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Problem 395

AMC 12 late, AIME early

Number theory Difficulty 4.8 Find the answer

2. Let $A_{n}$ denote the product of multiples of $n$ not exceeding 100, where $n$ is a positive integer. For example, $A_{3}=3 \times 6 \times 9 \times \cdots \times 99$ . Then the greatest common divisor of $A_{2}, A_{3}, A_{4}, \cdots, A_{17}$ is ( ).
(A)I
(B) 6
(C) 30
(D) 120

Your answer

Official solution

2. D.
$\begin{array}{l} A_{2}=2 \times 4 \times 6 \times \cdots \times 100=2^{50}(1 \times 2 \times 3 \times \cdots \times 50), \\ A_{3}=3 \times 6 \times 9 \times \cdots \times 99=3^{33}(1 \times 2 \times 3 \times \cdots \times 33), \\ \cdots \cdots \\ A_{16}=16 \times 32 \times 48 \times 64 \times 80 \times 96 \\ =16^{6}(1 \times 2 \times 3 \times 4 \times 5 \times 6), \\ A_{17}=17 \times 34 \times 51 \times 68 \times 85=17^{5}(1 \times 2 \times 3 \times 4 \times 5), \end{array}$

Therefore, the greatest common divisor of $A_{2}, A_{3}, A_{4}, \cdots, A_{17}$ is
$1 \times 2 \times 3 \times 4 \times 5=120 \text {. }$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.