What is the smallest four-digit palindrome that is the sum of two different three-digit palindromic numbers?
There are different answers.
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$\begingroup$ If the puzzle is not your own creation, you should include the source in your question. $\endgroup$– BubblerCommented Oct 23, 2020 at 1:42
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2 Answers
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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.
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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 $B+D = 11$.
