, 2022

A palindrome greater than 10 000 and less than 100 000 is a
5-digit positive integer of the form $abcba$, where $a, b$ and $c$ are digits and $a\neq0$.

A positive integer is a multiple of 18 if it is a multiple of both 2 and
9 (and a positive integer that is a multiple of both 2 and 9 is a
multiple of 18).

A positive integer is a multiple of 2 if it is even, and thus the digit
$a$ is equal to $2,4,6$ or 8 (recall $a\neq0$).

A positive integer is a multiple of 9 exactly when the sum of its digits
is a multiple of 9, and thus $a+b+c+b+a$ or $2a+2b+c$ is a multiple of 9.

Next we consider four possible cases, one case for each of the possible
values of $a$.

Case 1: $a=2$

When $a=2$, we require that $2a+2b+c=4+2b+c$ be a multiple of 9.

Since $4+2b+c\geq4$, then the
smallest possible multiple of 9 that $4+2b+c$ can equal is 9.

Since $b\leq 9$ and $c\leq9$, then $4+2b+c$ is at most $4+2(9)+9=31$.

Thus, $4+2b+c$ can equal $9, 18$ or 27, which gives $2b+c$ equal to $5, 14$ or 23 respectively.

Next, we determine the possible values of $b$ and $c$ so that $2b+c$ is equal to $5, 14$ or 23.

$2b+c=5$
$2b+c=14$
$2b+c=23$

$b=2,
c=1$
$b=7,
c=0$
$b=9,
c=5$

$b=1,
c=3$
$b=6,
c=2$
$b=8,
c=7$

$b=0,
c=5$
$b=5,
c=4$
$b=7,
c=9$

$b=4,
c=6$

$b=3,
c=8$

Thus when $a=2$, there are $3+5+3=11$ such palindromes.

Case 2: $a=4$

When $a=4$, we require that $2a+2b+c=8+2b+c$ be a multiple of 9.

Since $8+2b+c\geq8$, then the
smallest possible multiple of 9 that $8+2b+c$ can equal is 9.

Since $b\leq 9$ and $c\leq9$, then $8+2b+c$ is at most $8+2(9)+9=35$.

Thus, $8+2b+c$ can equal $9, 18$ or 27, which gives $2b+c$ equal to $1, 10$ or 19 respectively.

Next, we determine the possible values of $b$ and $c$ so that $2b+c$ is equal to $1, 10$ or 19.

$2b+c=1$
$2b+c=10$
$2b+c=19$

$b=0,
c=1$
$b=5,
c=0$
$b=9,
c=1$

$b=4,
c=2$
$b=8,
c=3$

$b=3,
c=4$
$b=7,
c=5$

$b=2,
c=6$
$b=6,
c=7$

$b=1,
c=8$
$b=5,
c=9$

Thus when $a=4$, there are $1+5+5=11$ such palindromes.

Case 3: $a=6$

When $a=6$, we require that $2a+2b+c=12+2b+c$ be a multiple of 9.

Since $12+2b+c\geq12$ and $12+2b+c\leq12+2(9)+9=39$, then $12+2b+c$ can equal 18, 27 or 36, which
gives $2b+c$ equal to 6, 15 or 24,
respectively.

$2b+c=6$
$2b+c=15$
$2b+c=24$

$b=3,
c=0$
$b=7,
c=1$
$b=9,
c=6$

$b=2,
c=2$
$b=6,
c=3$
$b=8,
c=8$

$b=1,
c=4$
$b=5,
c=5$

$b=0,
c=6$
$b=4,
c=7$

$b=3,
c=9$

Thus when $a=6$, there are $4+5+2=11$ such palindromes.

Case 4: $a=8$

When $a=8$, we require that $2a+2b+c=16+2b+c$ be a multiple of 9.

Since $16+2b+c\geq16$ and $16+2b+c\leq16+2(9)+9=43$, then $16+2b+c$ can equal 18, 27 or 36, which
gives $2b+c$ equal to 2, 11 or 20,
respectively.

$2b+c=2$
$2b+c=11$
$2b+c=20$

$b=1,
c=0$
$b=5,
c=1$
$b=9,
c=2$

$b=0,
c=2$
$b=4,
c=3$
$b=8,
c=4$

$b=3,
c=5$
$b=7,
c=6$

$b=2,
c=7$
$b=6,
c=8$

$b=1,
c=9$

Thus when $a=8$, there are $2+5+4=11$ such palindromes.

Therefore, the number of palindromes that are greater than 10 000 and
less than 100 000 and that are multiples of 18 is $11+11+11+11=44$.