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'Weekly Problem 14 - 2013' printed from http://nrich.maths.org/

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All six digits are different $360$ times between $10$:$00$:$00$ and $11$:$00$:$00$.

To satisfy the stated condition, the display will have the form $10$:$m_{1}m_{2}$:$s_{1}s_{2}$.

The values of both $m_{1}$ and $s_{1}$ have to be chosen from $2$, $3$, $4$, $5$. So there are four ways of choosing $m_{1}$ and three choices for $s_{1}$. Since four digits have been chose, $m_{2}$ and $s_{2}$ are selected from the remaining six.

Thus the total number of times is $4 \times 3 \times 6 \times 5 = 360$.

This problem is taken from the UKMT Mathematical Challenges.

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