155 If $a, b \in \mathbf{N}, 1 \leqslant a<b \leqslant 100, a^{2}+b^{2}+a b$ can be divided by 9, then the number of such positive integer pairs $a, b$ is
A. 508
B. 528
C. 1056
D. 1089
155 B. If $a, b$ are both divisible by 3, then $a^{2} \mid a b+b^{2}$ is divisible by 9. There are $C_{33}^{2}=$ 528 such pairs of $a, b$.
If $a, b$ when divided by 3, yield different remainders, then $a^{2}+a b+b^{2}$ is not divisible by 3, and thus not by 9.
Let $a \equiv b \equiv 1(\bmod 3)$, then $a \equiv 1,4,7(\bmod 9), b \equiv 1,4,7(\bmod 9) . a, b$ when divided by 9, yield remainders in the combinations $(1,1),(4,4),(7,7),(1,4),(1,7),(4,7)$, 6 in total, all of which have $a^{2}+a b+b^{2} \equiv 3(\bmod 9)$. Therefore, only the 528 pairs mentioned earlier satisfy the requirement.