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 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 $(13a+1)^ {p_1} \equiv 1 \pmod{169}$. In other words, the $p_1 \equiv 0 \pmod{13}$. 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 $3^{3k} \equiv 1 \pmod{13}$. 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 (BigBash)

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 $5$ an $39$ in modulo $121$ and modulo $169$, respectively. The answer is $lcm(5, 39) = \boxed{195}$.

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.