Recent content by Almutairi

I might have misunderstood the question, but Based on what you made a=2 and b=0. Can you give them any value you want? In that case can't you always make a=0, b= (6K 2 +4k)? So that 6k2+4k is always of the form 6ab+5a+b, where K is any rational number. eg. for k=1, 6k2+4k=10=6x0x10+5x0+10...

Thank you for the confidence boost, and also thanks for the mod operator info it really helped. Unfortunately, every one who knows me will count this as a life achievement:smile:.

[SIZE="3"]Proof Given that [(n3)/3 + (n2)/2 + n/6] is an integer, then [ 2n3 + 3n2+ n] is an integer that is a multiple of 6 and therefore, [2n3+ n] is a multiple of 3 and it can be written as n[2n3 + 1] so either n or [2n2 + 1] is a multiple of 3, or both. To rollout the...

I am sorry I didn't understand what you were trying to do could you please explain it differently.

thanx for the welcome, and sorry for the confusion:smile: but I meant 2n2, and also you used 2 as a value of n but 2 is not divisible by 3 (i.e 2/3 is not an integer). So for example let n=7( seven is not zero and not divisible by 3) so (72)*2=98 and 98+1=99 which is divisible by 3. also the...

Hi, while trying to answer a home work question I came up with an unrelated equation, although it is very trivial it seemed interesting to me: If n is an integer not divisible by 3 and not equal to 0, then [(2*(n^2)+1] MUST be divisible by 3. I tried to proof it using the fact that...