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Difference between revisions of "2018 AIME I Problems/Problem 11"

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Find the least positive integer <math>n</math> such that when <math>3^n</math> is written in base <math>143</math>, its two right-most digits in base <math>143</math> are <math>01</math>.
 
  
−
==Solutions==
+
==Problem==
 +
Find the least positive integer <math>n</math> such that when <math>3^n</math> is written in base <math>143^2</math>, its two right-most digits in base <math>143</math> are <math>01</math>.
  
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==Modular Arithmetic Solution- Strange (MASS)==
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==Solution 1==
−
Note that the given condition is equivalent to <math>3^n \equiv 1 \pmod{143^2}</math> and <math>143=11\cdot 13</math>. Because <math>gcd(11^2, 13^2) = 1</math>, the desired condition is equivalent to <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>.
+
Note that the given condition is equivalent to <math>3^n \equiv 1 \pmod{143^2}</math> and <math>143=11\cdot 13</math>. Because <math>\gcd(11^2, 13^2) = 1</math>, the desired condition is equivalent to <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>.
  
 
If <math>3^n \equiv 1 \pmod{121}</math>, one can see the sequence <math>1, 3, 9, 27, 81, 1, 3, 9...</math> so <math>5|n</math>.
 
If <math>3^n \equiv 1 \pmod{121}</math>, one can see the sequence <math>1, 3, 9, 27, 81, 1, 3, 9...</math> so <math>5|n</math>.
  
−
Now if <math>3^n \equiv 1 \pmod{169}</math>, it is harder. But we do observe that <math>3^3 \equiv 1 \pmod{13}</math>, therefore <math>3^3 = 13a + 1</math> for some integer <math>a</math>. So our goal is to find the first number <math>p_1</math> such that <math>(13a+1)^ {p_1} \equiv 1 \pmod{169}</math>. In other words, the <math>a \equiv 0 \pmod{13}</math>. It is not difficult to see that the smallest <math>p_1=13</math>, so ultimately <math>3^{39} \equiv 1 \pmod{169}</math>. Therefore, <math>39|n</math>.
+
Now if <math>3^n \equiv 1 \pmod{169}</math>, it is harder. But we do observe that <math>3^3 \equiv 1 \pmod{13}</math>, therefore <math>3^3 = 13a + 1</math> for some integer <math>a</math>. So our goal is to find the first number <math>p_1</math> such that <math>(13a+1)^ {p_1} \equiv 1 \pmod{169}</math>. Then, <math>p_1 \equiv 0 \pmod{13}, </math> which follows from the binomial theorem. It is not difficult to see that the smallest <math>p_1=13</math>, so ultimately <math>3^{39} \equiv 1 \pmod{169}</math>. Therefore, <math>39|n</math>.
  
 
The first <math>n</math> satisfying both criteria is thus <math>5\cdot 39=\boxed{195}</math>.
 
The first <math>n</math> satisfying both criteria is thus <math>5\cdot 39=\boxed{195}</math>.
Line 17: Line 17:
 
Note that Euler's Totient Theorem would not necessarily lead to the smallest <math>n</math> and that in this case that <math>n</math> is greater than <math>1000</math>.  
 
Note that Euler's Totient Theorem would not necessarily lead to the smallest <math>n</math> and that in this case that <math>n</math> is greater than <math>1000</math>.  
  
−
We wish to find the least <math>n</math> such that <math>3^n \equiv 1 \pmod{143^2}</math>. This factors as <math>143^2=11^{2}*13^{2}</math>. Because <math>gcd(121, 169) = 1</math>, we can simply find the least  <math>n</math> such that <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>.
+
We wish to find the least <math>n</math> such that <math>3^n \equiv 1 \pmod{143^2}</math>. This factors as <math>143^2=11^{2}*13^{2}</math>. Because <math>\gcd(121, 169) = 1</math>, we can simply find the least  <math>n</math> such that <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>.
  
