Difference between revisions of "2020 CIME I Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | {{ | + | 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 | ||
==See also== | ==See also== |
Latest revision as of 11:47, 2 February 2021
Problem 10
Let be the divisors of a positive integer . Let be the sum of all positive integers satisfying Find the remainder when is divided by .
Solution
Note that is even, as if is odd then the RHS is even and the LHS is odd. This implies that the first two divisors of are and . Now, there are three cases (note that represents a prime):
When . Then, is always odd, so this is a contradiction.
When . This implies that or . Then, .
When . If then . If then . This means as , however, this fails.
So, the sum of all possible values of are , so the remainder is . ~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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