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Difference between revisions of "2020 CIME I Problems/Problem 10"
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==Solution==
 
==Solution==
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Note that <math>n</math> is even, as if <math>n</math> is odd then the RHS is even and the LHS is odd. This implies that the first two divisors of <math>n</math> are <math>1</math> and <math>2</math>. Now, there are three cases (note that <math>p_i</math> represents a prime):
  
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<math>\bullet</math> When <math>\{ d_3, d_4 \} = \{ p_1, p_2 \}</math>. Then, <math>1+2^2+p_1^3+p_2^4</math> is always odd, so this is a contradiction.
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<math>\bullet</math> When <math>\{ d_3, d_4 \} = \{ p_1, 2p_1 \}</math>. This implies that <math>p_1 | 1+2^2</math> or <math>p_1 = 5</math>. Then, <math>n = 1+2^2+5^3+10^4 = 10130</math>.
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<math>\bullet</math> When <math>\{ d_3, d_4 \} = \{ 4, p_1 \}</math>. If <math>p_1 < 4 \implies p_1 = 3</math> then <math>n = 288</math>. If <math>p_1 \geq 5</math> then <math>4p_1 | 1+2^3+4^3+p_1^4 \implies p_1 | 1+2^2+4^3 = 69</math>. This means <math>p_1 = 23</math> as <math>p_1 \geq 5</math>, however, this fails.
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So, the sum of all possible values of <math>n</math> are <math>10130 + 288 = 10418</math>, so the remainder is <math>\boxed{418}</math>. ~rocketsri
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==See also==
 
{{CIME box|year=2020|n=I|num-b=9|num-a=11}}
 
{{CIME box|year=2020|n=I|num-b=9|num-a=11}}
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
{{MAC Notice}}
 
{{MAC Notice}}

Latest revision as of 10:47, 2 February 2021

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 (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

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