Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 6(LTE))
m (added a solution for this problem)
Line 83: Line 83:
  
 
-bradleyguo
 
-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>
 +
[b]Proof:[/b]
 +
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 <math>13n + 2 \cdot 13 + 1 \mod 13^2 \implies </math>13(n+2) + 1 \mod 13^2<math>. 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}.$
 +
 +
-fidgetboss_4000
  
 
==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 10:58, 10 April 2020

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

See Also

2018 AIME I (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_11&oldid=120781"