# I Find With My Little Mind

• Aug 14th 2008, 09:25 AM
Obsidantion
I Find With My Little Mind
I'm thinking of a number between 1-100...

1/5 and 1/3 of the number are both integers.
there are 9 other integers in between 1/5 and 1/3 (excluding them).

(Easy question for most of you)
• Aug 14th 2008, 09:40 AM
Quick
Quote:

Originally Posted by Obsidantion
I'm thinking of a number between 1-100...

1/5 and 1/3 are both integers.
there are 9 other integers in between 1/5 and 1/3 (excluding them).

(Easy question for most of you)

By definition a fraction is not an integer...
• Aug 14th 2008, 09:48 AM
o_O
I think the OP means 1/3 and 1/5 of the integer.

Well, set up your equation. Let n be this number you're trying to find. So, the difference between $\displaystyle \frac{1}{3}n$ and $\displaystyle \frac{1}{5}n$ should equal to 10 (you want 9 integers between (1/3)n and (1/5)n but when you subtract 2 numbers, you include the number you start with which adds 1 to the number of integers in between them. For example, if you want the number of integers between 6 and 3, it's 2. But 6 - 3 = 3 (one more than 2) because you include 6 ).

So you have your equation: $\displaystyle \frac{1}{3}n - \frac{1}{5}n = 10$
• Sep 28th 2008, 09:46 PM
Sweed
n(2/15) = 10

2n = 150

n = 75
• Sep 29th 2008, 01:57 PM
Obsidantion
Quote:

Originally Posted by Sweed
n(2/15) = 10

2n = 150

n = 75

Right again! (Star)