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 $v_5(149^{4n}-2^{4n})=1$ 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 (For people who don't know like crazy stuff)

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*7^2, 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 mod (5) and 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 * 5^4.

Calculating the LCM of all these, one gets 2^2 * 3^2 * 5^4 * 7^5. Using the factor counting formula, the answer is 3*3*5*6 = $\boxed{270}$ . Can someone please format better for me?

-Solution by thanosaops

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2020 AIME I Problems/Problem 12

Contents

Problem

Solution 1

Solution 2 (For people who don't know like crazy stuff)

See Also