2020 CIME I Problems/Problem 10

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

2020 CIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2020 CIME I Problems/Problem 10

Problem 10

Solution

See also