Find all positive integers $n$ which are divisible by 11 and satisfy the following condition: all the numbers which are generated by an arbitrary rearrangement of the digits of $n$, are also divisible by 11.
Problem
Source:
Tags: number theory, Digits, Divisibility
cobbler
30.03.2015 23:06
All numbers of the form $aaa...a$ with an even number of digits.
I am happy to give the proof on request.
bobthesmartypants
31.03.2015 04:32
Think divisibility rule for $11$.
What happens when we switch two distinct digits?
primesarespecial
21.12.2020 13:47
Take an example of a five five digit number that has digit reperesentation abcde.Now by the divisibility rule,a+c+e-b-d=11k assuming k not equal to 0. As per the condition a permutation stil satisfies ,b+d+a-c-e=11r .Add both equations to get 2a is a multiple of 11 and the least we can get is 22 since k is not zero.But that gives a as 11 which is not possible as a is a digit.That implies all differences of digits at odd and even places irrespective of the permutation is 0.But we can easily see this is not working for numbers with odd number of digits. We can run this argument for numbers with even number of digits.The first case doesnt work here as well but the second case works here and it works when all digits are distinct.I dont know how to use Latex or would have shown you.But this question rather than proving was more on intuition. .