−
Quick inspection yields <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^3 \equiv 1 \pmod{13}</math>. Now we must find the smallest <math>k</math> such that <math>3^{3k} \equiv 1 \pmod{13}</math>. Euler's gives <math>3^{156} \equiv 1 \pmod{169}</math>. So <math>3k</math> is a factor of <math>156</math>. This gives <math>k=1,2, 4, 13, 26, 52</math>. Some more inspection yields <math>k=13</math> is the smallest valid <math>k</math>. So <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^{39} \equiv 1 \pmod{169}</math>. The least <math>n</math> satisfying both is <math>lcm(5, 39)=\boxed{195}</math>. (RegularHexagon)
+
Quick inspection yields <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^3 \equiv 1 \pmod{13}</math>. Now we must find the smallest <math>k</math> such that <math>3^{3k} \equiv 1 \pmod{169}</math>. Euler's gives <math>3^{156} \equiv 1 \pmod{169}</math>. So <math>3k</math> is a factor of <math>156</math>. This gives <math>k=1,2, 4, 13, 26, 52</math>. Some more inspection yields <math>k=13</math> is the smallest valid <math>k</math>. So <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^{39} \equiv 1 \pmod{169}</math>. The least <math>n</math> satisfying both is <math>lcm(5, 39)=\boxed{195}</math>. (RegularHexagon)
  
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==Solution 3 (BigBash)==
+
==Solution 3 (Big Bash)==
 
Listing out the powers of <math>3</math>, modulo <math>169</math> and modulo <math>121</math>, we have:
 
Listing out the powers of <math>3</math>, modulo <math>169</math> and modulo <math>121</math>, we have:
 
<cmath>\begin{array}{c|c|c}
 
<cmath>\begin{array}{c|c|c}
Line 67: Line 67:
 
\end{array}</cmath>
 
\end{array}</cmath>
  
−
The powers of <math>3</math> repeat in cycles of <math>5</math> an <math>39</math> in modulo <math>121</math> and modulo <math>169</math>, respectively. The answer is <math>lcm(5, 39) = \boxed{195}</math>.
+
The powers of <math>3</math> repeat in cycles of <math>5</math> and <math>39</math> in modulo <math>121</math> and modulo <math>169</math>, respectively. The answer is <math>\text{lcm}(5, 39) = \boxed{195}</math>.
 +
 
 +
~Minor edit by Yiyj1
 +
 
 +
==Solution 4 (Order+Bash)==
 +
We have that <cmath>3^n \equiv 1 \pmod{143^2}.</cmath>Now, <math>3^{110} \equiv 1 \pmod{11^2}</math> so by the Fundamental Theorem of Orders, <math>\text{ord}_{11^2}(3)|110</math> and with some bashing, we get that it is <math>5</math>. Similarly, we get that <math>\text{ord}_{13^2}(3)=39</math>. Now, <math>\text{lcm}(39,5)=\boxed{195}</math> which is our desired solution.
 +
 
 +
==Solution 5 (Easy Binomial Theorem)==
 +
We wish to find the smallest <math>n</math> such that <math>3^n\equiv 1\pmod{143^2}</math>, so we want <math>n\equiv 1\pmod{121}</math> and <math>n\equiv 1\pmod{169}</math>. Note that <math>3^5\equiv 1\pmod{121}</math>,  so <math>3^n</math> repeats <math>121</math> with a period of <math>5</math>, so <math>5|n</math>. Now, in order for <math>n\equiv 1\pmod{169}</math>, then <math>n\equiv 1\pmod{13}</math>. Because <math>3^3\equiv 1\pmod{13}</math>, <math>3^n</math> repeats with a period of <math>3</math>, so <math>3|n</math>.
 +
Hence, we have that for some positive integer <math>p</math>, <math>3^n\equiv (3^3)^p\equiv (26+1)^p\equiv \binom{p}{0}26^p+\binom{p}{1}26^{p-1}....+\binom{p}{p-2}26^2+\binom{p}{p-1}26+\binom{p}{p}\equiv 26p+1\equiv 1\pmod{169}</math>, so <math>26p\equiv 0\pmod{169}</math> and <math>p\equiv 0\pmod{13}</math>. Thus, we have that <math>5|n</math>, <math>3|n</math>, and <math>13|n</math>, so the smallest possible value of <math>n</math> is <math>3\times5\times13=\boxed{195}</math>.
 +
-Stormersyle
 +
 
