A three-digit palindrome reads the same forwards and backwards, like 353 or 909. How many three-digit palindromes are multiples of 3?

A palindrome "$aba$" has digit sum $a+b+a=2a+b$, and it is a multiple of 3 exactly when $2a+b$ is. Since $2a$ leaves the same remainder as −a, the condition is simply that $b$ and a leave the SAME remainder on division by 3. If a is 3, 6 or 9 then $b$ may be 0, 3, 6 or 9: that is $3\times 4=12$ palindromes. If a is 1, 4 or 7 then $b$ must be 1, 4 or 7: $3\times 3=9$. If a is 2, 5 or 8 then $b$ must be 2, 5 or 8: another 9. Total 12 + 9 + 9 = 30. The trap is 90 — that is ALL the three-digit palindromes, and exactly one third of them survive.