#1**+1 **

\(7^n\div\frac{1000!}{(500!)^2}=N \qquad N\in Z\\ \)

500! = 7*14* ....(71*7) * k where k is some integer that is not divisale by 7 (each k is a different interger)

500! = 7(1*2*3*.....*71)k = 7*71! * k

(500!)^2 = 49*(71)!^2 *k

1000! = 7*14* ...... 142*7 *k

\(\frac{1000!}{(500!)^2}\\ =\frac{7(142!)}{7*71!*7*71!}k\\ =\frac{(142!)}{71!*7*71!}k\\ =\frac{(7*14*21*....20*7)}{7(7*14*21* .... 10*7)*(7*14*21* .... 10*7)}k\\ =\frac{(7*20!)}{7(7*10!)*(7*10!)}k\\ =\frac{(20!)}{7*7*(10!)*(10!)}k\\ =\frac{k}{(10!)*(10!)}\\ =\frac{k}{49}\\\)

So I think the biggest n must be less than or equal to 2

I could easily have made stupid mistakes though.

Actually I think my answer could easily be rubbish.

Melody Apr 6, 2019