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anonymous
 4 years ago
need help fast!
anonymous
 4 years ago
need help fast!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we have k(k+1)...(k+r1) = m * r!; m,k being positive integers. if k doesnt divide r, then k divides m. prove it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what do mean when you put the periods?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k(k+1)(k+2)(k+3)(k+4)....(k+r1) = m*r!. k,m,r are positive integers.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what kind of anwser do you need?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a proof. see the 1st post

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh. Yeah i have no clue. Sorry to waste your time :/ try looking on google? :)

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0I could be wrong in interpreting your question, but I seem to have found a counterexample. k=21 r=3 => r! = 3!=6 so (k+r1)=21+31=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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