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1. Since it must be divisible by 5, and all of the digits are 4 or less, then the ones digit must be a 0. Next, since it must be divisible by 3, the sum of the digits must be a multiple of 3. 4 must be in the number, in either the hundreds place or tens place. To make it so that the sum of digits is a multiple of 3, the last digit must be 2 or 5, but it can’t be 5 since it is greater than 4. As a result, there are 2 numbers that satisfy this condition: 240, 420

2.There can only be 8 rooks, since each rook attacks one row and one column. We can place one bishop on one edge, since that limits the pieces the most, and they take up the least space possible. By doing this, we can fit 6 more bishops on the bottom edge, for a total of 14 bishops.

3. We can do casework:

1×1: 25

2×2: 16, We look at which spots the top left corner can be in the square

3×3: 9

4×4: 4

5×5: 1

The sum of these is 55