60.3. What is the remainder when the sum
$$
1^{5}+2^{5}+3^{5}+\cdots+2007^{5}
$$
is divided by 5 ?
60.3. 3
By Fermat's Little Theorem, we have $a^{5} \equiv a(\bmod 5)$ for any integer a. Modulo 5, we have
$$
1^{5}+2^{5}+3^{5}+\cdots+2007^{5} \equiv 1+2+3+\cdots+2007=2007 \cdot 1004 \equiv 2 \cdot 4 \equiv 3 \text {. }
$$

Thus, the desired remainder is 3 .