Difference between revisions of "2021 AIME I Problems/Problem 14"

Revision as of 05:21, 31 July 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 $a$ . Find the sum of the prime factors in the prime factorization of $n$ .

Solution 1

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 $2021=43\cdot 47$ , 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 $a \not\equiv 0,1 \pmod{47}$ .

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 $e_i \ge 1$ , so $\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in}),$ 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 $p_i \not\equiv 1 \pmod{43}$ , 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

Solution 2 (Along the Lines of Solution 1)

$n$ only needs to satisfy $\sigma(a^n)\equiv 1 \pmod{43}$ and $\sigma(a^n)\equiv 1 \pmod{47}$ for all $a$ . Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$ . If $a>1$ , let $a$ 's prime factorization be $a=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ . The following three statements are the same:

For all $a$ , we have $\sigma(a^n)=\prod_{i=1}^k(1+p_i+\cdots+p_i^{ne_i})\equiv1\pmod{43}$ .

For all $e>0$ and prime $p$ , we have $1+p+\cdots+p^{ne}\equiv1\pmod{43}$ .

For all prime $p$ , we have $p+p^2+\cdots+p^n \equiv 0 \pmod{43}$ .

We can show this by casework on $p$ :

For $p\equiv0\pmod{43}$ , no matter what $n$ is, it is true that $p+p^2+\cdots+p^n \equiv 0 \pmod{43}.$

For all $p\equiv1\pmod{43}$ , it is true that $p+p^2+\cdots+p^n \equiv n \equiv 0 \pmod{43}.$ One can either use brute force or Dirichlet's Theorem to show such $p$ exists. Therefore, $43|n$ .

For all $p\not\equiv0,1\pmod{43}$ , it is true that $p+p^2+\cdots+p^n \equiv 0 \pmod{43} \Leftrightarrow p^n-1\equiv0\pmod{43}.$ According to Fermat's Little Theorem, $42|n$ is sufficient. To show it's necessary we just need to show $43$ has a prime primitive root. This can be done either by brute force or as follows. $43$ is prime and it must have a primitive root $t\neq 1$ that's coprime to $43$ . All numbers of the form $43k+t$ are also primitive roots of $43$ . According to Dirichlet's Theorem there must be primes among them.

Similar arguments for modulo $47$ lead to $46|n$ and $47|n$ . Therefore, we get $n=\text{lcm}[42,43,46,47]$ . Following the last paragraph of Solution 1 gives the answer $\boxed{125}$ .

~Mryang6859 (Solution)

~MRENTHUSIASM (Reformatting)

Solution 2 (cheap and not very reliable)

Since the problem works for all positive integers $a$ , let's plug in $a=2$ and see what we get. Since $\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \equiv 2 \pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem, we get that $2^n \equiv 0 \pmod{42}$ and $2^n \equiv 0 \pmod{46}.$ Then, we can look at $a=2022$ just like in Solution 1 to find that $43$ and $47$ also divide $n$ . There don't seem to be any other odd "numbers" to check, so we can hopefully assume that the answer is the sum of the prime factors of $\text{lcm(42, 43, 46, 47)}.$ From here, follow solution 1 to get the final answer.

-PureSwag

Solution 3 (Kind of similar to Sol 2 and kind of what cosmicgenius said)

It says the problem implies that it works for all positive integers $a$ , we basically know that If $p \equiv 1 \pmod q$ , than from "USEMO 2019 Problem 4 which was $p^n + p^{n-1} + ... + 1 \equiv 1$ (mod $q$ ) we have that $\frac{p^{en+1}-1}{p-1} \equiv p^{en} + p^{en-1} + \dots + 1 \equiv en + 1 \equiv 1 \pmod q,$ . From here we can just let $\sigma(2^n)$ or be a power of 2 which we can do $\sigma(2^n)=1+2+2^2+2^3+2^4+\cdots+2^n=2^{n+1}-1$ which is a geo series. We can plug in $a=2^a$ like in Solution 2 and use CRT which when we prime factorize we get that $2021 = 43 \cdot 47$ which like everyone knows. We use CRT to find that \begin{align*} 2^n &\equiv 1 \pmod{43}, \\ 2^n &\equiv 1 \pmod{47}. \end{align*}. We see that this is just FLT which is $a^{p-1} \equiv 1 \pmod p$ we see all multiples of 42 will work for first and 46 for the second. We can figure out that it is just $\text{lcm}(43-1,47-1)\cdot43\cdot47$ which when we add up we get that it's just the sum of the prime factors of lcm(42,43,46,47) which you can just look at Solution 1 to find out the sum of the prime factors and get the answer.

Remarks: This was kind of used in what cosmicgenius and Solution 2 said and what mathisawesome2169 said and It "Reminded me of 2019 USEMO Problem 4 when solving the $p^n + p^{n-1} + ... + 1 \equiv 1$ which gave $p \equiv 1$ $q$ . Sorry if it's bad I need to do something else so I kind of rushed.

@@ Line 21: / Line 21: @@
 ~scrabbler94
-==Solution 2 (Along the line of Solution 1)==
+==Solution 2 (Along the Lines of Solution 1)==
 <math>n</math> only needs to satisfy <math>\sigma(a^n)\equiv 1 \pmod{43}</math> and <math>\sigma(a^n)\equiv 1 \pmod{47}</math> for all <math>a</math>. Let's work on the first requirement (mod 43) first. All <math>n</math> works for <math>a=1</math>. If <math>a>1</math>, let <math>a</math>'s prime factorization be <math>a=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}</math>. The following three statements are the same:
 <ol style="margin-left: 1.5em;">
@@ Line 36: / Line 36: @@
 </ol>
 Similar arguments for modulo <math>47</math> lead to <math>46|n</math> and <math>47|n</math>. Therefore, we get <math>n=\text{lcm}[42,43,46,47]</math>. Following the last paragraph of Solution 1 gives the answer <math>\boxed{125}</math>.
+~Mryang6859 (Solution)
+~MRENTHUSIASM (Reformatting)
 ==Solution 2 (cheap and not very reliable)==

2021 AIME I (Problems • Answer Key • Resources)
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