Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 CIME I Problems/Problem 10"
(→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
−
the only such positive integer is <math>288</math> i put <math>288</math> confirmed?
+
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):
 +
 
 +
<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.
 +
 
 +
<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>.
 +
 
 +
<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.
 +
 
 +
 
 +
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
  
 
==See also==
 
==See also==

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

The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_CIME_I_Problems/Problem_10&oldid=144639"