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 $n.$
As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$.
[Lifting the Exponent](https://artofproblemsolving.com/wiki/index.php/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 \( p \mid x-y \).
Obviously, \( 149^n \equiv 2^n \pmod{3} \) implies \( 149^n - 2^n \equiv 0 \pmod{3} \), so LTE works here.
Furthermore, \( 149^n \equiv 2^n \pmod{7} \) implies \( 149^n - 2^n \equiv 0 \pmod{7} \), so LTE works here.
However, when we get to the case of 5, we see that \( 149^n \equiv 2^n \pmod{5} \) 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