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IMO Shortlist 1991

Posted: Fri Sep 19, 2014 10:10 am
by mutasimmim
Find the largest power of $1991$ that divides$1990^{1991^{1992}}+1992^{1991^{1990}}$.

Re: IMO Shortlist 1991

Posted: Fri Sep 19, 2014 9:10 pm
by Nirjhor
Apply LTE twice: on \(1990^{1991^{1992}}+1\) and on \(1992^{1991^{1990}}-1\). The answer is \(1991\).
\(1991=11\times 181\). So \[v_{11} 1990^{1991^{1992}}+1=v_{11} 1991 + v_{11} 1991^{1992}=1993\] and \[v_{11} 1992^{1991^{1990}}-1=v_{181} 1991+v_{181} 1991^{1990}=1991.\] From these two we have \(v_{11}\left(1990^{1991^{1992}}+1992^{1991^{1990}}\right)=1991.\) The similar approach above gives \(v_{181}\left(1990^{1991^{1992}}+1992^{1991^{1990}}\right)=1991.\) So \(1991\) is the answer.

Re: IMO Shortlist 1991

Posted: Fri Sep 19, 2014 10:05 pm
by mutasimmim
Nirjhor, elaborate please. And.. one more thing: $1991$ is not a prime.

Re: IMO Shortlist 1991

Posted: Fri Sep 19, 2014 10:55 pm
by Nirjhor
\(1991=11\times 181\). Work with each of these primes. Solution added to the previous comment.

Re: IMO Shortlist 1991

Posted: Sat Sep 20, 2014 10:26 am
by mutasimmim
$1990^{{1991}^{1992}}+1992^{{1991}^{1990}}=[{1990}^{{1991}^2}]^{{1991}^{1990}}+1992^{{1991}^{1990}}$
Now apply LTE twice.

Re: IMO Shortlist 1991

Posted: Wed Jul 12, 2023 9:15 am
by otis
No power of 1991 is divisible by 199019911992+199219911990.
retro bowl