Question

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

Answer 1

46

Explanation:

The smallest value of n (found by trial and error) is 13. The next larger value is 4·5 = 20 more than that, 33.

The sum of these values is 13 +33 = 46.

_____

Actually, we looked for solutions to ...

n = 4k +1 = 5m +3

4k -5m = 2 . . . . . . subtract 5m+1

We found the smallest solution to be k=3, m=2 giving 12 -10 = 2. Then further solutions are of the form ...

k = 5p+3, m = 4p+2 . . . . . for any integer p

The corresponding values of n are ...

n = 4(5p+3)+1 = 20p+13

The two smallest values correspond to p=0 and p=1, so are 13, 33.