2020 AIME I Problems/Problem 12

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 1

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 $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4n}-2^{4n})=1+v_5(n)$ 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 $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$ .

~kevinmathz

Solution 2 (Simpler, just basic mods and Fermat's theorem)

BY THE WAY, please feel free to correct my formatting. I don't know latex.

Note that for all $n$ , $149^n - 2^n$ is divisible by $149-2 = 147$ because that is a factor. That is $3\cdot7^2$ , so now we can clearly see that the smallest $n$ to make the expression divisible by $3^3$ is just $3^2$ . Similarly, we can reason that the smallest $n$ to make the expression divisible by $7^7$ is just $7^5$ .

Finally, for $5^5$ , take $(\text{mod} 5)$ and $(\text{mod} 25)$ of each quantity (They happen to both be $-1$ and $2$ respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum $n$ for divisibility by $5$ is $4$ , and other values are factors of $4$ . Testing all of them(just $1$ , $2$ , $4$ using mods-not too bad), $4$ is indeed the smallest value to make the expression divisible by $5$ , and this clearly is NOT divisible by $25$ . Therefore, the smallest $n$ to make this expression divisible by $5^5$ is $2^2 \cdot 5^4$ .

Calculating the LCM of all these, one gets $2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$ . Using the factor counting formula, the answer is $3\cdot3\cdot5\cdot6$ = $\boxed{270}$ . Can someone please format better for me?

-Solution by thanosaops -formatted by MY-2

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

Page

Toolbox

Search

2020 AIME I Problems/Problem 12

Contents

Problem

Solution 1

Solution 2 (Simpler, just basic mods and Fermat's theorem)

See Also