jmc

Note that $n$ is even, since the $LHS$ consists of two odd and two even numbers. By Fermat's Little Theorem, we know $n^5$ is congruent to $n$ modulo 5. Hence, $3+0+4+2\equiv n(\mathrm{mod}5)$ $4\equiv n(\mathrm{mod}5)$ Continuing, we examine the equation modulo 3, $1-1+0+0\equiv n(\mathrm{mod}3)$ $0\equiv n(\mathrm{mod}3)$ Thus, $n$ is divisible by three and leaves a remainder of four when divided by 5. It's obvious that $n>133$, so the only possibilities are $n=144$ or $n\ge 174$. It quickly becomes apparent that 174 is much too large, so $n$ must be $\boxed{144}$.