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

As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$. Lifting the Exponent shows that \[3=v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\] so thus, $3^2$ divides $n$. It also shows that \[7=v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\] so thus, $7^5$ divides $n$.


Now, setting $n = 4c$ (necessitated by $149^n \equiv 2^n \pmod 5$ in order to set up LTE), we see \[v_5(149^{4c}-2^{4c}) = v_5(149^{4c}-16^{c})\] and since $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4c}-2^{4c})=v_5(149^4-16)+v_5(c)=1+v_5(c)$ meaning that we have that by LTE, $5^4 | c$ and $4 \cdot 5^4$ 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

clarified by another user

notation note from another user

Note

We were able to use LTE with 3 and 7 but not 5 because in order to use LTE, we need pxy.

Obviously, 149n2n(mod3) implies 149n2n0(mod3), so LTE works here.

Furthermore, 149n2n(mod7) implies 149n2n0(mod7), so LTE works here.

However, when we get to the case of 5, we see that 149n2n(mod5) doesn't always hold; specifically, this is only valid when n is a multiple of 4, which is why we let n=4c in the solution.

mathboy282

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

Note that for all $n$, $149^n - 2^n$ is divisible by $149-2 = 147$ by difference of $n$th powers. 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 $\pmod {5}$ and $\pmod {25}$ 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 $3\cdot3\cdot5\cdot6$ = $\boxed{270}$.

~Solution by thanosaops

~formatted by MY-2 and pandyhu2001

Solution 3 (Elementary and Thorough)

As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$. For divisibility by $3^3$, notice that $v_3(149^3 - 2^3) = 2$ as $149^3 - 2^3 =$ $(147)(149^2 + 2\cdot149 + 2^2)$, and upon checking mods, $149^2 + 2\cdot149 + 2^2$ is divisible by $3$ but not $9$. In addition, $149^9 - 2^9$ is divisible by $3^3$ because $149^9 - 2^9 = (149^3 - 2^3)(149^6 + 149^3\cdot2^3 + 2^6)$, and the rightmost factor equates to $1 + 1 + 1 \pmod{3} \equiv 0 \pmod{3}$. In fact, $n = 9 = 3^2$ is the least possible choice to ensure divisibility by $3^3$ because if $n = a \cdot 3^b$, with $3 \nmid a$ and $b < 2$, we write \[149^{a \cdot 3^b} - 2^{a \cdot 3^b} =  (149^{3^b} - 2^{3^b})(149^{3^b(a - 1)} + 149^{3^b(a - 2)}\cdot2^{3^b}+\cdots2^{3^b(a - 1)}).\] Then, the rightmost factor is equivalent to $\pm a \pmod{3} \not\equiv 0 \pmod{3}$, and $v_3(149^{3^b} - 2^{3^b}) = b + 1 < 3$.

For divisibility by $7^7$, we'll induct, claiming that $v_7(149^{7^k} - 2^{7^k}) = k + 2$ for whole numbers $k$. The base case is clear. Then, \[v_7(149^{7^{k+1}} - 2^{7^{k+1}}) = v_7(149^{7^k} - 2^{7^k}) + v_7(149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k}).\] By the induction hypothesis, $v_7(149^{7^k} - 2^{7^k}) = k + 2$. Then, notice that \[S(k) = 149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{7} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{49}.\] This tells us that $S(k)$ is divisible by $7$, but not $49$ so that $v_7\left(S(k)\right) = 1$, completing our induction. We can verify that $7^5$ is the least choice of $n$ to ensure divisibility by $7^7$ by arguing similarly to the $3^3$ case.

