PART 2- REMAINDERS - MONEY DIVISION WITH WHOLE NUMBERS LESSON PLAN

DIVISION LESSON CONTINUED WITH REMAINDERS

Remainders

Sometimes, division problems don’t have exact whole numbers as quotients, or answers.  For example, 44÷ 9 = ? Inverting this problem to a multiplication problem, we ask “what times 9 = 44?  There is not an exact, even number.  We know that 5 x 9 = 45, which is close, but not exactly 44.  That is when we have to write the quotient with a remainder.  A remainder is what is leftover after we’ve solved the problem with the closest quotient, and then subtract the product of the quotient and divisor from the dividend.  To solve, write the problem this way:

                                                       ?

                                                9 ) 44

We know that 5 x 9 = 45, but that is higher than the dividend 44. So we take the next lower number, 4 x 9 = 36.  Insert 4 as the quotient, multiply the divisor (9) by this quotient (4), and then subtract this product from the dividend (44) to find the remainder.

                        4  R8         (quotient with remainder)

(divisor)      9)  44              (dividend)

                  -   36               (product of 9 x 4)

                        8               (remainder)

When you do a division problem like this, make the quotient as large as you can.  To check your work, make sure the remainder is less than the divisor.  If the remainder is more, you need to try again with a larger quotient.  For example, if the problem is 26 divided by 4, would your quotient be 5 or 6?  If you use 5 as the quotient, we know that

5 x 4 = 20, and 26 – 20 = 6 left over.  Since the remainder (6) is more than the divisor (4), we know that the quotient can be 1 greater.

So we try again, with a 6 as the quotient:      6 x 4 = 24, leaving 2 leftover.  Because the remainder (2) is less than the divisor (4), we know we’ve found the largest possible quotient. (See below)

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