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Revision as of 08:44, 22 February 2022 by Bxiao31415 (talk | contribs) (AOPS Contributions)

AOPS Contributions

Observe that if $1 \le n \le 60$ such that n is a solution to the desired equation, so is $n + 60\cdot m$, where m is an integer, $0 \le m \le 9$. \\So we only need to consider n from 1 to 60. As shown in Solution 2, there are 4 cases which we will split into 2 main cases: \\ \\Case 1: $4 \mid n$ or $6 \mid n$, $5 \mid (n+1)$ \\Case 2: $4 \mid (n+1)$ or $6 \mid (n+1)$, $5 \mid n$ \\There are 4 values of n where $1 \le n \le 12$ satisfying $4 \mid n$ or $6 \mid n$. \\ \\I claim that there are 4 values of $n \le 60$ satisfying Case 1. Suppose x is one value of n satisfying $4 \mid n$ or $6 \mid n$, and $n \le 12$. \\Hence the solutions satisfying $4 \mid n$ or $6 \mid n$, $n \le 60$ are of the form $x + 12m$, so the values of $n + 1$ are $x + 12m + 1 \equiv x + 2m + 1 \equiv 0$ (mod 5), so $2m \equiv 4 + 4x$ (mod 5) and hence the value of m is unique since $0 \le m \le 4$ to satisfy $1 \le n \le 60$ and 2 and 5 are relatively prime. \\ \\A similar approach can be used to show the same for Case 2, that there are 4 values of $n \le 60$. \\ \\Hence our answer is $(4+4)*10 = \fbox{080}$.