Positive integer $m$ shall be called anagram of positive $n$ if every digit $a$ appears as many times in the decimal representation of $m$ as it appears in the decimal representation of $n$ also. Is it possible to find $4$ different positive integers such that each of the four to be anagram of the sum of the other $3$? (Bulgaria)
Problem
Source: Balkan BMO Shortlist N7
Tags: number theory, Digits, decimal representation, IMO Shortlist
05.08.2019 21:25
Hmm... seems like an interesting problem, haven't solved it yet. Just putting out ideas you get looking at the problem- Obviously each number has the same number of digits. Modulo $9$ invariance $=>$ all are divisble by $9$... don't know how it'd be too useful though, unless maybe to try and hunt for a solution... Maybe rephrase the problem into does there exist some number $S$ such that it can be expressed as both the sum of $4$ numbers and the sum of each one and one of its anagrams? If you're sure there isn't a solution ,you could try a method reminiscent of FMID- if an $n-digit$ combination works, then some $n-1-digit$ combination must work too, and so on and so forth=> contradiction... doesn't look too promising though...
27.06.2020 22:55
There is four different integer satisfy the condition of the problem. It took me about 90 minutes to find them. 28103409 28103418 28103427 28103445
24.09.2024 17:29
Marvelous problem by Nikolai Beluhov! No idea why did I not appreciate it back in the days when it was at Bulgaria IMO TST just after Balkan MO.