What is the smallest positive value of 36^k5^l ?
k, l are positive integers.
Credit due: Victors Linis, Ottawa University
Consider the expression mod 4, 5, and 6
Mod 4: the k term is a multiple of 4 and the l lterm = 1 mod 4 (since 5 == 1 mod 4) and so the expression is 1 mod 4
Mod 5: the k term is 1 mod 5 (since 36 = 1 mod 5) and the l term is a multiple of 5 so the expression is 1 mod 5
Mod 6: the k term is 0 mod 6, but the l term is more complex  it's either 1 mod 6 when l is odd or +1 mod 6 when l is even (since 5 = 1 mod 6)
So the overall expression must be either:
1) 1 mod 4, +1 mod 5, +1 mod 6 OR
2) 1 mod 4, +1 mod 5, 1 mod 6
In either case, there's exactly one possible residue mod LCM(4,5,6) = 60
For 1) the expression must be 31 mod 60 since 31 = 1 mod 4, +1 mod 5, and +1 mod 6
For 2) the expression must be 11 mod 60 since 11 = 1 mod 4, +1 mod 5, and 1 mod 6
So overall the expression must be either 11 or 31 mod 60. If it's also positive, that means the smallest value it can possibly have is 11. Fortunately, by inspection it's easy to see that when k = 1, l = 2, the value of 36  25 is indeed 11. Since we have a concrete example where the expression's value is 11 and proof that it can't be smaller, this is the minimum.

Posted by Paul
on 20230912 11:32:10 