A palindrome is a positive integer whose digits are the same when read forwards or backwards. For example, 32523 is a palindrome. How many palindromes greater than 10000 and less than 100000 are multiples of 18 ?
(A) 41
(B) 42
(C) 43
(D) 44
(E) 45
A palindrome greater than 10000 and less than 100000 is a 5 -digit positive integer of the form $a b c b a$, where $a, b$ and $c$ are digits and $a \neq 0$.

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 $a \neq 0$ ).

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 $2 a+2 b+c$ 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 $2 a+2 b+c=4+2 b+c$ be a multiple of 9 .

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

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

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

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

| $2 b+c=5$ | $2 b+c=14$ | $2 b+c=23$ |
| :---: | :---: | :---: |
| $b=2, c=1$ | $b=7, c=0$ | $b=9, c=5$ |
| $b=1, c=3$ | $b=6, c=2$ | $b=8, c=7$ |
| $b=0, c=5$ | $b=5, c=4$ | $b=7, c=9$ |
|  | $b=4, c=6$ |  |
|  | $b=3, c=8$ |  |

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

Case 2: $a=4$

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

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

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

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

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

| $2 b+c=1$ | $2 b+c=10$ | $2 b+c=19$ |
| :---: | :---: | :---: |
| $b=0, c=1$ | $b=5, c=0$ | $b=9, c=1$ |
|  | $b=4, c=2$ | $b=8, c=3$ |
|  | $b=3, c=4$ | $b=7, c=5$ |
|  | $b=2, c=6$ | $b=6, c=7$ |
|  | $b=1, c=8$ | $b=5, c=9$ |

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

Case 3: $a=6$

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

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

| $2 b+c=6$ | $2 b+c=15$ | $2 b+c=24$ |
| :---: | :---: | :---: |
| $b=3, c=0$ | $b=7, c=1$ | $b=9, c=6$ |
| $b=2, c=2$ | $b=6, c=3$ | $b=8, c=8$ |
| $b=1, c=4$ | $b=5, c=5$ |  |
| $b=0, c=6$ | $b=4, c=7$ |  |
|  | $b=3, c=9$ |  |

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

Case 4: $a=8$

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

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

| $2 b+c=2$ | $2 b+c=11$ | $2 b+c=20$ |
| :---: | :---: | :---: |
| $b=1, c=0$ | $b=5, c=1$ | $b=9, c=2$ |
| $b=0, c=2$ | $b=4, c=3$ | $b=8, c=4$ |
|  | $b=3, c=5$ | $b=7, c=6$ |
|  | $b=2, c=7$ | $b=6, c=8$ |
|  | $b=1, c=9$ |  |

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

Therefore, the number of palindromes that are greater than 10000 and less than 100000 and that are multiples of 18 is $11+11+11+11=44$.

ANSWER: (D)