ^{1}⁄_{3} cannot be obtained from ^{2}⁄_{3} by repeated addition. But when the multiplier is a positive integer, the product is always the sum of the appropriate number of copies of the multiplicand. And using repeated addition to compute the product of two positive integers has no practical value for bigger numbers. If you try to compute 59*12, by adding 59 copies of 12, it is too much work and it to easy to make an error. Key Observation On the counting board we use, adding ten copies of any number is achieved by shifting the number one position to the left. So in order to compute 59*12 we compute 5*12 and 9*12 separately and then add them, shifting the first number one location to the left. Remark 2 So far, the procedure is the same as in written "long multiplication" that is taught in school. But now we need to figure out how to compute 5*12 and 9*12. In teaching a written method of computation, it usually requires that children memorize “multiplication facts” and that they learn how to "carry". Here we can handle this part of the procedure differently, because counting boards provide more flexibility. Task: Compute 59*12 Using Counting Boards Children work in small groups. Each group has three boards, one with 2 columns, to hold the multiplicand 12, and two 3-column boards, one for holding the final answer, and one to compute the partial sums, first 5*12 and next 9*12. Each group is told to figure out how to compute 5*12 and 9*12, and after everyone is finished, to tell how they did it. Examples of Explanations
Example with Pictures On the 5 by 2 board, the multiplicand 12 is represented. To begin, the first partial sum 5*12 must be determined. Because 12 is the double of 6, 5*12 is 5*2*6 = 10*6. Knowing this, 10*6 is represented on the lower left counting board. Now that the first partial sum has been calculated, it must be added to the final answer board shown at the top left of the image. Because ten copies of the partial sum need to be added, the partial sum must be shifted one location to the left. The result of this addition can be seen on the final answer board. Now, the second partial sum, 9*12, must be found. 9 is 10 – 1, so 9*12 is 10*12 + -12. Knowing this, 10*12 - 12 is represented on the lower left counting board. At this point, the second partial sum must be added to the final answer board. Because only one copy of 9*12 needs to be added, the partial sum can be added to the final answer board without shifting. The result of this addition can be seen on the final answer board. Using regrouping rules, the final answer board can be simplified to show one counter on the 500 square, two counters on the 100 square, 1 counter on the 5 square, and 1 counter on the 3 square. This representation shows 500 + 100 + 100 + 5 + 3 = 708. This means that 59*12 = 708! Number Board index |