Question

The sum of three different positive integers is 12. show that one of them must be greater than 4.

Answer 1

Let n = 1st positive interger
n+1 = 2nd
n + 2 = 3rd

n+n+1+n+2 = 12
3n + 3 = 12
Minus 3 from 3 and 12

3n = 9
Divide 3n and 9 by 3

n = 3
So, the three intergers are 3, 4, and 5.

Answer 2

if one is 4 , then other may be 3 and so other 2......this will get you only 9 .

so it is understood that one must be greater than 4