Problem 379

Solution 1

The are 10 palindromes between 100 and 200: $101, 111, 121, 131, 141, 151, 161, 171, 181$, and 191.

The are 10 palindromes between 200 and 300: $202,212,222,232,242,252,262,272,282$, and 292.

Similary, there are 10 palindromes between each of 300 and 400, 400 and 500, 500 and 600, 600 and 700, 700 and 800, 800 and 900, 900 and 1000.

That is, there are 10 palindromes between each of the 9 pairs of consecutive multiples of 100 from 100 to 1000.

The number of palindromes between 100 and 1000 is $10\times9=90$.

Solution 2

Each palindrome between 100 and 1000 is a 3-digit number of the form $aba$, where $a$ is a digit from 1 to 9 inclusive and $b$ is a digit from 0 to 9 inclusive, and $a$ and $b$ are not necessarily different digits.

Thus, there are 9 choices for the first digit $a$ and each of those choices determines the third digit which must also be $a$.

There are 10 choices for the second digit $b$, and so there are $9\times10=90$ choices for $a$ and $b$.

Thus, there are 90 palindromes between 100 and 1000.