Prove that the value of [M!/((M+1)(M+2))] is always even, for any given positive integer M, where [x] denotes the greatest integer ≤ x.

** For purposes of the problem, treat zero as an even number.

There are 3 cases:
1. m={1,2,3,4}; By computation, these are equal to 0.
2. Both (m+1) and (m+2) are compound, m>4; 2 divides one of them, with some other factors to be found in (m!) as the remaining factors of (m+1) and (m+2).But since m>4, there is at least surplus factor of 2 left over in m! so the result, which is a whole number, is even.
3. Either (m+1) or (m+2) is prime, m>4; they cannot both be prime. The result is an irreducible fraction over the prime denominator, P, comprising a whole number part, W, and a fractional part, F.
(a) Because P is relatively prime to each and every (remaining after cancelling) number in the numerator, F has to be (P-1)/P.
(b) Now the whole number part, W, is divisible by P, and P-1 is even, so W=(2a-2b)/P or 2(a-b)/P. Since P does not divide 2, W must be an even number.
(c) The required result then follows at once.
4. As an aside, rounding down in this way is a pretty dreadful approximation, since the nearest whole number by far to (W+F)/P is odd, not even.

Edited on February 6, 2011, 5:58 am

Posted by broll on 2011-02-06 05:37:27