Answer:
30
Step-by-step explanation:
zero cannot be the first digit of the palindromes. There are nine other numbers that can be.
9 * 10
The last digit also cannot be zero so it has the same value as the first digit. (Otherwise the number cannot be palindromic) So you are really looking at only 90 numbers total.
That's not the question, but it does put a limit on how many there can be.
Let's find the first number (closest to 100 ) that is palindromic and divisible by 3.
I think 111 is the smallest.
Now what about the largest? That would be 999.
So what is the next number after 111? Wouldn't that be 141?
The difference between 111 and 141 is 30
It turns out that any first digit that is divisible by 3, has 4 palindromic numbers because for them, the second digit is a 0. So for 3 for example, you get
303
333
363
393
All the other numbers only have 3 entries for example, 1 gives you
111
141
171
So making a table we get
First digit Number of palindromes divisible by 3
1 3
2 3
3 4
4 3
5 3
6 4
7 3
8 3
9 4
18 12
Total
