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)
Wrong:

Correct: Try again with
higher quotient:

5 R6
4 ) 26
20
6
(remainder is higher
than the divisor) 
6 R2
4 ) 26
24
2 (less than
divisor) 
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