Find all positive integers n such that 10n+3n+2 is a palindrom number.
Problem
Source: Kosovo Mathematical Olympiad 2022, Grade 11, Problem 3
Tags: number theory
06.03.2022 17:57
10n+3n+2=100...00⏟n−12+3n we must have 2+ the last digit of the 3n together 1. So n=4k+2, but must have k=0, because else we have too much zero after 1, and no zeros at the end. So only siruation it is n=2, the palindrom number it is 111
06.03.2022 18:00
Mr.Math.Boy wrote: because else we have too much zero after 1 True but that has to be proven. Otherwise the solution is not complete.
06.03.2022 18:15
The numbers of 0 on the LHS i creasing fast with every n power, much that the 0 in the pover of 3,where we have not consecutive 0 etc.
06.03.2022 18:22
Mr.Math.Boy wrote: The numbers of 0 on the LHS i creasing fast with every n power, much that the 0 in the pover of 3,where we have not consecutive 0 etc. You have to prove that mathematically.
06.03.2022 18:33
First, n=2 works and n=1 does not. For n≥3, we must have n−⌊log10(3n)⌋−1≤⌊log10(3n)⌋−1 where the LHS is the number of zeroes before the first digit of 3n+2 and the RHS is an upper bound on the number of zeroes in the number 3n+2. However, this is false because n>2nlog10(3)=2log10(3n)≥2⌊log10(3n)⌋
06.03.2022 18:42
The number begins with a 1, so it must end with a 1. 3^n+2 \equiv 1 \pmod{10} 3^n \equiv 9 \pmod{10} n \equiv 2 \pmod{4} for n=2, 10^2+3^2+2=111, which is a palindrome. However, for n>2, we would also need 3^n+2 \equiv 1 \pmod{100} to account for the 0s contributed by 10^n 3^n \equiv 99 \pmod{100} This implies that 3^n \equiv 99 \pmod{4}. 3^n \equiv 3 \pmod{4} implies n is odd But n \equiv 2 \pmod{4}, so it is even, contradiction. Only \boxed{2} works.