Bariz - Problem

Source: Caucasus 2015 9.5

Tags: number theory, combinatorics, palindromes, sum of digits

parmenides51

26.04.2019 15:49

Let's call a natural number a palindrome, the decimal notation of which is equally readable from left to right and right to left (decimal notation cannot start from zero; for example, the number $1221$ is a palindrome, but the numbers $1231, 1212$ and $1010$ are not). Which palindromes among the numbers from $10,000$ to $999,999$ have an odd sum of digits, which have an one even, and how many times are the ones with odd sum more than the ones with the even sum?

I think I got it now. The range from 10000 to 999999 includes all 5 digit numbers and 6 digit numbers. Since the 6 digit numbers are in the form abccba, their sum is 2(a+b+c) which is always even For 5 digit number abcba, the sum of digits is 2(a+b)+c which is even when c is even and odd when c is odd. therefore half of the possibilities for c are odd and the other half even. The number of 5 digit palindromes is the same as the number of 6 digit palindromes (since they are both the number of ways to pick digits a b and c which is 9*10*10). Since half of the five digit palindromes are odd, 1/4 of the 5 digit and 6 digit palindromes are odd. So the ratio of odd sum of digits to even sum of digits is 1:3