Find minimal number of divisors that can number $|2016^m-36^n|$ have,where $m$ and $n$ are natural numbers.
Problem
Source: Serbian junior TST 2nd
Tags: Divisors, number theory
tenplusten
21.05.2016 22:21
Hi dude where can I find other Serbian Junior TST problems?
Garfield
21.05.2016 22:26
Here http://www.dms.rs/DMS/data/takmicenja/matematika/osnovna/2016/JSMO16.pdf but on Serbian language.
mihajlon
21.05.2016 22:58
Murad.Aghazade wrote: Hi dude where can I find other Serbian Junior TST problems? You can find here other Serbia JTST's, but there are not many of those...
I am probably gonna post few more on summer...
MilosMilicev
15.10.2016 14:10
Let x=our number. For m,n>1, we have that 81Ix, 16Ix, 5Ix, it means that x has at least (4+1)(4+1)(1+1)=50 positive divisors, and for m=1, n=2,x=720(30 positive divisors).It is not hard to prove that it is the minimum number...
IvanMitrovic
08.03.2020 16:51
MilosMilicev wrote: Let x=our number. For m,n>1, we have that 81Ix, 16Ix, 5Ix, it means that x has at least (4+1)(4+1)(1+1)=50 positive divisors, and for m=1, n=2,x=720(30 positive divisors).It is not hard to prove that it is the minimum number... You did not check all of the cases