# Thread: Dividing unevenly divisible numbers in the head

1. ## Dividing unevenly divisible numbers in the head

Ok this sounds pathetic, and is, but I'm too embarrassed to ask anyone and google was not fruitful.

Say I want 11/5 and need to do it in my head.

What formula can I follow?

2. ## Re: Dividing unevenly divisible numbers in the head

maybe 10/5 + 1/5 = 2+1/5 ??

3. ## Re: Dividing unevenly divisible numbers in the head

10/5=2
1/5=.2
2+.2=2.2

Calculator says: 2.2

Good, so: take the number into an evenly divisible term, divide that by the previous divisor, them divide your remainder by the same divisor and add them?

4. ## Re: Dividing unevenly divisible numbers in the head

the general idea is this:

say we want to know m/n.

we first write m = qn + r, where q is the largest integer that "goes into m", when multiplying by n.

then m/n = (qn + r)/n = (qn)/n + r/n = q + r/n.

for example, say we want to know 43/7.

7*6 = 42, and 7*7 is 49, so q = 6. the remainder, r, is 1, so 43/7 = 6 and 1/7.