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

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== Solution ==
 
== Solution ==
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Lifting the Exponent shows that <math>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</math> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <math>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</math> so thus, <math>7^5</math> divides <math>n</math>.  
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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>.  
  
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Now, multiplying <math>n</math> by <math>4</math>, we see v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})<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>.  
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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>.  
  
 
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}$.

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$ (Error compiling LaTeX. Unknown error_msg)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

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