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Berkeley Math Circle: Monthly Contest 5

Number theory Difficulty 4.9 AIME Prove it United States

Problem:
Each of the positive integers a1,a2,…,ana_{1}, a_{2}, \ldots, a_{n} is less than 20162016, and the least common multiple of any two is greater than 20162016. Show that
1a1+⋯+1an<1+n2016. \frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}<1+\frac{n}{2016}.

Solution

Solution:
By considering multiples of the aia_{i} which are less than 20162016 (these don't overlap by condition) we derive
∑⌊2016ai⌋≤2016 \sum\left\lfloor\frac{2016}{a_{i}}\right\rfloor \leq 2016
Upon using the fact that ⌊x⌋>x−1\lfloor x\rfloor>x-1, we then obtain
∑(2016ai−1)<2016. \sum\left(\frac{2016}{a_{i}}-1\right)<2016 \text{.}
which rearranges to the desired conclusion.

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