Finally, for $5^5$, we take the powers of $149$ and $2$ in mod $5$ and mod $25$. Writing out these mods, we have that $149^n \equiv 2^n \pmod{5}$ if and only if $4 | n$, in which $149^n \equiv 2^n \equiv 1 \pmod{5}$. So here we claim that $v_5(149^{4\cdot5^k} - 2^{4\cdot5^k}) = k + 1$ and perform yet another induction. The base case is true: $5 | 149^4 - 2^4$, but $149^4 - 2^4 \equiv 1 - 16 \pmod{25}$. Now then, assuming the induction statement to hold for some $k$, \[v_5(149^{4\cdot5^{k+1}} - 2^{4\cdot5^{k+1}}) = (k+1) + v_5(149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}).\] Note that $S'(k) =  149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}$ equates to $S''(k) = 1 + 2^{4\cdot5^k} + \cdots + 2^{16\cdot5^k}$ in both mod $5$ and mod $25$. We notice that $S''(k) \equiv 0 \pmod{5}$. Writing out the powers of $2$ mod $25$, we have $S''(0) \equiv 5 \pmod{25}$. Also $2^n \equiv 1 \pmod{25}$ when $n$ is a multiple of $20$. Hence for $k > 0$, $S''(k) \equiv 5 \mod{25}$. Thus, $v_5\left(S'(k)\right) = 1$, completing our induction. Applying the same argument from the previous two cases, $4\cdot5^4$ is the least choice to ensure divisibility by $5^5$.

Our answer is the number of divisors of $\text{lcm}(3^2, 7^5, 2^2\cdot5^4) = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. It is $(2 + 1)(2 + 1)(4 + 1)(5 + 1) = \boxed{270}$.

~hnkevin42

Solution 4 (Official MAA)

Analyze each prime power separately. Start with the case of $3^3$. By the Binomial Theorem, \begin{align*} 		149^n - 2^n &= (147+2)^n - 2^n \\ 		&= \binom n1 \cdot 147 \cdot 2^{n-1} 		+ \binom n2 \cdot 147^2 \cdot 2^{n-2}\\ 		&\qquad+ \binom n3 \cdot 147^3 \cdot 2^{n-3} 		+ \cdots. 	\end{align*}Because $147$ is divisible by $3$, all terms after the first two are divisible by $3^3$, and the exponent of $3$ in the first term is less than that in the second term. Hence it is necessary and sufficient that $3^3 \mid 147n$, that is, $3^2 \mid n$. For the $7^7$ case, consider the same expansion as in the previous case. Because $147$ is divisible by $49 = 7^2$, all terms after the first three are divisible by $7^7$, and the exponent of $7$ in the first term is less than that in the second and third term. Hence it is necessary and sufficient that $7^7 \mid 147n$, that is, $7^5 \mid n$. For the $5^5$ case, working modulo $5$ gives $149^n - 2^n \equiv 4^n - 2^n = 2^n(2^n-1) \pmod 5$, so it must be that $4 \mid n$. Let $n = 4m$, and let $c = 149^4 - 2^4 = (149^2-2^2)(149^2+2^2) = 147 \cdot 151 \cdot 22205$. Note that $\frac c5$ is an integer not divisible by $5$. Expand by the Binomial Theorem again to get \begin{align*} 		(149^4)^m - (2^4)^m &= (c+16)^m - (16)^m \\ 		&= \binom m1 \cdot c \cdot 16^{m-1} 		+ \binom m2 \cdot c^2 \cdot 16^{m-2} \\ 		&\qquad+ \binom m3 \cdot c^3 \cdot 16^{m-3} 		+ \binom m4 \cdot c^4 \cdot 16^{m-4} 		+ \cdots. 	\end{align*}All terms after the first four are divisible by $5^5$, and the exponent of $5$ in the first term is less than that in the second, third, or fourth term. Hence it is necessary and sufficient that $5^5 \mid cm$. Thus $5^4 \mid m$, and it follows that $4 \cdot 5^4 \mid n$. Therefore the least $n$ is $3^2 \cdot (2^2 \cdot 5^4) \cdot 7^5$. The requested number of divisors is $(1+2)(1+2)(1+4)(1+5) = 270$.

The results of the above cases can be generalized using the following lemma.

Lifting the Exponent Lemma: Let $p$ be an odd prime, and let $a$ and $b$ be integers relatively prime to $p$ such that $p \mid (a-b)$. Let $n$ be a positive integer. Then the number of factors of $p$ that divide $a^n - b^n$ is equal to the number of factors of $p$ that divide $a-b$ plus the number of factors of $p$ that divide $n$.

Video Solution

https://www.youtube.com/watch?v=O0BprEOVkjo ~ Math Gold Medalist

See Also

2020 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png