prove that the product of three consecutive positive integers is divisible by 2. (This has to be proved using Euclid's Division Algorithm). I hv solved it, but just want to reconfirm....

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prove that the product of three consecutive positive integers is divisible by 2. (This has to be proved using Euclid's Division Algorithm). I hv solved it, but just want to reconfirm....

Mathematics
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? why three integer ? two is enough
Well, the question has been put up like that....
the product of an odd integer and even will always be even and the product of an even integer with any other integer is always even and an even integer is divisible by 2 by definition (2n)*(2n+1)*(2n) = 2*[n*(2n+1)(2n)] id have to look up euclids algorithm though

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yeah, u hv given a trimmed version of what i did basically u hv used Euclid's div lemma only
oh cool, i didnt know if i remembered it or not you are probably correct then
Euclid's division lemma says "Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r where \[0\le r < b\]
Therefore, taking b as 2 we get a = 2q + r and \[0\le r < 2\]
So, if r = 0, then a = 2q + 0 = 2q If r = 1, then a = 2q + 1
Hence all positive integers are of the form 2a and 2q+1
right and if a number can be represented in the 2q form it can be said it is divisible by 2
Yes, so for my question I hv two possibilities 1) 2q, 2q+1 and 2q 2) 2q+1, 2q, 2q+1
In first case product is 8q^3 + 4q^2 = 2(4q^3 + 2q^2) and since it is a multiple of 2, it is divisible by 2
In second case the product is 2(4q^3 + 4q^2 + q) which is again a multiple of 2 and hence divisible by 2
correct
Thanks for confirming..U get a medal !!

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