How do you reduce 45 over 63?

Answer 1

Actually, this is quite an interesting question, since it is not as complicated as you might think ....

First, calculate the sum of digits of both numbers.

Its nine in both cases , which is an interesting coincidence ...

If the cross sum is divisible by 3 ( nine divided by 3 is three) , the number itself is divisible by 3.

So the first reduction step would be 45 / 3 = 15 over 63 / 3 = 21

If you look closely, both numbers are again divisible by 3.

Which means, they actually were divisible by nine, but I don´t want to make it too complicated.

After the second reduction, you`ll get 15 / 3 = 5 over 21 / 3 = 7

Now you are stuck.

5 is a Prime number and seven is a prime number, too.

Remember, a prime number is a number divisible only by itself and one (1).

So you cant reduce further and the answer to your question is :

Divide both numbers by nine and you get 5 over 7.

Answer 2

The fraction 41/55 cannot be reduced any more.