Prove that the product of any r consecutive numbers is divisible by r!.
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Answered by
6
Step-by-step explanation:
The product of some r consecutive integers can be represented asr consective integers⏞(n+r)(n+r−1)⋯(n+1)=(n+r)!n!where n is the number one less than the smallest of the consecutive integers. Now, if it is true that primes in (n+r)! appear just as frequently or more as in n!r!, then you are saying that for some integer k (likely big) that (n+r)!=k⋅n!r!. So your product of n consecutive integers is(n+r)!n!=k⋅n!r!n!=k⋅r!
and is therefore divisible by r!.
Answered by
7
Step-by-step explanation:
Step-by-step explanation:
The product of some r consecutive integers can be represented asr consective integers⏞(n+r)(n+r−1)⋯(n+1)=(n+r)!n!where n is the number one less than the smallest of the consecutive integers. Now, if it is true that primes in (n+r)! appear just as frequently or more as in n!r!, then you are saying that for some integer k (likely big) that (n+r)!=k⋅n!r!. So your product of n consecutive integers is(n+r)!n!=k⋅n!r!n!=k⋅r!
and is therefore divisible by r!.
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