2018 AIME I Problems/Problem 11

Problem

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

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.

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 $3^{3(k+1)} \equiv (13(n+2) + 1) \mod 13^2.$ [b]Proof:[/b] 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

2018 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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