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, 2003

Geometry Difficulty 5.5 AIME, harder Prove it Canada

Problem:

Consider a standard twelve-hour clock whose hour and minute hands move continuously. Let $m$ be an integer, with $1 \leq m \leq 720$ . At precisely $m$ minutes after 12:00, the angle made by the hour hand and minute hand is exactly $1^{\circ}$ . Determine all possible values of $m$ .

Solution

Solution:

The minute hand makes a full revolution of $360^{\circ}$ every 60 minutes, so after $m$ minutes it has swept through $\frac{360}{60} m = 6m$ degrees. The hour hand makes a full revolution every 12 hours (720 minutes), so after $m$ minutes it has swept through $\frac{360}{720} m = m/2$ degrees. Since both hands started in the same position at 12:00, the angle between the two hands will be $1^{\circ}$ if $6m - m/2 = \pm 1 + 360k$ for some integer $k$ . Solving this equation we get
$m = \frac{720k \pm 2}{11} = 65k + \frac{5k \pm 2}{11}$
Since $1 \leq m \leq 720$ , we have $1 \leq k \leq 11$ . Since $m$ is an integer, $5k \pm 2$ must be divisible by 11, say $5k \pm 2 = 11q$ . Then
$5k = 11q \pm 2 \Rightarrow k = 2q + \frac{q \pm 2}{5}$
It is now clear that only $q = 2$ and $q = 3$ satisfy all the conditions. Thus $k = 4$ or $k = 7$ and substituting these values into the expression for $m$ we find that the only possible values of $m$ are 262 and 458.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.