Suppose there was one extra imaginary student in the production. Then, when they lined up in threes, the extra student could go with the left-over students to make another line. The same would happen with fours, fives and sixes.

Therefore, when there is one extra student, the number is divisible by $3$, $4$, $5$ and $6$. This means it is a common multiple of these numbers. The lowest common multiple of $3$, $4$, $5$ and $6$ is $60$, so there must be $60$ students with the extra.

Therefore the number of students (without the imaginary student) is $59$.

This problem is taken from the UKMT Mathematical Challenges.
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