2020 CIME I Problems/Problem 10

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Problem 10

Let $1=d_1<d_2<\cdots<d_k=n$ be the divisors of a positive integer $n$. Let $S$ be the sum of all positive integers $n$ satisfying \[n=d_1^1+d_2^2+d_3^3+d_4^4.\] Find the remainder when $S$ is divided by $1000$.

Solution

Note that $n$ is even, as if $n$ is odd then the RHS is even and the LHS is odd. This implies that the first two divisors of $n$ are $1$ and $2$. Now, there are three cases (note that $p_i$ represents a prime):

$\bullet$ When $\{ d_3, d_4 \} = \{ p_1, p_2 \}$. Then, $1+2^2+p_1^3+p_2^4$ is always odd, so this is a contradiction.

$\bullet$ When $\{ d_3, d_4 \} = \{ p_1, 2p_1 \}$. This implies that $p_1 | 1+2^2$ or $p_1 = 5$. Then, $n = 1+2^2+5^3+10^4 = 10130$.

$\bullet$ When $\{ d_3, d_4 \} = \{ 4, p_1 \}$. If $p_1 < 4 \implies p_1 = 3$ then $n = 288$. If $p_1 \geq 5$ then $4p_1 | 1+2^3+4^3+p_1^4 \implies p_1 | 1+2^2+4^3 = 69$. This means $p_1 = 23$ as $p_1 \geq 5$, however, this fails.


So, the sum of all possible values of $n$ are $10130 + 288 = 10418$, so the remainder is $\boxed{418}$. ~rocketsri

See also

2020 CIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All CIME Problems and Solutions

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