 +
==Solution 6 (LTE)==
 +
We can see that <math>3^n-1 = 143^2*x</math>, which means that <math>v_{11}(3^n-1) \geq 2</math>, <math>v_{13}(3^n-1) \geq 2</math>. <math>v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})</math>, <math>v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})</math> by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as <math>v_{13}(3^n-1) \geq 2</math>. Therefore the minimum possible value of n is <math>3\times5\times13=\boxed{195}</math>.
 +
 
 +
-bradleyguo
 +
 
 +
==Solution 7==
 +
Note that the problem is basically asking for the least positive integer <math>n</math> such that <math>11^2 \cdot 13^2 | 3^n - 1.</math> It is easy to see that <math>n = \text{lcm } (a, b),</math> where <math>a</math> is the least positive integer satisfying <math>11^2 | 3^a - 1</math> and <math>b</math> the least positive integer satisfying <math>13^2 | 3^b - 1</math>. Luckily, finding <math>a</math> is a relatively trivial task, as one can simply notice that <math>3^5 = 243 \equiv 1 \mod 121</math>. However, finding <math>b</math> is slightly more nontrivial. The order of <math>3^k</math> modulo <math>13</math> (which is <math>3</math>) is trivial to find, as one can either bash out a pattern of remainders upon dividing powers of <math>3</math> by <math>13</math>, or one can notice that <math>3^3 = 27 \equiv 1 \mod 13</math> (the latter which is the definition of period/orders by FLT). We can thus rewrite <math>3^3</math> as <math>(2 \cdot 13 + 1) \mod 13^2</math>. Now suppose that <cmath>3^{3k} \equiv (13n + 1) \mod 13^2. </cmath> I claim that <math>3^{3(k+1)} \equiv (13(n+2) + 1) \mod 13^2. </math>
 +
 
 +
Proof:
 +
To find <math>3^{3(k+1)}, </math> we can simply multiply <math>3^{3k}</math> by <math>3^3, </math> which is congruent to <math>2 \cdot 13 + 1</math> modulo <math>13^2</math>.
 +
By expanding the product out, we obtain <cmath>3^{3(k+1)} \equiv (13n + 1)(2 \cdot 13 + 1) = 13^2 \cdot 2n + 13n + 2 \cdot 13 + 1 \mod 13^2, </cmath> and since the <math>13^2</math> on the LHS cancels out, we're left with <cmath>13n + 2 \cdot 13 + 1 \mod 13^2 \implies 13(n+2) + 1 \mod 13^2</cmath>. Thus, our claim is proven.
 +
Let <math>f(n)</math> be the second to last digit when <math>3^{3k}</math> is written in base <math>13^2</math>.
 +
Using our proof, it is easy to see that <math>f(n)</math> satisfies the recurrence <math>f(1) = 2</math> and <math>f(n+1) = f(n) + 2</math>. Since this implies <math>f(n) = 2n, </math> we just have to find the least positive integer <math>n</math> such that <math>2n</math> is a multiple of <math>13</math>, which is trivially obtained as <math>13</math>.
 +
The least integer <math>n</math> such that <math>3^n - 1</math> is divisible by <math>13^2</math> is <math>3 \cdot 13 = 39, </math> so our final answer is <math>\text{lcm } (5, 39) = \boxed{195}.</math>
 +
 
 +
-fidgetboss_4000
 +
-minor edits made by srisainandan6
 +
 
 +
==Solution 8 (Official MAA)==
 +
The requested positive integer is the least value of <math>n>0</math> such that <math>3^n\equiv 1\pmod{143^2}.</math> Note that <math>143=11\cdot 13.</math> The least power of <math>3</math> that is congruent to <math>1</math> modulo <math>11^2</math> is <math>3^5=243=2\cdot 11^2+1.</math> It follows that. <math>3^n\equiv 1\pmod {11^2}</math> if and only if <math>n=5j</math> for some positive integer <math>j</math>.
 +
 
