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Berkeley Math Circle: Monthly Contest 4

Number theory Difficulty 5.6 AIME, harder Prove it United States

Problem:

Aerith timed how long it took herself to solve a BMC monthly contest. She writes down the elapsed time as days:hours:minutes:seconds, and also simply as seconds. For example, if she spent $1,000,000$ seconds, she would write down $11:13:46:40$ and $1,000,000$ .

Bob sees her numbers and subtracts them, ignoring punctuation; in this case he would get

$11134640-1000000=10134640.$

What is the largest number that always must divide his result?

Solution

Solution:

Say that Aerith took $d$ days, $h$ hours, $m$ minutes, and $s$ seconds. Bob would then get

$\begin{aligned} \Delta & =\left(100^{3} d+100^{2} h+100 m+s\right)-(24 \cdot 60 \cdot 60 d+60 \cdot 60 h+60 m+s) \\ & =\left(100^{3}-24 \cdot 60 \cdot 60\right) d+\left(100^{2}-60 \cdot 60\right) h+(100-60) m+(1-1) s \\ & =\left(2^{6} \cdot 5^{6}-2^{7} \cdot 3^{3} \cdot 5^{2}\right) d+\left(2^{4} \cdot 5^{4}-2^{4} \cdot 3^{2} \cdot 5^{2}\right) h+40 m \\ & =\left(2^{3} \cdot 5\right)\left[\left(2^{3} \cdot 5^{5}-2^{4} \cdot 3^{3} 5\right) d+\left(2 \cdot 5^{3}-2 \cdot 3^{2} \cdot 5\right) h+m\right], \end{aligned}$

so $2^{3} \cdot 5=40$ must divide his number. However, no larger number works, because if Aerith spent $60$ seconds $=1:00$ , Bob would get $100-60=40$ seconds.

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