Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.5 AIME Find the answer Italy

Problem:

A natural number is called a palindrome if it is equal to the number obtained by reading the digits of its base-ten representation from right to left (for example, 68386 and 44 are palindromes, 220 is not). We know that the natural number $x$ and the number $x+312$ are both palindromes; $x$ has four digits, while $x+312$ has five. What is the sum of the digits of $x$ ?

Pick one

Solution

Solution:

The answer is $(\mathbf{E})$ . If $x$ is a 4-digit number (hence less than 10000), $x+312$ , which must have 5 digits, lies between 10000 and 10311 (endpoints included). This means that the first two digits of $x+312$ are 1 and 0; since it is a palindromic number, it must therefore be of the form $10a01$ for some digit $a$ . On the other hand, $x=10a01-312$ ends in 89; it follows that the only possibility for $x$ is 9889, and indeed $9889+312=10201$ has 5 digits and is a palindrome. In conclusion, the sum of the digits of $x$ is $9+8+8+9=34$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.