2
$\begingroup$

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

There are different answers.

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

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.

$\endgroup$
2
$\begingroup$

The answer is

The smallest sum is 1111 = 505 + 606 (among others, e.g. 202 + 909 also works), and the only other possible sum is 1221 = 555 + 666.

Reasoning:

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 $B+D = 11$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.