GCD(ab,c) = GCD(a,c) * GCD(b,c)

so a counterexample proves it false since it is for all a,b,c.

]]>C25

]]>GCD(1024,4)=4

GCD(512,4)=4

GCD(2,4)=2

2*4<>4

Sorry,I took b=512. You can see that we have a counterexample so the claim is false.

]]>GCD(8,1024) = 8

GCD(2,1024) = 2

GCd(4,1024) = 4

hence 8 = 4*2

]]>i am guessing we show, a,b,c have some prime factors since all of them have the same prime factor, the statement should be true.

EDIT:

This is what i have so far:

Let a = p1 * p2 where pN is an integer and a prime.

Let b = p3 * p4 " " " """

Let c = p5* p6 " " " " "

Now i am not sure how to show that a,b,c have the same unique prime factor.

]]>Prove GCD(ab,c) = GCD(a,c) * GCD(b,c)

]]>I need help with one more proof.

Prove that GCD(ab,c) = GCD(a,c) * GCD(b,c)

]]>Well if the su mof the first n numbers is.n(n+1)/2 then the sum of the.first n+1 elements is n(n+1)/2 + n+1 =(n^2+n+2n-2)/2=(n^2+3n+2)/2=(n+1)(n+2)/2

Qed

Oh and your formula is.not correct. It is n(n+1)/2 like I wrote up there.

Stefy

]]>I am sure this has to be proved by induction.

I am guessing my basis step would be, for n rocks where n = 2, then k*m = 1 = (n(n-1))/2 = 1.

induction hypothesis: basis step holds true for n >= 2

Now I am stuck, because k*m could break apart easily into many different possibilities and if i try to keep it such that k decreases and m is always 1, this doesnt take into account all the possibilities.

Can someone give me a little push in the right direction?

Thanks!

]]>