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"

m (→‎Modular Arithmetic Solution- Strange (MASS))
m (→‎Modular Arithmetic Solution- Strange (MASS))
Line 4: Line 4:
  
 
==Modular Arithmetic Solution- Strange (MASS)==
 
==Modular Arithmetic Solution- Strange (MASS)==
−
Note that <math>3^n \cong 1 (\mod 143^2)</math>. And <math>143=11*13</math>. Because <math>gcd(11^2, 13^2) = 1</math>, <math>3^n \cong 1 (\mod 121 = 11^2)</math> and <math>3^n \cong 1 (mod 169=13^2)</math>.
+
Note that <math>3^n \equiv 1 (\mod 143^2)</math>. And <math>143=11*13</math>. Because <math>gcd(11^2, 13^2) = 1</math>, <math>3^n \equiv 1 (\mod 121 = 11^2)</math> and <math>3^n \equiv 1 (mod 169=13^2)</math>.
  
−
If <math>3^n \cong 1 (\mod 121)</math>, one can see the sequence <math>1, 3, 9, 27, 81, 1, 3, 9...</math> so <math>n \mid 5</math>.
+
If <math>3^n \equiv 1 (\mod 121)</math>, one can see the sequence <math>1, 3, 9, 27, 81, 1, 3, 9...</math> so <math>n \mid 5</math>.
  
−
Now if <math>3^n \cong 1 (\mod 169)</math>, it is harder. But we do observe that <math>3^3 \cong 1 (\mod 13)</math>, therefore <math>3^3 = 13a + 1</math> for some integer <math>a</math>. So our goal is to find the first number <math>p_1</math> such that <math>(13a+1)^ {p_1} \cong 1 (\mod 169)</math>. In other words, the <math>a</math> coefficient must be <math>0 (\mod 169)</math>. It is not difficult to see that this first <math>p_1=13</math>, so ultimately <math>3^39 \cong 1 (\mod 169)</math>. Therefore, <math>n \mid 39</math>.
+
Now if <math>3^n \equiv 1 (\mod 169)</math>, it is harder. But we do observe that <math>3^3 \equiv 1 (\mod 13)</math>, therefore <math>3^3 = 13a + 1</math> for some integer <math>a</math>. So our goal is to find the first number <math>p_1</math> such that <math>(13a+1)^ {p_1} \equiv 1 (\mod 169)</math>. In other words, the <math>a</math> coefficient must be <math>0 (\mod 169)</math>. It is not difficult to see that this first <math>p_1=13</math>, so ultimately <math>3^39 \equiv 1 (\mod 169)</math>. Therefore, <math>n \mid 39</math>.
  
 
The first <math>n</math> satisfying both criteria is <math>5*39=\boxed{195}</math>.
 
The first <math>n</math> satisfying both criteria is <math>5*39=\boxed{195}</math>.
  
 
-expiLnCalc
 
-expiLnCalc

Revision as of 17:45, 7 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 $n \mid 5$.

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, $n \mid 39$.

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

-expiLnCalc

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_11&oldid=92891"