ratindasgupta wrote:If n is a positive integer, is n^3 - n divisible by 4?
1) n=2k + 1, where k is an integer.
2) n^2 + n is divisible by 6.
(1) okay, first off, we are able to know that K
>0. This is because of "if n is a positive integer" means that n>0. If k<0, then n would equal a negative number. we also know that n is odd. this is because of the "+1" and the facts that k is multiple by 2, and any number multiplied by an even number will result in an even number. adding one insures that n will be odd, because an odd and an even added together create an odd number. However, this alone is insufficient
(2) gives us a couple of ideas about what n maybe be. a good way of thinking about this is breaking 6 into it's factors or 3 and 2. we know that 2 would work for 6, but so would 3, and so would 4 and so would 6, and so forth. so insufficient as well.
together we know that n is odd, so n cannot be a multiple of 2, however, it could be an odd multiple of of 3.
is n^3 - n divisible by 4?
look at it as an (oddxoddxodd)-odd
three odds multiplied by each other = an odd-odd=an even. I would plug in 3,9,15
3^3-3=24/4=6
9^3-9=261-9=252/4-=63
i would probably stop here and feel comfortable with saying C
I see what I did wrong here, looking at A, i should have quickly plugged in 1,2,3 for K just to check. and seeing that n would equal to 3,5,7, each of those would be sufficient.
