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

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We wish to find the least <math>n</math> such that <math>3^n \equiv 1(\mod 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(\mod 121)</math> and <math>3^n \equiv 1(\mod 169)</math>.
 
We wish to find the least <math>n</math> such that <math>3^n \equiv 1(\mod 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(\mod 121)</math> and <math>3^n \equiv 1(\mod 169)</math>.
  
Quick inspection yields <math>3^5 \equiv 1(\mod 121)</math> and <math>3^3 \equiv 1(\mod 13)</math>. Now we must find the smallest <math>k</math> such that <math>3^3k \equiv 1(\mod 13)</math>. Euler's gives <math>3^156 \equiv 1(\mod 169)</math>. So <math>3k</math> is a factor of <math>156</math>. This gives <math>k=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(\mod 121)</math> and <math>3^39 \equiv 1(\mod 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(\mod 121)</math> and <math>3^3 \equiv 1(\mod 13)</math>. Now we must find the smallest <math>k</math> such that <math>3^{3k} \equiv 1(\mod 13)</math>. Euler's gives <math>3^{156} \equiv 1(\mod 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(\mod 121)</math> and <math>3^{39} \equiv 1(\mod 169)</math>. The least <math>n</math> satisfying both is <math>lcm[5,39]=\boxed{195}</math>. (RegularHexagon)
  
 
==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}}

Revision as of 10:59, 8 March 2018

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

Solutions

Modular Arithmetic Solution- Strange (MASS)

Note that $3^n \equiv 1 (\mod 143^2)$. And $143=11*13$. Because $gcd(11^2, 13^2) = 1$, $3^n \equiv 1 (\mod 121 = 11^2)$ and $3^n \equiv 1 (mod 169=13^2)$.

If $3^n \equiv 1 (\mod 121)$, one can see the sequence $1, 3, 9, 27, 81, 1, 3, 9...$ so $5 \mid n$.

Now if $3^n \equiv 1 (\mod 169)$, it is harder. But we do observe that $3^3 \equiv 1 (\mod 13)$, therefore $3^3 = 13a + 1$ for some integer $a$. So our goal is to find the first number $p_1$ such that $(13a+1)^ {p_1} \equiv 1 (\mod 169)$. In other words, the $a$ coefficient must be $0 (\mod 169)$. It is not difficult to see that this first $p_1=13$, so ultimately $3^{39} \equiv 1 (\mod 169)$. Therefore, $39 \mid n$.

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

-expiLnCalc

Solution

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

Quick inspection yields $3^5 \equiv 1(\mod 121)$ and $3^3 \equiv 1(\mod 13)$. Now we must find the smallest $k$ such that $3^{3k} \equiv 1(\mod 13)$. Euler's gives $3^{156} \equiv 1(\mod 169)$. 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 $3^5 \equiv 1(\mod 121)$ and $3^{39} \equiv 1(\mod 169)$. The least $n$ satisfying both is $lcm[5,39]=\boxed{195}$. (RegularHexagon)

See Also

2018 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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