 +
The least power of <math>3</math> that is congruent to <math>1</math> modulo <math>13</math> is <math>3^3=27=2\cdot 13+1.</math> It follows that <math>3^n\equiv 1\pmod{13}</math> if and only if <math>n=3k</math> for some positive integer <math>k</math>. Additionally, for some positive integer <math>k</math>, the Binomial Theorem shows that <math>3^{3k}=(26+1)^k=26\cdot k+1 \pmod{13^2}</math>. In particular, <math>3^n=3^{3k}\equiv 1\pmod {13^2}</math> if and only if <math>k=13m</math> for some positive integer <math>m</math>, that is, if and only if <math>n=39m.</math>
 +
 
 +
Because <math>11^2</math> and <math>13^2</math> are relatively prime, <math>3^n\equiv 1\pmod {143^2}</math> if and only if <math>3^n\equiv 1\pmod{11^2}</math> and <math>3^n \equiv 1\pmod {13^2}</math>. This occurs if and only if <math>n</math> is a multiple of both of the relatively prime integers <math>5</math> and <math>39</math>, so the least possible value of <math>n</math> is <math>5\cdot 39=\boxed{195}.</math>
 +
 
 +
==Solution 9 (Motivation and LTE)==
 +
 
 +
We first note that we wish to find <math>3^n \equiv 1 \pmod{11^2}</math> and <math>3^n \equiv 1 \pmod{13^2}.</math> Not thinking of anything else, we try a few numbers for the first condition to get that <math>5 \mid n.</math> For the second condition, upon testing up to 729, we find that it doesn't yield a solution readily, so we use Lifting the Exponent from our toolkit to get that
 +
<cmath>v_{13}(27^m-1^m)=v_{13}(26)+v(m)=1+v(m), 3m = n</cmath> which clearly implies <math>m=13</math> and <math>39 | n.</math> Our answer is then obviously <math>39 \cdot 5 = \boxed{195}.</math>
 +
 
 +
~Dhillonr25
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2018|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2018|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 05:30, 21 April 2024

Problem

Find the least positive integer $n$ such that when $3^n$ is written in base $143^2$, its two right-most digits in base $143$ are $01$.

Solution 1

Note that the given condition is equivalent to $3^n \equiv 1 \pmod{143^2}$ and $143=11\cdot 13$. Because $\gcd(11^2, 13^2) = 1$, the desired condition is equivalent to $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$.

If $3^n \equiv 1 \pmod{121}$, one can see the sequence $1, 3, 9, 27, 81, 1, 3, 9...$ so $5|n$.

Now if $3^n \equiv 1 \pmod{169}$, it is harder. But we do observe that $3^3 \equiv 1 \pmod{13}$, therefore $3^3 = 13a + 1$ for some integer $a$. So our goal is to find the first number $p_1$ such that $(13a+1)^ {p_1} \equiv 1 \pmod{169}$. Then, $p_1 \equiv 0 \pmod{13},$ which follows from the binomial theorem. It is not difficult to see that the smallest $p_1=13$, so ultimately $3^{39} \equiv 1 \pmod{169}$. Therefore, $39|n$.

The first $n$ satisfying both criteria is thus $5\cdot 39=\boxed{195}$.

-expiLnCalc

Solution 2

Note that Euler's Totient Theorem would not necessarily lead to the smallest $n$ and that in this case that $n$ is greater than $1000$.

We wish to find the least $n$ such that $3^n \equiv 1 \pmod{143^2}$. This factors as $143^2=11^{2}*13^{2}$. Because $\gcd(121, 169) = 1$, we can simply find the least $n$ such that $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$.

