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: 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 $(2 + 1)(2 + 1)(4 + 1)(5 + 1) = \boxed{270}$

Answer:  270