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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.
  
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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 $a, \sigma(a)$ denotes the sum of the positive integer divisors of $a$. Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$. What is the sum of the prime factors in the prime factorization of $n$?

Solution

We first claim that $n$ must be divisible by 42. Since $\sigma(a^n)-1$ is divisible by 2021 for all positive integers $a$, we can first consider the special case where $a \neq 0,1 \pmod{43}$.

Then $\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)$. In order for this expression to be divisible by $2021=43\times 47$, a necessary condition is $a^n - 1 \equiv 0 \pmod{43}$. By Fermat's Little Theorem, $a^{42} \equiv 1 \pmod{43}$. Moreover, if $a$ is a primitive root modulo 43, then $\text{ord}_{43}(a) = 42$, so $n$ must be divisible by 42.

By similar reasoning, $n$ must be divisible by 46, by considering $a \neq 0,1 \pmod{47}$.

We next claim that $n$ must be divisible by 43 and 47. Consider the case $a=2022$. Then $\sigma(a^n) \equiv n \pmod{2021}$, so $\sigma(2022^n)-1$ is divisible by 2021 if and only if $n$ is divisible by 2021.

Lastly, we claim that if $n = \text{lcm}(42, 46, 43, 47)$, then $\sigma(a^n) - 1$ is divisible by 2021 for all positive integers $a$. The claim is trivially true for $a=1$ so suppose $a>1$. Let $a = p_1^{e_1}\ldots p_k^{e_k}$ be the prime factorization of $a$. Since $\sigma$ is multiplicative, we have \[\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.\] We can show that $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for all primes $p_i$ and integers $e_i \ge 1$, where \[\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in})\] where each expression in parentheses contains $n$ terms. It is easy to verify that if $p_i = 43$ or $p_i = 47$ then $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for this choice of $n$, so suppose $p_i \not\equiv 0 \pmod{43}$ and $p_i \not\equiv 0 \pmod{47}$. Each expression in parentheses equals $\frac{p_i^n - 1}{p_i - 1}$ multiplied by some power of $p_i$. If $p_i \neq 1 \pmod{43}$, then FLT implies $p_i^n - 1 \equiv 0 \pmod{43}$, and if $p_i \equiv 1 \pmod{43}$, then $p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}$ (since $n$ is also a multiple of 43, by definition). Similarly, we can show $\sigma(p_i^{e_in}) \equiv 1 \pmod{47}$, and a simple CRT argument shows $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$. Then $\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}$.

Then the prime factors of $n$ are $2$, $3$, $7$, $23$, $43$, and $47$, and the answer is $2+3+7+23+43+47 = \boxed{125}$. ~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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