555 wkwkwk

By the Chinese Remainder Theorem, we require the expression to be divisible by each of 3, 5, and 37. Taking the expression modulo each of these gives the following:

$\begin{aligned} 37^n + 97^nx &\equiv 0 \! \! \! \! \pmod{37} \\ 23^nx &\equiv 0 \! \! \! \! \pmod{37} \\ \\ 37^n + 97^nx &\equiv 0 \! \! \! \! \pmod{5} \\ 2^n + 2^nx &\equiv 0 \! \! \! \! \pmod{5} \\ 2^n(1 + x) &\equiv 0 \! \! \! \! \pmod{5} \\ \\ 37^n + 97^nx &\equiv 0 \! \! \! \! \pmod{3} \\ 1 + x &\equiv 0 \! \! \! \! \pmod{3}. \end{aligned}$

For the first expression, we can divide both sides by $23^n$ and get $x \equiv 0 \! \pmod{37},$ since any power of 23 will be relatively prime to 37. For the second expression, we can divide both sides by $2^n$ and get $x + 1 \equiv 0 \! \pmod{5},$ since a power of two cannot divide five. Therefore, we seek the smallest positive integer $x$ such that

$\begin{aligned} x &\equiv 0 \! \! \! \! \pmod{37} \\ x &\equiv -1 \! \! \! \! \pmod{5} \\ x &\equiv -1 \! \! \! \! \pmod{3}. \end{aligned}$

Quick trial-and-error calculations with multiples of 37 yields $x = \boxed{74}$ as the desired number.

Since $103$ and $555$ are coprime and $97 \times 103 \equiv 1 \pmod{555}$ , and since $37 \times 103 \equiv 481 \pmod{555}$ , multiplying the original equation by $103^n$ means that we need to find the smallest positive integer $x$ such that $481^n + x \equiv 0 \pmod{555}$ for all positive integers $n$ . Since $555$ divides $481 \times 480$ we deduce that $481^2 \equiv 481 \pmod{555}$ , and hence $481^n \equiv 481 \pmod{555}$ for all positive integers $n$ . Thus we simply need to find the smallest positive integer $x$ such that $481 + x \equiv 0 \pmod{555}$ , which is $\boxed{74}$ .