A palindrome is a positive integer whose digits are the same when read forwards or backwards. For example, 2882 is a four-digit palindrome and 49194 is a five-digit palindrome. There are pairs of fourdigit palindromes whose sum is a five-digit palindrome. One such pair is 2882 and 9339. How many such pairs are there?
(A) 28
(B) 32
(C) 36
(D) 40
(E) 44

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-136.jpg?height=52&width=201&top_left_y=2229&top_left_x=932)
A palindrome is a positive integer whose digits are the same when read forwards or backwards. For example, 2882 is a four-digit palindrome and 49194 is a five-digit palindrome. There are pairs of four-digit palindromes whose sum is a five-digit palindrome. One such pair is 2882 and 9339. How many such pairs are there?
(A) 28
(B) 32
(C) 36
(D) 40
(E) 44

## Solution

Our first observation is that since we are adding two four-digit palindromes to form a five-digit palindrome then the following must be true,

$$
\begin{gathered}
a b b a \\
c d d c \\
\hline e f e 1
\end{gathered}
$$

(i.e. the first digit of the 5-digit palindrome is 1.)

From this, we can see that $a+c=11$ since $a+c$ has a units digit of 1 and $10<a+c<20$. We first note that there are four possibilities for $a$ and $c$. We list the possibilities:

| $a$ | 2 | 3 | 4 | 5 |
| :--- | :--- | :--- | :--- | :--- |
| $c$ | 9 | 8 | 7 | 6 |

Note that there are only four possibilities here.

(If we extended this table, we would get an additional four possibilities which would be duplicates of these four, with $a$ and $c$ reversed.)

Let us consider one case, say $a=2$ and $c=9$.

## $2 b b 2$

$9 d d 9$

1 e fe 1

From this, we can only get palindromes in two ways. To see this we note that $e$ is either 1 or 2 depending on whether we get a carry from the previous column (we see this looking at the thousands digit of 1 e fe 1). If $e=1$, then $b+d$ has no carry and so looking at the tens digit $e=1$, we see that $b+d=0$ to get this digit.

If $e=2$, we do get a carry from $b+d$, so looking again at the tens digit $e=2$, we see that $b+d=11$.

Possibility $1 \quad b=d=0$

Since there are only four possibilities for $a$ and $c$ and just one way of selecting $b$ and $d$ so that $b+d=0$ for each possibility, there are just four possibilities.

Possibility $2 \quad b+d=11$

For each of the four possible ways of choosing $a$ and $c$, there are eight ways of choosing $b$ and $d$ so that $b+d=11$ thus giving 32 possibilities.

This gives a total of $4+32=36$ possibilities.

ANSWER: (C)