Quantcast
Channel: GMAT Club Forum - latest posts
Viewing all articles
Browse latest Browse all 420781

Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest

$
0
0
10! = 2^8*3^4*5^2*7= 2 zeros (10*5*2)

after rearranging 10! = 100*2^6*3^4*7

2*5!^2 = 2^7*3^2*5^2=2 zeroes(5*2*5*2)

after rearranging 100*2^5*3^2

now 10!-2*5!^2 = 100(2^6*3^4*7 - 2^5*3^2)
= 100*2^5*3^2(2*3^2*7 - 1)
=100*2^5*3^2*125
=100 *2^5*3^2 *5^3
= 10^5 *2^2*3^2

OA:E

Viewing all articles
Browse latest Browse all 420781

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>