### PROBLEM LINKS

### DIFFICULTY

EASY

### EXPLANATION

The problem here is divide all numbers by some constant so that the divisions have no remainder. We produce the smallest result by dividing by a number that is as large as possible, that is the greatest common divisor. The greatest common divisor can be computed efficiently by Euclid’s algorithm, but in this case it was fast enough to simply check all numbers from 1 to 1000 for divisibility.

### SETTER’S SOLUTION

Can be found here.

### TESTER’S SOLUTION

Can be found here.