How do you find an integer n such that #((n+7)!)/(n!(n+6)(n+4))#?

1 Answer
Cesareo R.
Dec 16, 2016

See below.

Explanation:

#((n+7)!)/(n!(n+6)(n+4))=((n+1)(n+2)cdots(n+7))/((n+6)(n+4))=#
#=(n+1)(n+2)(n+3)(n+5)(n+7)#

As we can see

#((n+7)!)/(n!(n+6)(n+4))=(n+1)(n+2)(n+3)(n+5)(n+7)# and the division always gives an integer as result. In other words

#((n+7)!)/(n!(n+6)(n+4))# is integer for any #n in NN, n ge 0#

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