# Smallest four-digit palindromic numbers

What is the smallest four-digit palindrome that is the sum of two different three-digit palindromic numbers?

• If the puzzle is not your own creation, you should include the source in your question. Oct 23 '20 at 1:42

The palindromes:

202 + 909 = 1111, 303 + 808 = 1111, 404 + 707 = 1111, 505 + 606 = 1111

The process of getting this answer:

First, we note that the palindrome we're looking for has 4 digits. Ideally, we'd like something that is less than 2000. So, we'd have first digit and last digit 1. We note that palindromes of length 3 that consist of only the same number (111, 222, ..., 999), when added together, result in a palindrome of the desired form. This leads us to things like 222 + 999 = 1221, 333 + 888 = 1221, 444 + 777 = 1221, and 555 + 666 = 1221. Finally, we note that the "insides" of each of these numbers can be hollowed out (zero'd). This subtracts 110 from each of the above sums, resulting in the smallest 4-digit palindrome with the 2 3-digit palindrome sum property, 1111.

Since the sum of two three-digit numbers cannot exceed 2000, the equation should look like $$ABA + CDC = 1EE1$$. Now we know $$A+C \ge 9$$ from the highest digit and $$A+C$$ ends in 1 from the lowest digit, so $$A+C=11$$. Then the minimum possible sum is 1111 where $$B=D=0$$, which happens to be a palindrome. One possible sum is $$505 + 606 = 1111$$. The only other sum that meets the condition is 1221, where $$B+D = 11$$.