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Palindromes and elevens

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A palindrome reads the same forwards and backwards. How many palindromes with $5$ digits and no leading zero, such as $70407$ , are divisible by $11$ ?

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Hint

Write the palindrome as $abcba$ and reduce its value modulo $11$ to a condition on $a$ , $b$ and $c$ .

Solution

The palindrome $abcba$ equals $10001 a + 1010 b + 100 c$ , and modulo $11$ this is $2 a + 9 b + c$ .
For each of the $9 \times 10 = 90$ choices of the digits $a$ and $b$ , exactly one residue of $c$ modulo $11$ makes the sum $0$ , and it is a usable digit unless that residue is $10$ .
The residue $10$ needs $2 a + 9 b \equiv 1$ , that is $b \equiv 5 (1 - 2 a)$ modulo $11$ ; for $a = 5$ this gives $b = 10$ , which is not a digit, while each of the other $8$ values of $a$ gives exactly one digit $b$ .
So $8$ pairs are unusable and the count is $90 - 8 = 82$ .

Answer: $82$

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