Bunuel wrote:
aalriy wrote:
If \(10! - 2*(5!)^2\) is divisible by \(10^n\), what is the greatest value of n?
A. 1
B. 2
C. 3
D. 4
E. 5
We should determine the # of trailing zeros of \(10! - 2*(5!)^2\). Trailing zeros are a sequence of 0s in the decimal representation of a number, after which no other digits follow. For example, 2000 has 3 trailing zeros.
Note that every 5 and 2 in prime factorization will give one more trailing zero.
\(10!=5!*6*7*8*9*10=5!*(2*3)*7*(4*2)*9*(5*2)=5!*(2*3*4*5)*(7*2*9*2)*=252*(5!)^2\);
\(252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=5^2*12*12*1,000\) --> 5^2 with two 2-s from 12 will provide with 2 more zeros, so \(10! - 2*(5!)^2\) has total of 5 trailing zeros.
So, max value of n for which above is divisible by 10^n is 5.
Answer: E.
P.S. Final value is \(252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=1,000*25*12*12=1,000*300*12=12*300,000=3,600,000\).
Hope it's clear.
Hi Bunuel,
Though I understood your approach completely. when I try to simplify the expression myself in a different manner, I don't arrive at the right conclusion.
Could you please explain what is the flaw in what I am trying to do :
10! = 5! * 10*9*8*7*6
= 10^2 * 2^6 * 9^2 * 7
2*(5!)^2 = 5^2 * 2^7 * 9
10! - 2*(5!)^2 = (10^2 * 2^6 * 9^2 * 7) - (10^2 * 2^5 * 9)
= 10^2 * 2^5 (2*81*7 -9)
At this point, we have only 2 0s (contributed by 10^2) . We have 2^5 but no additional 5s (in 2, 81 o 7)
Then how do we get 36000000 as the final result. (I verified : 10^2 * 2^5 (2*81*7 -9) = 3,600,000 which means I am not making any calculation errors)
I am sure there is something silly that I am missing but I am not able to understand. Please explain.