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

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==Solution 1==
 
==Solution 1==
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We first claim that <math>n</math> must be divisible by 42. Since <math>\sigma(a^n)-1</math> is divisible by 2021 for all positive integers <math>a</math>, we can first consider the special case where <math>a \neq 0,1 \pmod{43}</math>.
+
We first claim that <math>n</math> must be divisible by <math>42</math>. Since <math>\sigma(a^n)-1</math> is divisible by <math>2021</math> for all positive integers <math>a</math>, we can first consider the special case where <math>a</math> is prime and <math>a \neq 0,1 \pmod{43}</math>. By Dirichlet's Theorem (Refer to the <b>Remark</b> section.), such <math>a</math> always exists.
  
−
Then <math>\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)</math>. In order for this expression to be divisible by <math>2021=43\times 47</math>, a necessary condition is <math>a^n - 1 \equiv 0 \pmod{43}</math>. By [[Fermat's Little Theorem]], <math>a^{42} \equiv 1 \pmod{43}</math>. Moreover, if <math>a</math> is a primitive root modulo 43, then <math>\text{ord}_{43}(a) = 42</math>, so <math>n</math> must be divisible by 42.
+
Then <math>\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)</math>. In order for this expression to be divisible by <math>2021=43\cdot 47</math>, a necessary condition is <math>a^n - 1 \equiv 0 \pmod{43}</math>. By [[Fermat's Little Theorem]], <math>a^{42} \equiv 1 \pmod{43}</math>. Moreover, if <math>a</math> is a primitive root modulo <math>43</math>, then <math>\text{ord}_{43}(a) = 42</math>, so <math>n</math> must be divisible by <math>42</math>.
  
−
By similar reasoning, <math>n</math> must be divisible by 46, by considering <math>a \neq 0,1 \pmod{47}</math>.
+
By similar reasoning, <math>n</math> must be divisible by <math>46</math>, by considering <math>a \not\equiv 0,1 \pmod{47}</math>.
  
−
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 <math>43</math>. By Dirichlet, let <math>a</math> be a prime that is congruent to <math>1 \pmod{43}</math>. Then <math>\sigma(a^n) \equiv n+1 \pmod{43}</math>, so since <math>\sigma(a^n)-1</math> is divisible by <math>43</math>, <math>n</math> is a multiple of <math>43</math>.
  
−
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
+
Alternatively, since <math>\left(\frac{a(a^n - 1^n)}{a-1}\right)</math> must be divisible by <math>43,</math> by LTE, we have <math>v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \geq 1,</math> which simplifies to <math>v_{43}(n) \geq 1,</math> which implies the desired result.
 +
 
 +
Similarly, <math>n</math> is a multiple of <math>47</math>.
 +
 
 +
Lastly, we claim that if <math>n = \text{lcm}(42, 46, 43, 47)</math>, then <math>\sigma(a^n) - 1</math> is divisible by <math>2021</math> 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(n)</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>, so
−
<cmath>\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})</cmath>
+
<cmath>\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}),</cmath>
−
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 \not\equiv 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 <math>43</math>, 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>2,3,7,23,43,</math> and <math>47,</math> and the answer is <math>2+3+7+23+43+47 = \boxed{125}</math>.
 +
 
 +
~scrabbler94, Revised by wzs26843545602
  
−
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
+
==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;">
 +
  <li>For all <math>a</math>, we have <math>\sigma(a^n)=\prod_{i=1}^k(1+p_i+\cdots+p_i^{ne_i})\equiv1\pmod{43}</math>.</li><p>
 +
  <li>For all <math>e>0</math> and prime <math>p</math>, we have <math>1+p+\cdots+p^{ne}\equiv1\pmod{43}</math>.</li><p>
 +
  <li>For all prime <math>p</math>, we have <math>p+p^2+\cdots+p^n \equiv 0 \pmod{43}</math>.</li><p>
 +
</ol>
 +
We can show this by casework on <math>p</math>:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>For <math>p\equiv0\pmod{43}</math>, no matter what <math>n</math> is, it is true that <cmath>p+p^2+\cdots+p^n \equiv 0 \pmod{43}.</cmath></li><p>
 +
  <li>For all <math>p\equiv1\pmod{43}</math>, it is true that <cmath>p+p^2+\cdots+p^n \equiv n \equiv 0 \pmod{43}.</cmath> One can either use brute force or Dirichlet's Theorem to show such <math>p</math> exists. Therefore, <math>43|n</math>.</li><p>
 +
  <li>For all <math>p\not\equiv0,1\pmod{43}</math>, it is true that
 +
<cmath>p+p^2+\cdots+p^n \equiv 0 \pmod{43} \Leftrightarrow p^n-1\equiv0\pmod{43}.</cmath> According to Fermat's Little Theorem, <math>42|n</math> is sufficient. To show it's necessary, we just need to show <math>43</math> has a prime primitive root. This can be done either by brute force or as follows. <math>43</math> is prime and it must have a primitive root <math>t\neq 1</math> that's coprime to <math>43</math>. All numbers of the form <math>43k+t</math> are also primitive roots of <math>43</math>. According to Dirichlet's Theorem there must be infinitely many primes among them.</li><p>
 +
</ol>
 +
Similar arguments for modulo <math>47</math> lead to <math>46|n</math> and <math>47|n</math>. Therefore, we get <math>n=\operatorname{lcm}[42,43,46,47]</math>. Following the last paragraph of Solution 1 gives the answer <math>\boxed{125}</math>.
  
