2021 JMC 10 Problems/Problem 11

Problem

There exist positive integers $k$ that satisfy $k = 3\gcd(20,k).$ What is the sum of all possible values of $k?$

$\textbf{(A) } 84 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 126 \qquad\textbf{(E) } 132$

Solution

Note that $k$ must be of the form $3 \cdot 2^a \cdot 5^b$ where $a = 0,1,2$ and $b = 0,1.$ To find this, observe that $3$ must divide $k.$ Suppose that $k=3x.$ This implies that $x = \gcd(k,20),$ so $x$ must be a divisor of $20,$ confirming what we noted. The sum of all $k$ equals $3(1+2+4)(1+5) = 126.$

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2021 JMC 10 Problems/Problem 11

Problem

Solution