10. $1^{1}+2^{2}+3^{3}+4^{4}+5^{5}+6^{6}+7^{7}+8^{8}+9^{9}$ divided by 3 has a remainder of what? Why?
10.【Solution】$3^{3}, 6^{6}, 9^{9}$ when divided by 3, the remainder is 0, so we only need to look at the expression $1^{1}+2^{2}+4^{4}+5^{5}+7^{7}+8^{8}$ modulo 3.

Note: If $a$ divided by 3 leaves a remainder of $a_{1}$, and $b$ divided by 3 leaves a remainder of $b_{1}$, then $a \times b$ divided by 3 will have the same remainder as $a_{1} \times b_{1}$ divided by 3.

Since $4, 7$ divided by 3 leave a remainder of 1, $4^{4}, 7^{7}$ divided by 3 will also leave a remainder of 1.
Since $5, 8$ divided by 3 leave a remainder of 2, $5^{5}, 8^{8}$ divided by 3 will have the same remainder as $2^{5}, 2^{8}$ divided by 3. And $2^{4}=16$ divided by 3 leaves a remainder of 1, so $2^{5}=2^{4} \times 2$ divided by 3 leaves a remainder of 2, and $2^{8}=2^{4} \times 2^{4}$ divided by 3 leaves a remainder of 1 (since $1 \times 1 = 1$).

Thus, $1^{1}+2^{2}+4^{4}+5^{5}+7^{7}+8^{8}$ divided by 3 will have the same remainder as $1+4+1+2+1+1$ divided by 3, which is a remainder of 1.