−
==Solution 2 (cheap and not very reliable)==
+
~Mryang6859 (Solution)
−
Since the problem works for all positive integers <math>a</math>, let's plug in <math>a=2</math> and see what we get. Since <math>\sigma({2^n}) = 2^{n+1}-1,</math> we have <math>2^{n+1} \equiv 2 \pmod{2021}.</math> Simplifying using CRT and [[Fermat's Little Theorem]], we get that <math>2^n \equiv 0 \pmod{42}</math> and <math>2^n \equiv 0 \pmod{46}.</math> Then, we can look at <math>a=2022</math> just like in Solution 1 to find that <math>43</math> and <math>47</math> also divide <math>n</math>. 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 <math>\text{lcm(42, 43, 46, 47)}.</math> From here, follow solution 1 to get the final answer.
 
  
−
-PureSwag
+
~MRENTHUSIASM (Reformatting)
  
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==Solution 3 (Kind of similar to Sol 2 and kind of what cosmicgenius said)==
+
==Solution 3 (Casework)==
−
It says the problem implies that it works for all positive integers <math>a</math>, we basically know that If <math>p \equiv 1 \pmod q</math>, than from "USEMO 2019 Problem 4 which was <math>p^n + p^{n-1} + ... + 1 \equiv 1</math> (mod <math>q</math>) we have that <cmath>\frac{p^{en+1}-1}{p-1} \equiv p^{en} + p^{en-1} + \dots + 1 \equiv en + 1 \equiv 1 \pmod q,</cmath>. From here we can just let <math>\sigma(2^n)</math> or be a power of 2 which we can do <cmath>\sigma(2^n)=1+2+2^2+2^3+2^4+\cdots+2^n=2^{n+1}-1</cmath> which is a geo series. We can plug in <math>a=2^a</math> like in Solution 2 and use CRT which when we prime factorize we get that <math>2021 = 43 \cdot 47</math> 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 <math>a^{p-1} \equiv 1 \pmod p</math> we see all multiples of 42 will work for first and 46 for the second. We can figure out that it is just <math>\text{lcm}(43-1,47-1)\cdot43\cdot47</math> 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.  
+
We perform casework on <math>a:</math>
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>a=1.</math> <p>
 +
For all positive integers <math>n,</math> we have <math>\sigma\left(a^n\right)-1=0.</math>
 +
</li>
 +
  <li><math>a</math> is a prime number. <p>
 +
For all prime numbers <math>p,</math> note that
 +
<cmath>\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}</cmath>
 +
by geometric series. <p>
 +
Recall that <math>2021=43\cdot47.</math> Applying the Chinese Remainder Theorem, we get the following system of congruences:
 +
<cmath>\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*}</cmath>
 +
We perform casework on <math>(1):</math>
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<ol style="margin-left: 1.5em;" type="A">
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  <li><math>p-1\not\equiv0\pmod{43}.</math> <p>
 +
We need <math>p\left(p^n-1\right)\equiv0\pmod{43}.</math> <p>
 +
If <math>p=43,</math> then this congruence is true for all positive integers <math>n.</math> <p>
 +
If <math>p\neq43,</math> then this congruence becomes <math>p^n-1\equiv0\pmod{43}.</math> By Fermat's Little Theorem, we obtain <math>n\equiv0\pmod{42}.</math></li>
 +
  <li><math>p-1\equiv0\pmod{43}.</math> <p>
 +
We need <math>p^n-1</math> to contain more factors of <math>43</math> than <math>p-1</math> does. More generally, for all positive integers <math>k,</math> if <math>p-1\equiv0 \ \left(\operatorname{mod} \ 43^k\right),</math> then <math>p^n-1\equiv0 \ \left(\operatorname{mod} \ 43^{k+1}\right).</math> <p>
 +
Let <math>p=43^km+1</math> for some positive integer <math>m</math> such that <math>m\not\equiv0\pmod{43}.</math> We substitute this into <math>p^n-1\equiv0 \ \left(\operatorname{mod} \ 43^{k+1}\right)</math> and apply the Binomial Theorem:
 +
<cmath>\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*}</cmath>
 +
from which <math>n\equiv0\pmod{43}.</math>
 +
</li>
 +
</ol>
 +
For <math>(1),</math> we find that <math>n\equiv0\pmod{42}</math> and <math>n\equiv0\pmod{43}.</math> <p>
 +
For <math>(2),</math> we find that <math>n\equiv0\pmod{46}</math> and <math>n\equiv0\pmod{47}</math> by a similar argument. <p>
 +
Together, we conclude that <math>n=\operatorname{lcm}(42,43,46,47)</math> is the least such positive integer <math>n</math> for this case.
 +
</li><p>
 +
  <li><math>a</math> is a composite number. <p>
 +
Let <math>a=\prod_{i=1}^{u}p_i^{e_i}</math> be the prime factorization of <math>a.</math> Note that <math>\sigma(a)</math> is a multiplicative function: <cmath>\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),</cmath> as this product of geometric series spans all positive divisors of <math>a.</math> <p>
 +
At <math>n=\operatorname{lcm}(42,43,46,47),</math> it follows that
 +
<cmath>\begin{align*}
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\sigma\left(a^n\right)-1&=\left(\prod_{i=1}^{u}\sigma\left(p_i^{e_in}\right)\right)-1 \\
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&\equiv\left(\prod_{i=1}^{u}1\right)-1 &&\pmod{2021} \\
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&\equiv0 &&\pmod{2021}.
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\end{align*}</cmath>
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</li><p>
 +
</ol>
 +
Finally, the least such positive integer <math>n</math> for all cases is
 +
<cmath>\begin{align*}
 +
n&=\operatorname{lcm}(42,43,46,47) \\
 +
&=\operatorname{lcm}(2\cdot3\cdot7,43,2\cdot23,47) \\
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&=2\cdot3\cdot7\cdot23\cdot43\cdot47,
 +
\end{align*}</cmath>
 +
so the sum of its prime factors is <math>2+3+7+23+43+47=\boxed{125}.</math>
  
