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Combinatorics Difficulty 5.1 AIME, harder Prove it United States

Problem:

A 24-hour digital clock shows times $h: m: s$ , where $h, m$ , and $s$ are integers with $0 \leq h \leq 23$ , $0 \leq m \leq 59$ , and $0 \leq s \leq 59$ . How many times $h: m: s$ satisfy $h+m=s$ ?

Solution

Solution:

Answer: $1164$

We are solving $h+m=s$ in $0 \leq s \leq 59$ , $0 \leq m \leq 59$ , and $0 \leq h \leq 23$ .

If $s \geq 24$ , each $h$ corresponds to exactly 1 solution, so we get $24(59-23)=24(36)$ in this case.

If $s \leq 23$ , we want the number of nonnegative integer solutions to $h+m \leq 23$ , which by lattice point counting (or balls and urns) is $\binom{23+2}{2}=(23+2)(23+1)/2=25 \cdot 12$ .

Thus our total is $12(72+25)=12(100-3)=1164$ .

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