Selected Problems from Open Contests

Note that for every $n$, $n!+(n+1)!=n!+n!\cdot (n+1)=n!\cdot (n+2)\le (n+2)!$. Thus if $n!+(n+1)!=k!+120$, then due to $120=5!$ we have $k!+5!\le (n+2)!$. This inequality in turn implies $k<n+2$ and $5<n+2$. Hence $5\le n+1$, leading to $0\le (n+1)!-5!=k!-n!$. Consequently, $k\ge n$, i.e., the cases to be considered are $k=n$ and $k=n+1$. If $k=n$, then the initial equation leads to $(n+1)!=120$, giving $n=4$, $k=4$. If $k=n+1$, then analogously $n=5$, $k=6$.

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Alternative solution.

If $k<n$, then $n!>k!$. If, additionally, $n\ge 4$, then $(n+1)!\ge 120$, giving $n!+(n+1)!>k!+120$. But if $n<4$, then $n!+(n+1)!\le 30<k!+120$. The desired equality can hold in neither of the cases. If $k>n+1$, then $k!-(n+1)!$ is positive and is also divisible by $(n+1)!$, hence $k!-(n+1)!\ge (n+1)!$. On the other hand, $n!-120<(n+1)!$, giving $k!-(n+1)!>n!-120$. Thus, there is no solution in this case either. Hence $n=k$ or $n+1=k$, leading to two solutions $(n,k)=(4,4),(5,6)$.

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Alternative solution.

The equation implies that $120$ is divisible by the minimum of $n!$ and $k!$. As $120=5!$, either $n\le 5$ or $k\le 5$. Consider both cases. If $n\in \{1,2,3\}$, then $n!+(n+1)!-k!<120$, whence the equation has no solution. If $n=4$, then $120=144-k!$, whence $k!=24$ and $k=4$. If $n=5$, then $120=840-k!$, whence $k!=720$ and $k=6$. If $k\in \{1,2,3,4,5\}$, then the initial equation implies that $n!+(n+1)!$ lies between 121 and 240. This is possible only if $n=4$, since if $n=3$, then $n!+(n+1)!=30$, and if $n=5$, then $n!+(n+1)!=840$. If $n=4$, then $n!+(n+1)!=144$, which corresponds to $k=4$.