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anonymous
 5 years ago
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....
anonymous
 5 years ago
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....

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0? why three integer ? two is enough

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, the question has been put up like that....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, u hv given a trimmed version of what i did basically u hv used Euclid's div lemma only

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh cool, i didnt know if i remembered it or not you are probably correct then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Euclid's division lemma says "Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r where \[0\le r < b\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Therefore, taking b as 2 we get a = 2q + r and \[0\le r < 2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, if r = 0, then a = 2q + 0 = 2q If r = 1, then a = 2q + 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hence all positive integers are of the form 2a and 2q+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right and if a number can be represented in the 2q form it can be said it is divisible by 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, so for my question I hv two possibilities 1) 2q, 2q+1 and 2q 2) 2q+1, 2q, 2q+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In second case the product is 2(4q^3 + 4q^2 + q) which is again a multiple of 2 and hence divisible by 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for confirming..U get a medal !!
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