What is the highest power of 7 in 5000! ?

Can someone explain this further? I'm not following the solutions here. Thanks!

[quote="Bunuel"]What is the highest positive integer power of 7 in 5000!?

(A) 714
(B) 816
(C) 832
(D) 835
(E) 4998




This concept is tested here
https://gmatclub.com/forum/everything-a ... 85592.html

5000/7 + 5000/7^2 + 5000/7^3 + 5000/7^4

C

Posted from my mobile device

OK so we are looking for maximum value of n when \(7^n<5000!\).
This basically means the number of times 7 comes in the product 1*2*3*4*...*5000.

Now when we divide by 7 what happens..

\(\frac{5000}{7}=714.xy\), so 714 gives us the number which have 7 in it.. 7, 14, 21, ......4998.
\(\frac{5000}{49}=102.04\)~120 gives us the number which are div by 49 as there is an extra 7 in these numbers... 49, 343....
\(\frac{5000}{49*7}=14.57\)~14 gives us the number which are div by 49*7 as there is an extra 7 in these numbers over and above what we have counted till now... 343....
similarly \(\frac{5000}{49*7*7}=2.08\), so 2 numbers more 2401 and 4802
Next 49*7*7*7 goes beyond 5000 and will give us an answer less than 1..

Now we have calculated all 7s whether as 49, 343 etc = 714+102+14+2=832
So our Ans 10+3+1=14

While the concept has been explained in earlier posts,

\(\frac{5000}{7}\) = 714 ;


5000/7^2 = 5000/7 x 1/7 = 714/ 7 = 102 ; (714 = 5000/7 from previous step)

Similarly

5000/7^3 = 102/7 = 14 ;
5000/7^4 = 14/7 = 2;

Add 714+102+14+2 = 832