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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 <math>p^n + p^{n-1} + ... + 1 \equiv 1</math> which gave <math>p \equiv 1</math> <math>q</math>. Sorry if it's bad I need to do something else so I kind of rushed.
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~MRENTHUSIASM (inspired by Math Jams's <b>2021 AIME I Discussion</b>)
  
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==Solution 4 (Along the line of Solution 1)==
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==Solution 4 (Cheap)==
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<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 3 statements are the same.
+
Since the problem works for all positive integers <math>a</math>, let's plug in <math>a=2</math> and see what we get. Since <math>\sigma({2^n}) = 2^{n+1}-1,</math> we have <math>2^{n+1} \equiv 2 \pmod{2021}.</math> Simplifying using CRT and [[Fermat's Little Theorem]], we get that <math>n \equiv 0 \pmod{42}</math> and <math>n \equiv 0 \pmod{46}.</math> Then, we can look at <math>a</math> being a <math>1\pmod{43}</math> prime and a <math>1\pmod{47}</math> prime, just like in Solution 1, to find that <math>43</math> and <math>47</math> also divide <math>n</math>. 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 <math>\text{lcm(42, 43, 46, 47)}.</math> From here, follow solution 1 to get the final answer <math>\boxed{125}</math>.
−
<cmath>\text{for all }a, \sigma(a^n)=\prod_{i=1}^k(1+p_i+\cdots+p_i^{ne_i})\equiv1\pmod{43}</cmath>
 
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<cmath>\text{for all }e>0\text{ and prime }p, 1+p+\cdots+p^{ne}\equiv1\pmod{43}</cmath>
 
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<cmath>\text{for all prime }p, p+p^2+\cdots+p^n \equiv 0 \pmod{43}</cmath>
 
  
 +
~PureSwag
  
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(1) For <math>p\equiv0\pmod{43}</math>, no matter <math>n</math>, it is true
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==Solution 5 (Similar to Solution 4 and USEMO 2019 Problem 4)==
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<cmath>p+p^2+\cdots+p^n \equiv 0 \pmod{43}</cmath>
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Warning: This solution doesn't explain why <math>43*47\mid n</math>.
  
