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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2020 AIME I Problems/Problem 12"

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
Lifting the Exponent shows that <cmath>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</cmath> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <cmath>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</cmath> so thus, <math>7^5</math> divides <math>n</math>.  
 
Lifting the Exponent shows that <cmath>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</cmath> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <cmath>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</cmath> so thus, <math>7^5</math> divides <math>n</math>.  
  
−
Now, multiplying <math>n</math> by <math>4</math>, we see <math>v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})</math><math> and since </math>145^{4} \equiv 1 \pmod{25}<math> and </math>16^1 \equiv 16 \pmod{25}<math> then </math>v_5(149^{4n}-2^{4n})=1<math> meaning that we have that by LTE, </math>4 \cdot 5^4<math> divides </math>n<math>.  
+
Now, multiplying <math>n</math> by <math>4</math>, we see <cmath>v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})</cmath> and since <math>145^{4} \equiv 1 \pmod{25}</math> and <math>16^1 \equiv 16 \pmod{25}</math> then <math>v_5(149^{4n}-2^{4n})=1</math> meaning that we have that by LTE, <math>4 \cdot 5^4</math> divides <math>n</math>.  
  
−
Since </math>3^2<math>, </math>7^5<math> and </math>4\cdot 5^4<math> all divide </math>n<math>, the smallest value of </math>n<math> working is their LCM, also </math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5<math>. Thus the number of divisors is </math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.
+
Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:14, 12 March 2020

Note: Please do not post problems here until after the AIME.

Problem

Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$

Solution

Lifting the Exponent shows that \[v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\] so thus, $3^2$ divides $n$. It also shows that \[v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\] so thus, $7^5$ divides $n$.

Now, multiplying $n$ by $4$, we see \[v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})\] and since $145^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4n}-2^{4n})=1$ meaning that we have that by LTE, $4 \cdot 5^4$ divides $n$.

Since $3^2$, $7^5$ and $4\cdot 5^4$ all divide $n$, the smallest value of $n$ working is their LCM, also $3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.

See Also

2020 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_12&oldid=119175"