Competition problem — Number theory

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Number theory Difficulty 5.0 AIME, harder Find the answer

1. Arrange the numbers $1,2, \cdots, 13$ in a row $a_{1}, a_{2}$ , $\cdots, a_{13}$ , where $a_{1}=13, a_{2}=1$ , and ensure that $a_{1}+a_{2}+$ $\cdots+a_{k}$ is divisible by $a_{k+1}(k=1,2, \cdots, 12)$ . Then the value of $a_{4}$ $+a_{5}+\cdots+a_{12}$ is $\qquad$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

II. 1.68.
Since $a_{1}+a_{2}+\cdots+a_{12}$ is divisible by $a_{13}$ , then $a_{1}+a_{2}+\cdots+a_{12}+a_{13}$ is also divisible by $a_{13}$ , meaning $a_{13}$ is a factor of $a_{1}+a_{2}+\cdots+a_{12}+a_{13}=13 \times 7$ . Given that $a_{1}=13$ , $a_{2}=1$ , we have $a_{13}=7$ .

Also, since $a_{1}+a_{2}$ is divisible by $a_{3}$ , then $a_{3}$ is a factor of $a_{1}+a_{2}=2 \times 7$ . Therefore, $a_{3}=2$ . We can sequentially obtain a series of numbers that satisfy the conditions: $13,1,2,8,3,9,4,10,5,11,6,12,7$ . Thus,
$a_{4}+a_{5}+\cdots+a_{12}=13 \times 7-13-1-2-7=68 .$

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