Question 1067319
While the middle digit could be any of the 10 digits, including 0,
he first and last digits cannot be zero.
They must be one of the {{{9}}} choices from 1 though 9.
So, there are {{{9*10=90}}} 3-digit palindromes,
For a number to be divisible by 3 , the sum of the digits must be divisible by 3.
Of the 9 choices for first (and last) digit, 3 are multiples of 3,
another 3 have a remainder of 1 when divided by 3,
and the other 3 have a remainder of 2.
For any middle digit, there will be {{{3}}} out of the {{{9}}} choices of first and last digits that will make the sum of digits a multiple of 3.
If the middle digit is has a remainder of 0 when divided by 3 (as happens for 0, 3, or 9),
a first/last digit that is a multiple of 3 makes the sum a multiple of 3.
If you the middle digit has a remainder of 1 when divided by 3,
a first/last digit with a reminder of 1,
will make the 3 remainders add up to 3,
for a sum that is a multiple of 3.
If you the middle digit has a remainder of 2 when divided by 3,
a first/last digit with a reminder of 2,
will make the 3 remainders add up to 6,
for a sum that is a multiple of 3.
Of the {{{90}}} palindromes, {{{1/3}}} or {{{90/3=highlight(30)}}}
will be divisible by 3.