Question

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....

Answer 1

? why three integer ? two is enough

Answer 2

Well, the question has been put up like that....

Answer 3

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

Answer 4

yeah, u hv given a trimmed version of what i did
basically u hv used Euclid's div lemma only

Answer 5

oh cool, i didnt know if i remembered it or not
you are probably correct then

Answer 6

Euclid's division lemma says "Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r where \[0\le r < b\]

Answer 7

Therefore, taking b as 2 we get a = 2q + r and \[0\le r < 2\]

Answer 8

So, if r = 0, then a = 2q + 0 = 2q
If r = 1, then a = 2q + 1

Answer 9

Hence all positive integers are of the form 2a and 2q+1

Answer 10

right and if a number can be represented in the 2q form it can be said it is divisible by 2

Answer 11

Yes, so for my question I hv two possibilities
1) 2q, 2q+1 and 2q
2) 2q+1, 2q, 2q+1

Answer 12

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

Answer 13

In second case the product is
2(4q^3 + 4q^2 + q) which is again a multiple of 2 and hence divisible by 2

Answer 14

correct

Answer 15

Thanks for confirming..U get a medal !!