ffirs.indd

Simple Division 139
Remainders
Let’s go back to our problem at the beginning of the chapter. How
would we divide 33 math books among 4 students? You couldn’t
really say that each student is given 8.25 or 8¼ books each, unless
you want to destroy one of the books! Each student receives 8 books
and there is 1 book left over. You can then decide what to do with
the extra book. We would write the answer as 8 r1, not 8.25 or 8¼.
If we were dividing up money, we could write the answer as 8.25,
because this is 8 dollars and 25 cents.
Some problems in division require a whole remainder to make sense,
others need the remainder expressed as a decimal.
BONUS: SHORTCUT FOR DIVISION BY 9
Th ere is an easy shortcut for division by 9. When you divide a
two-digit number by 9, the fi rst digit of the number is the answer
and adding the digits gives you the remainder. For instance, dividing
42 by 9, the fi rst digit, 4, is the answer, and the sum of the digits,
4 + 2, is the remainder.
42 ÷ 9 = 4 r6
61 ÷ 9 = 6 r7
23 ÷ 9 = 2 r5
Th ese are easy.
What do we do if the sum of the digits is 9 or higher? For instance, if
we divide 65 by 9, the fi rst digit, 6, is our answer and the remainder
is the sum of the digits, 6 + 5 = 11. Our answer is 6 r11. But that
doesn’t make sense, because you can’t have a remainder larger than
your divisor. Nine divides into 11 one more time, so we add 1 to
the answer (6 + 1), and the 2 left over from 11 becomes the new
remainder. So the answer becomes 7 r2.