IMO Shortlist 1991

[mutasimmim]

Find the largest power of $1991$ that divides$1990^{1991^{1992}}+1992^{1991^{1990}}$.

[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.

[mutasimmim]

Nirjhor, elaborate please. And.. one more thing: $1991$ is not a prime.

[Nirjhor]

\(1991=11\times 181\). Work with each of these primes. Solution added to the previous comment.

[mutasimmim]

$1990^{{1991}^{1992}}+1992^{{1991}^{1990}}=[{1990}^{{1991}^2}]^{{1991}^{1990}}+1992^{{1991}^{1990}}$
Now apply LTE twice.

[otis]

No power of 1991 is divisible by 199019911992+199219911990.
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