2021 AIME I Problems/Problem 14

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 $\sigma(n)$ 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:

  1. For all $a$, we have $\sigma(a^n)=\prod_{i=1}^k(1+p_i+\cdots+p_i^{ne_i})\equiv1\pmod{43}$.
  2. For all $e>0$ and prime $p$, we have $1+p+\cdots+p^{ne}\equiv1\pmod{43}$.
  3. For all prime $p$, we have $p+p^2+\cdots+p^n \equiv 0 \pmod{43}$.

We can show this by casework on $p$:

  1. For $p\equiv0\pmod{43}$, no matter what $n$ is, it is true that \[p+p^2+\cdots+p^n \equiv 0 \pmod{43}.\]
  2. 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, $43|n$.
  3. 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 $n=\operatorname{lcm}[42,43,46,47]$. Following the last paragraph of Solution 1 gives the answer $\boxed{125}$.

~Mryang6859 (Solution)

~MRENTHUSIASM (Reformatting)

Solution 3 (Casework)

We perform casework on $a:$

  1. $a=1.$

    For all positive integers $n,$ we have $\sigma\left(a^n\right)-1=0.$

  2. $a$ is a prime number.

    For all prime numbers $p,$ note that \[\sigma\left(p^n\right)-1=\left(\sum_{i=0}^{n}p^i\right)-1=\sum_{i=1}^{n}p^i=\frac{p^{n+1}-p}{p-1}=\frac{p\left(p^n-1\right)}{p-1}\] by geometric series.

    Recall that $2021=43\cdot47.$ Applying the Chinese Remainder Theorem, we get the following system of congruences: \begin{align*} \frac{p\left(p^n-1\right)}{p-1}&\equiv0\pmod{43}, &(1)\\ \frac{p\left(p^n-1\right)}{p-1}&\equiv0\pmod{47}. &(2) \end{align*} We perform casework on $(1):$

    1. $p-1\not\equiv0\pmod{43}.$

      We need $p\left(p^n-1\right)\equiv0\pmod{43}.$

      If $p=43,$ then this congruence is true for all positive integers $n.$

      If $p\neq43,$ then this congruence becomes $p^n-1\equiv0\pmod{43}.$ By Fermat's Little Theorem, we obtain $n\equiv0\pmod{42}.$

    2. $p-1\equiv0\pmod{43}.$

      We need $p^n-1$ to contain more factors of $43$ than $p-1$ does. More generally, for all positive integers $k,$ if $p-1\equiv0 \ \left(\operatorname{mod} \ 43^k\right),$ then $p^n-1\equiv0 \ \left(\operatorname{mod} \ 43^{k+1}\right).$

      Let $p=43^km+1$ for some positive integer $m$ such that $m\not\equiv0\pmod{43}.$ We substitute this into $p^n-1\equiv0 \ \left(\operatorname{mod} \ 43^{k+1}\right)$ and apply the Binomial Theorem: \begin{align*} \left(43^km+1\right)^n-1&\equiv0 \ \left(\operatorname{mod} \ 43^{k+1}\right) \\ \Biggl[\binom{n}{0}\left(43^km\right)^0+\binom{n}{1}\left(43^km\right)^1+\phantom{ }\underbrace{\binom{n}{2}\left(43^km\right)^2+\cdots+\binom{n}{n}\left(43^km\right)^n}_{0 \ \left(\operatorname{mod} \ 43^{k+1}\right)}\phantom{ }\Biggr]-1 &\equiv0 \ \left(\operatorname{mod} \ 43^{k+1}\right) \\ \left[1+43^kmn\right]-1&\equiv0 \ \left(\operatorname{mod} \ 43^{k+1}\right) \\ 43^kmn&\equiv0 \ \left(\operatorname{mod} \ 43^{k+1}\right), \end{align*} from which $n\equiv0\pmod{43}.$

    For $(1),$ we find that $n\equiv0\pmod{42}$ and $n\equiv0\pmod{43}.$

    For $(2),$ we find that $n\equiv0\pmod{46}$ and $n\equiv0\pmod{47}$ by a similar argument.

    Together, we conclude that $n=\operatorname{lcm}(42,43,46,47)$ is the least such positive integer $n$ for this case.

  3. $a$ is a composite number.

    Let $a=\prod_{i=1}^{u}p_i^{e_i}$ be the prime factorization of $a.$ Note that $\sigma(a)$ is a multiplicative function: \[\sigma(a)=\sigma\left(\prod_{i=1}^{u}p_i^{e_i}\right)=\prod_{i=1}^{u}\sigma\left(p_i^{e_i}\right)=\prod_{i=1}^{u}\left(\sum_{j=0}^{e_i}p_i^j\right),\] as this product of geometric series spans all positive divisors of $a.$

    At $n=\operatorname{lcm}(42,43,46,47),$ it follows that \begin{align*} \sigma\left(a^n\right)-1&=\left(\prod_{i=1}^{u}\sigma\left(p_i^{e_in}\right)\right)-1 \\ &\equiv\left(\prod_{i=1}^{u}1\right)-1 &&\pmod{2021} \\ &\equiv0 &&\pmod{2021}. \end{align*}

Finally, the least such positive integer $n$ for all cases is \begin{align*} n&=\operatorname{lcm}(42,43,46,47) \\ &=\operatorname{lcm}(2\cdot3\cdot7,43,2\cdot23,47) \\ &=2\cdot3\cdot7\cdot23\cdot43\cdot47, \end{align*} so the sum of its prime factors is $2+3+7+23+43+47=\boxed{125}.$

~MRENTHUSIASM (inspired by Math Jams's 2021 AIME I Discussion)

Solution 4 (Cheap)

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 $\boxed{125}$.

-PureSwag

Solution 5 (Similar to Solution 4 and USEMO 2019 Problem 4)

It says the problem implies that it works for all positive integers $a$, we basically know that If $p \equiv 1 \pmod q$, then from "USEMO 2019 Problem 4", if $p^n + p^{n-1} + \cdots + 1 \equiv 1 \pmod{q}$, then \[\frac{p^{en+1}-1}{p-1} = p^{en} + p^{en-1} + \cdots + 1 \equiv 1 \pmod{q}.\] 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 geometric series. We can plug in $a=2^a$ like in Solution 4 and use CRT. We have the prime factorization $2021 = 43 \cdot 47$. 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 that 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 $\text{lcm}(42,43,46,47)$ which you can just look at Solution 1 to find out the sum of the prime factors and get the answer $\boxed{125}$.

See Also

2021 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS