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Mathematics 7 Online
OpenStudy (anonymous):

need help fast!

OpenStudy (anonymous):

we have k(k+1)...(k+r-1) = m * r!; m,k being positive integers. if k doesnt divide r, then k divides m. prove it.

OpenStudy (anonymous):

anyone ideas

OpenStudy (anonymous):

what do mean when you put the periods?

OpenStudy (anonymous):

k(k+1)(k+2)(k+3)(k+4)....(k+r-1) = m*r!. k,m,r are positive integers.

OpenStudy (anonymous):

what kind of anwser do you need?

OpenStudy (anonymous):

a proof. see the 1st post

OpenStudy (anonymous):

Oh. Yeah i have no clue. Sorry to waste your time :/ try looking on google? :)

OpenStudy (mathmate):

I could be wrong in interpreting your question, but I seem to have found a counter-example. k=21 r=3 => r! = 3!=6 so (k+r-1)=21+3-1=23 k(k+1)(k+2)=21*22*23=10626 10626 = mr! = m*3! = 6m => m= 10626/6 =1771 so k(k+1)(k+2) = 10626=1771*3! But 21 does not divide 1771, nor 3 so the question cannot be proved.

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