October 2026 Problem of the Month

You have a bowl that contains from 50 to 100 raisins.

Scoop them out and divide them into groups of five.
Four raisins are left over which you eat.

Divide the rest into groups of four.
Three raisins are left over which you eat.

Divide the rest into groups of three.
Two raisins are left over which you eat.

Divide the rest into groups of two.
No raisins are left over.

How many raisins were in the bowl at the start?

Solution to the Problem:

Here is the solution:

The answer is 99 raisins.

The original number is at least 50 and no more than 100.

After you divide the raisins into groups of 5, there are 4 left over.
So the starting number must be four more than a multiple of 5.
Within the given range, this is true only of 54, 59, 64, 69, 74, 79, 84, 89, 94, and 99.
After you eat 4 raisins, 50, 55, 60, 65, 70, 75, 80, 85, 90, or 95 raisins remain.

After you divide these raisins into groups of 4, there are 3 left over.
So the number of raisins must be 3 more than a multiple of 4.
This is true only of 55 (originally 59), 75 (originally 79), and 95 (originally 99).
After you eat the 3 raisins, 52, 72, or 92 remain.

After you divide these raisins into groups of 3, there are 2 left over.
So the number of raisins must be 2 more than a multiple of 3.
This is true only of 92 (originally 99).
92 - 2 = 90, which is evenly divisible by 2, meeting the final requirement.

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Click here to see Dr. Hari Kishan's mathematical explanation