Quick inspection yields $3^5 \equiv 1 \pmod{121}$ and $3^3 \equiv 1 \pmod{13}$. Now we must find the smallest $k$ such that $3^{3k} \equiv 1 \pmod{169}$. Euler's gives $3^{156} \equiv 1 \pmod{169}$. So $3k$ is a factor of $156$. This gives $k=1,2, 4, 13, 26, 52$. Some more inspection yields $k=13$ is the smallest valid $k$. So $3^5 \equiv 1 \pmod{121}$ and $3^{39} \equiv 1 \pmod{169}$. The least $n$ satisfying both is $lcm(5, 39)=\boxed{195}$. (RegularHexagon)

Solution 3 (Big Bash)

Listing out the powers of $3$, modulo $169$ and modulo $121$, we have: \[\begin{array}{c|c|c} n & 3^n\mod{169} & 3^n\mod{121}\\ \hline 0 & 1 & 1\\ 1 & 3 & 3\\ 2 & 9 & 9\\ 3 & 27 & 27\\ 4 & 81 & 81\\ 5 & 74 & 1\\ 6 & 53\\ 7 & 159\\ 8 & 139\\ 9 & 79\\ 10 & 68\\ 11 & 35\\ 12 & 105\\ 13 & 146\\ 14 & 100\\ 15 & 131\\ 16 & 55\\ 17 & 165\\ 18 & 157\\ 19 & 133\\ 20 & 61\\ 21 & 14\\ 22 & 42\\ 23 & 126\\ 24 & 40\\ 25 & 120\\ 26 & 22\\ 27 & 66\\ 28 & 29\\ 29 & 87\\ 30 & 92\\ 31 & 107\\ 32 & 152\\ 33 & 118\\ 34 & 16\\ 35 & 48\\ 36 & 144\\ 37 & 94\\ 38 & 113\\ 39 & 1\\ \end{array}\]

The powers of $3$ repeat in cycles of $5$ and $39$ in modulo $121$ and modulo $169$, respectively. The answer is $\text{lcm}(5, 39) = \boxed{195}$.

~Minor edit by Yiyj1

Solution 4 (Order+Bash)

We have that \[3^n \equiv 1 \pmod{143^2}.\]Now, $3^{110} \equiv 1 \pmod{11^2}$ so by the Fundamental Theorem of Orders, $\text{ord}_{11^2}(3)|110$ and with some bashing, we get that it is $5$. Similarly, we get that $\text{ord}_{13^2}(3)=39$. Now, $\text{lcm}(39,5)=\boxed{195}$ which is our desired solution.

Solution 5 (Easy Binomial Theorem)

We wish to find the smallest $n$ such that $3^n\equiv 1\pmod{143^2}$, so we want $n\equiv 1\pmod{121}$ and $n\equiv 1\pmod{169}$. Note that $3^5\equiv 1\pmod{121}$, so $3^n$ repeats $121$ with a period of $5$, so $5|n$. Now, in order for $n\equiv 1\pmod{169}$, then $n\equiv 1\pmod{13}$. Because $3^3\equiv 1\pmod{13}$, $3^n$ repeats with a period of $3$, so $3|n$. Hence, we have that for some positive integer $p$, $3^n\equiv (3^3)^p\equiv (26+1)^p\equiv \binom{p}{0}26^p+\binom{p}{1}26^{p-1}....+\binom{p}{p-2}26^2+\binom{p}{p-1}26+\binom{p}{p}\equiv 26p+1\equiv 1\pmod{169}$, so $26p\equiv 0\pmod{169}$ and $p\equiv 0\pmod{13}$. Thus, we have that $5|n$, $3|n$, and $13|n$, so the smallest possible value of $n$ is $3\times5\times13=\boxed{195}$. -Stormersyle

Solution 6 (LTE)

We can see that $3^n-1 = 143^2*x$, which means that $v_{11}(3^n-1) \geq 2$, $v_{13}(3^n-1) \geq 2$. $v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})$, $v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})$ by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as $v_{13}(3^n-1) \geq 2$. Therefore the minimum possible value of n is $3\times5\times13=\boxed{195}$.

