1. Does 3 divide(3k+1)(3k+2)(3k+3)???
Two ways to show that. Factor 3 out of the third factor and we have
(3k+1)(3k+2)3(k+1), so one of its factors is 3, which means 3 divides it.
The other way. For any n consecutive integers, exactly one of them must be
divisible by n. Those factors are 3 consecutive integers, so one of them is
divisible by 3, and so is their product.
2. Is 6m(2m+10) divisible by 4?
6m(2m+10) = 6m(2)(m+5) = 12m(m+5) = 4[3(m+5)], So yes.
3. if n=4k+1, does 8 divide n2-1?
Substitute n=4k+1 in n2-1 and see:
(4k+1)2-1 = 16k2+8k+1-1 = 16k2+8k = 8(k2+k),
so yes since 8 is one of its factors.
Edwin
