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

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 $01$.

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)

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