2021 CIME I Problems/Problem 7

Problem

For unequal real numbers $k > 1$ and $x > 1$ , the function $f_{k}(x)$ is defined by $f_{k}(x) = \left|\frac{\log_{x}\left(\log_{x}k\right)}{\log_{k}\left(\log_{k}x\right)}\right|.$ For some positive integer $n$ , it is given that $\sum_{\substack{d \mid n \\ 1 < d < n}} f_{d}(n) = 9.$ Find the least possible value of $n$ .

Solution

Let $x = k^{a}$ : $f_{k}(x) = \left|\frac{\log_{k^{a}}\left(\log_{k^{a}}k\right)}{\log_{k}\left(\log_{k}k^{a}\right)}\right| = \left|\frac{\log_{k^{a}}\left(\frac{1}{a}\right)}{\log_{k}a}\right| = \left|- \frac{\log_{k^{a}}a}{\log_{k}a}\right| = \left|- \frac{\tfrac{\log a}{a \log k}}{\tfrac{\log a}{\log k}}\right| = \left|- \frac{1}{a}\right| = \left|\frac{1}{a}\right|.$ Thus, $f_{k}(x) = \frac{1}{\log_{k}x} = \log_{x}k$ . So we need to find the least positive integer $n$ for which $\sum_{\substack{d \mid n \\ 1 < d < n}} \log_{n}d = 9.$ That is to say, there are exactly $9$ unordered pairs of positive integers $\left(d, \tfrac{n}{d}\right) \neq (1, n)$ with $d \neq \tfrac{n}{d}$ , as each such pair increases the value of the total sum by $1$ . So $n$ has exactly $18$ positive integer divisors not equal to $1$ or itself, and thus $20$ in total. Note that $20 = 5 \cdot 2 \cdot 2$ . The least such $n$ , therefore, is obviously $2^{4} \cdot 3 \cdot 5 = \boxed{240}$ .

2021 CIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2021 CIME I Problems/Problem 7

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