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(2) For all <math>p\equiv1\pmod{43}</math>,
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It says the problem implies that it works for all positive integers <math>a</math>, we basically know that If <math>p \equiv 1 \pmod q</math>, then from "USEMO 2019 Problem 4", if <math>p^n + p^{n-1} + \cdots + 1 \equiv 1 \pmod{q}</math>, then <cmath>\frac{p^{en+1}-1}{p-1} = p^{en} + p^{en-1} + \cdots + 1 \equiv 1 \pmod{q}.</cmath> From here we can just let <math>\sigma(2^n)</math> or be a power of <math>2</math> which we can do <cmath>\sigma(2^n)=1+2+2^2+2^3+2^4+\cdots+2^n=2^{n+1}-1,</cmath> which is a geometric series. We can plug in <math>a=2</math> like in Solution 4 and use CRT. We have the prime factorization <math>2021 = 43 \cdot 47</math>. We use CRT to find that <cmath>\begin{align*} 2^n &\equiv 1 \pmod{43}, \\ 2^n &\equiv 1 \pmod{47}. \end{align*}</cmath> We see that this is just FLT which is <math>a^{p-1} \equiv 1 \pmod p</math> we see that all multiples of <math>42</math> will work for first and <math>46</math> for the second. We can figure out that it is just <math>\text{lcm}(43-1,47-1)\cdot43\cdot47</math> which when we add up we get that it's just the sum of the prime factors of <math>\text{lcm}(42,43,46,47)</math> which you can just look at Solution 1 to find out the sum of the prime factors and get the answer <math>\boxed{125}</math>.
−
<cmath>p+p^2+\cdots+p^n \equiv n \equiv 0 \pmod{43}</cmath>
 
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One can either use brute force or Dirichlet's Theorem to show such <math>p</math> exists. Therefore
 
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<cmath>43|n.</cmath>
 
  
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(3) For all <math>p\not\equiv0,1\pmod{43}</math>,
+
==Remark (Dirichlet's Theorem)==
−
<cmath>p+p^2+\cdots+p^n \equiv 0 \pmod{43} \Leftrightarrow p^n-1\equiv0\pmod{43}</cmath>
+
All solutions require Dirichlet's Theorem, which states that for any coprime positive integers <math>k</math> and <math>r</math>, there are infinitely many primes <math>p</math> congruent to <math>r\pmod{k}</math>.
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According to Fermat's Little Theorem, <math>42|n</math> 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 <math>t\neq 1</math> that's coprime to 43. All numbers of the form <math>43k+t</math> are also primitive roots of 43. According to Dirichlet's Theorem there must be primes among them.
 
  
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Similar arguments for (mod 47) lead to <math>46|n</math> and <math>47|n</math>. Therefore
+
==Video Solution==
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<cmath>n=\text{lcm}[42,43,46,47]</cmath>
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https://youtu.be/q0zUFAyGN4s  ~MathProblemSolvingSkills.com
  
 
==See Also==
 
==See Also==

Latest revision as of 01:28, 13 November 2023

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 $a$ is prime and $a \neq 0,1 \pmod{43}$. By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists.

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 $43$. By Dirichlet, let $a$ be a prime that is congruent to $1 \pmod{43}$. Then $\sigma(a^n) \equiv n+1 \pmod{43}$, so since $\sigma(a^n)-1$ is divisible by $43$, $n$ is a multiple of $43$.

Alternatively, since $\left(\frac{a(a^n - 1^n)}{a-1}\right)$ must be divisible by $43,$ by LTE, we have $v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \geq 1,$ which simplifies to $v_{43}(n) \geq 1,$ which implies the desired result.

Similarly, $n$ is a multiple of $47$.

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, Revised by wzs26843545602

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 infinitely many 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 $n \equiv 0 \pmod{42}$ and $n \equiv 0 \pmod{46}.$ Then, we can look at $a$ being a $1\pmod{43}$ prime and a $1\pmod{47}$ prime, 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)

Warning: This solution doesn't explain why $43*47\mid n$.

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 $a=2$ 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 $\boxed{125}$.

Remark (Dirichlet's Theorem)

All solutions require Dirichlet's Theorem, which states that for any coprime positive integers $k$ and $r$, there are infinitely many primes $p$ congruent to $r\pmod{k}$.

Video Solution

https://youtu.be/q0zUFAyGN4s ~MathProblemSolvingSkills.com

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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