-bradleyguo

Solution 7

Note that the problem is basically asking for the least positive integer $n$ such that $11^2 \cdot 13^2 | 3^n - 1.$ It is easy to see that $n = \text{lcm } (a, b),$ where $a$ is the least positive integer satisfying $11^2 | 3^a - 1$ and $b$ the least positive integer satisfying $13^2 | 3^b - 1$. Luckily, finding $a$ is a relatively trivial task, as one can simply notice that $3^5 = 243 \equiv 1 \mod 121$. However, finding $b$ is slightly more nontrivial. The order of $3^k$ modulo $13$ (which is $3$) is trivial to find, as one can either bash out a pattern of remainders upon dividing powers of $3$ by $13$, or one can notice that $3^3 = 27 \equiv 1 \mod 13$ (the latter which is the definition of period/orders by FLT). We can thus rewrite $3^3$ as $(2 \cdot 13 + 1) \mod 13^2$. Now suppose that \[3^{3k} \equiv (13n + 1) \mod 13^2.\] I claim that $3^{3(k+1)} \equiv (13(n+2) + 1) \mod 13^2.$

Proof: To find $3^{3(k+1)},$ we can simply multiply $3^{3k}$ by $3^3,$ which is congruent to $2 \cdot 13 + 1$ modulo $13^2$. By expanding the product out, we obtain \[3^{3(k+1)} \equiv (13n + 1)(2 \cdot 13 + 1) = 13^2 \cdot 2n + 13n + 2 \cdot 13 + 1 \mod 13^2,\] and since the $13^2$ on the LHS cancels out, we're left with \[13n + 2 \cdot 13 + 1 \mod 13^2 \implies 13(n+2) + 1 \mod 13^2\]. Thus, our claim is proven. Let $f(n)$ be the second to last digit when $3^{3k}$ is written in base $13^2$. Using our proof, it is easy to see that $f(n)$ satisfies the recurrence $f(1) = 2$ and $f(n+1) = f(n) + 2$. Since this implies $f(n) = 2n,$ we just have to find the least positive integer $n$ such that $2n$ is a multiple of $13$, which is trivially obtained as $13$. The least integer $n$ such that $3^n - 1$ is divisible by $13^2$ is $3 \cdot 13 = 39,$ so our final answer is $\text{lcm } (5, 39) = \boxed{195}.$

-fidgetboss_4000 -minor edits made by srisainandan6

Solution 8 (Official MAA)

The requested positive integer is the least value of $n>0$ such that $3^n\equiv 1\pmod{143^2}.$ Note that $143=11\cdot 13.$ The least power of $3$ that is congruent to $1$ modulo $11^2$ is $3^5=243=2\cdot 11^2+1.$ It follows that. $3^n\equiv 1\pmod {11^2}$ if and only if $n=5j$ for some positive integer $j$.

The least power of $3$ that is congruent to $1$ modulo $13$ is $3^3=27=2\cdot 13+1.$ It follows that $3^n\equiv 1\pmod{13}$ if and only if $n=3k$ for some positive integer $k$. Additionally, for some positive integer $k$, the Binomial Theorem shows that $3^{3k}=(26+1)^k=26\cdot k+1 \pmod{13^2}$. In particular, $3^n=3^{3k}\equiv 1\pmod {13^2}$ if and only if $k=13m$ for some positive integer $m$, that is, if and only if $n=39m.$

Because $11^2$ and $13^2$ are relatively prime, $3^n\equiv 1\pmod {143^2}$ if and only if $3^n\equiv 1\pmod{11^2}$ and $3^n \equiv 1\pmod {13^2}$. This occurs if and only if $n$ is a multiple of both of the relatively prime integers $5$ and $39$, so the least possible value of $n$ is $5\cdot 39=\boxed{195}.$

Solution 9 (Motivation and LTE)

We first note that we wish to find $3^n \equiv 1 \pmod{11^2}$ and $3^n \equiv 1 \pmod{13^2}.$ Not thinking of anything else, we try a few numbers for the first condition to get that $5 \mid n.$ For the second condition, upon testing up to 729, we find that it doesn't yield a solution readily, so we use Lifting the Exponent from our toolkit to get that \[v_{13}(27^m-1^m)=v_{13}(26)+v(m)=1+v(m), 3m = n\] which clearly implies $m=13$ and $39 | n.$ Our answer is then obviously $39 \cdot 5 = \boxed{195}.$

~Dhillonr25

See Also

2018 AIME I (Problems • Answer Key • Resources)
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Problem 12
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