Determine all possible triplets (p, q, n) of nonnegative integers that satisfy this equation:
2p + 2q = n!
Provide valid argument for your answer.
*** Adapted from a problem which appeared at the Harvard/MIT math Olympiad.
My approach was similar to Larry's:
WLOG assume p >= q. Then we can write:
n! = 2^q(1 + 2^r) where r = p-q
All factors of n! that aren't 2's must come from the 2nd expression.
But 1 + 2^r is a multiple of 3 only when r is odd, and is a multiple of 5 only when r is even (specifically when r = 2 mod 4), so this expression cannot simultaneously contribute both a factor of 3 and of 5. As a result, no n > 4 can be a solution, as those factorials require both factors.
Inspection quickly yields solutions for n = 2 (p = q = 1), n = 3 (p = 2, q = 1 or the reverse) and n = 4 (p = 4, q = 3 or the reverse) which must therefore be all of the solutions.
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Posted by Paul
on 2023-02-27 16:25:22 |
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