2021 JMC 10 Problems/Problem 19

Problem

Two distinct divisors of $6^4=1296$ are mutual if their difference divides their product. For instance, $(4,2)$ is mutual as $(4-2)\mid 4\cdot2.$ Suppose a mutual pair $(d_1,d_2)$ exists where $d_1 = kd_2$ for a positive integer $k.$ What is the sum of all possible $k?$

$\textbf{(A) } 14 \qquad\textbf{(B) } 18 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 23$

Solution

Observe that $(d_1-d_2) |d_1 d_2 \implies (k-1)d_2 | k\cdot (d_2)^2 \implies (k-1) | k d_2.$ Because $\gcd(k,k-1)=1$ , it follows that $(k-1)|d_2$ . Since $d_2|1296$ , we must also have $(k-1)|1296$ . Note that we cannot have $6|(k-1)$ , because this will result in $k$ being neither a multiple of $2$ nor $3$ . Thus, we need only to check $(k-1)|16$ and $(k-1)|81.$ Note that $k$ must be of the form $2^a\cdot 3^b$ for non-negative integers $a$ and $b,$ so our desired answer is $2+3+4+9=18$ .

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2021 JMC 10 Problems/Problem 19

Problem

Solution