Difference between revisions of "2021 AIME I Problems/Problem 14"
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We next claim that <math>n</math> must be divisible by 43 and 47. Consider the case <math>a=2022</math>. Then <math>\sigma(a^n) \equiv n \pmod{2021}</math>, so <math>\sigma(2022^n)-1</math> is divisible by 2021 if and only if <math>n</math> is divisible by 2021. | We next claim that <math>n</math> must be divisible by 43 and 47. Consider the case <math>a=2022</math>. Then <math>\sigma(a^n) \equiv n \pmod{2021}</math>, so <math>\sigma(2022^n)-1</math> is divisible by 2021 if and only if <math>n</math> is divisible by 2021. | ||
− | Lastly, we claim that if <math>n = \text{lcm}(42, 46, 43, 47)</math>, then <math>\sigma(a^n) - 1</math> is divisible by 2021 for all positive integers <math>a</math>. The claim is trivially true for <math>a=1</math> so suppose <math>a>1</math>. Let <math>a = p_1^{e_1}\ldots p_k^{e_k}</math> be the prime factorization of <math>a</math>. Since <math>\sigma</math> is multiplicative, we have | + | Lastly, we claim that if <math>n = \text{lcm}(42, 46, 43, 47)</math>, then <math>\sigma(a^n) - 1</math> is divisible by 2021 for all positive integers <math>a</math>. The claim is trivially true for <math>a=1</math> so suppose <math>a>1</math>. Let <math>a = p_1^{e_1}\ldots p_k^{e_k}</math> be the prime factorization of <math>a</math>. Since <math>\sigma</math> is [[multiplicative function|multiplicative]], we have |
<cmath>\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.</cmath> | <cmath>\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.</cmath> | ||
We can show that <math>\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}</math> for all primes <math>p_i</math> and integers <math>e_i \ge 1</math>, where | We can show that <math>\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}</math> for all primes <math>p_i</math> and integers <math>e_i \ge 1</math>, where | ||
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where each expression in parentheses contains <math>n</math> terms. It is easy to verify that if <math>p_i = 43</math> or <math>p_i = 47</math> then <math>\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}</math> for this choice of <math>n</math>, so suppose <math>p_i \not\equiv 0 \pmod{43}</math> and <math>p_i \not\equiv 0 \pmod{47}</math>. Each expression in parentheses equals <math>\frac{p_i^n - 1}{p_i - 1}</math> multiplied by some power of <math>p_i</math>. If <math>p_i \neq 1 \pmod{43}</math>, then FLT implies <math>p_i^n - 1 \equiv 0 \pmod{43}</math>, and if <math>p_i \equiv 1 \pmod{43}</math>, then <math>p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}</math> (since <math>n</math> is also a multiple of 43, by definition). Similarly, we can show <math>\sigma(p_i^{e_in}) \equiv 1 \pmod{47}</math>, and a simple [[Chinese Remainder Theorem|CRT]] argument shows <math>\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}</math>. Then <math>\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}</math>. | where each expression in parentheses contains <math>n</math> terms. It is easy to verify that if <math>p_i = 43</math> or <math>p_i = 47</math> then <math>\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}</math> for this choice of <math>n</math>, so suppose <math>p_i \not\equiv 0 \pmod{43}</math> and <math>p_i \not\equiv 0 \pmod{47}</math>. Each expression in parentheses equals <math>\frac{p_i^n - 1}{p_i - 1}</math> multiplied by some power of <math>p_i</math>. If <math>p_i \neq 1 \pmod{43}</math>, then FLT implies <math>p_i^n - 1 \equiv 0 \pmod{43}</math>, and if <math>p_i \equiv 1 \pmod{43}</math>, then <math>p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}</math> (since <math>n</math> is also a multiple of 43, by definition). Similarly, we can show <math>\sigma(p_i^{e_in}) \equiv 1 \pmod{47}</math>, and a simple [[Chinese Remainder Theorem|CRT]] argument shows <math>\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}</math>. Then <math>\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}</math>. | ||
− | Then the prime factors of <math>n</math> are <math>2</math>, <math>3</math>, <math>7</math>, <math>23</math>, <math>43</math>, and <math>47</math>, and the answer is <math>2+3+7+23+43+47 = \boxed{125}</math>. | + | Then the prime factors of <math>n</math> are <math>2</math>, <math>3</math>, <math>7</math>, <math>23</math>, <math>43</math>, and <math>47</math>, and the answer is <math>2+3+7+23+43+47 = \boxed{125}</math>. ~scrabbler94 |
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=13|num-a=15}} | {{AIME box|year=2021|n=I|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:23, 11 March 2021
Problem
For any positive integer denotes the sum of the positive integer divisors of
. Let
be the least positive integer such that
is divisible by
for all positive integers
. What is the sum of the prime factors in the prime factorization of
?
Solution
We first claim that must be divisible by 42. Since
is divisible by 2021 for all positive integers
, we can first consider the special case where
.
Then . In order for this expression to be divisible by
, a necessary condition is
. By Fermat's Little Theorem,
. Moreover, if
is a primitive root modulo 43, then
, so
must be divisible by 42.
By similar reasoning, must be divisible by 46, by considering
.
We next claim that must be divisible by 43 and 47. Consider the case
. Then
, so
is divisible by 2021 if and only if
is divisible by 2021.
Lastly, we claim that if , then
is divisible by 2021 for all positive integers
. The claim is trivially true for
so suppose
. Let
be the prime factorization of
. Since
is multiplicative, we have
We can show that
for all primes
and integers
, where
where each expression in parentheses contains
terms. It is easy to verify that if
or
then
for this choice of
, so suppose
and
. Each expression in parentheses equals
multiplied by some power of
. If
, then FLT implies
, and if
, then
(since
is also a multiple of 43, by definition). Similarly, we can show
, and a simple CRT argument shows
. Then
.
Then the prime factors of are
,
,
,
,
, and
, and the answer is
. ~scrabbler94
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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