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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 \(252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=1,000*25*12*12=1,000*300*12=12*300,000=3,600,000\).

If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]

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18 Oct 2014, 00:01

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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

The equation can be simplified as =>(5!)(10*9*8*7*6*5 - 2*(5!)) =>120(30240-240) =>(10^5) * K (No need to calculate the remaining part as we knew the power of 10)

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

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19 Sep 2015, 12:02

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saroshgilani wrote:

10! - 2*(5!)^2 5!(6*7*8*9*10 -(2*5!)) 5!(6*7*8*9*10- (2*(1*2*3*4*5)) 5!((6*7*8*9*10)-(2*4*6*8*10)) 5!*6*8*10(7*9-(2*4)) 5!*6*8*10(55) 5!*6*8*10*55 1*2*3*4*5 *6*7*8*9*2*5 *11*5 i found only three 5... can some one plz tell me where did i went wrong?Bunuel

Highlighted step is wrong. You multiplied 2 with every term in the bracket which are in multiplication, not addition. 5!(6*7*8*9*10- (2*(1*2*3*4*5)) = 5!((6*7*8*9*10)-(10*8*3)) = 5!*10*8*3(9*7*2 - 1) = 12*10*10*3*\(2^3\)(125) = 12*10*10*3*\(2^3\)(\(5^3\)) = 12*10*10*3*\(10^3\) =12*3*\(10^5\)

Can someone explain the bolded part to me? Maybe I'm missing something, but going from that first step to the 2nd step makes no sense to me. 10*9*8*7*6*5-2*5!*5*4*3*2*1, shouldn't that equal 10*9*8*7*6-2*5*4*3*2*1, since the second part of that equation contains - 2*5!, and there's also another 5! in the 5*4*3*2*1 portion? Can someone link me to a page explaining the logic here, because this factorial stuff is making no sense to me. I read the factorial guide post, but it doesn't seem to have anything that applies here. Desperate for help here. I have GMATs in about 6 weeks, and these factorial problems keep beating my head in! I can't find anything that explains or helps to solve these problems at all, and the explanations make no sense to me.

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, When I see these kinds of questions I immediatly go to the regular formula, where 10! has 2 trailing "0".

A. is there a way to do it with the formula above? B. if not, can you try and specify how you come to solve this kind of question? how did you know not to use the formula above?

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, When I see these kinds of questions I immediatly go to the regular formula, where 10! has 2 trailing "0".

A. is there a way to do it with the formula above? B. if not, can you try and specify how you come to solve this kind of question? how did you know not to use the formula above?

We need to find the number of trailing zeros of 10! - 2*(5!)^2, not 10!. You cannot find the number of trailing zeros of 10! - 2*(5!)^2 directly with that formula. To do that you should use algebraic manipulations as shown.
_________________

Thanks. But when I try to find the number of "10" in 250 I tried doing factorization of 250 (2*5*5*5) and only got one "10" out of it.... Is there another way it can be done if I don't see what you see the numbers?

Thanks. But when I try to find the number of "10" in 250 I tried doing factorization of 250 (2*5*5*5) and only got one "10" out of it.... Is there another way it can be done if I don't see what you see the numbers?

I try to understand what you mean there but fail to.

250 has one trailing zero, why to prime factorize it? Also, what does "but" in your post refer to?
_________________

So if I follow your logic, in 250*120*120 there should only be 3 trailing zeros. one in 250, one in 120 and one in 120 again.... This rises our count to 3, yet somewhere in there are hiding an additional 2 trailing zeros. If I were to arrive at this equation on test day, I would have selected "3". I am trying to see where in this form of the equation the additional zeros are hiding...

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

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18 Nov 2013, 18:12

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

If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of n?

=10!-2(5!*5!)/10^n =10*9*8*7*6*5!-2(5!*5!)/10^n =5!*10*8*3(9*7*2-1)/10^n 2400*5!/10^n 2400*5*4*3*2/10^n 2400*10*12/10^n max value of n for the above can be 5
_________________

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

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19 Nov 2013, 09:01

rango 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

If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of n?

=10!-2(5!*5!)/10^n =10*9*8*7*6*5!-2(5!*5!)/10^n =5!*10*8*3(9*7*2-1)/10^n 2400*5!/10^n 2400*5*4*3*2/10^n 2400*10*12/10^n max value of n for the above can be 5

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

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19 Nov 2013, 15:37

ronr34 wrote:

rango 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

If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of n?

=10!-2(5!*5!)/10^n =10*9*8*7*6*5!-2(5!*5!)/10^n =5!*10*8*3(9*7*2-1)/10^n 2400*5!/10^n 2400*5*4*3*2/10^n 2400*10*12/10^n max value of n for the above can be 5

How did you make this calculation?

in the above calculation; i have deliberately eliminated (9*7*2-1) after step 3 for the ease of calculation and understanding. Try to find how many 10 multiples can be required to divide the numerator. 2400 req 10^2; 10*12 req 10^2; also 9*7*2-1 req 10^2; so after adding it comes to be 10^6 ; but we have in options max n0. 5 ; so right answer.

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

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09 Dec 2013, 06:22

LOGIC: In fact this is a divisibility problem. In order 10! - 2*(5!)^2 to be divisible by 10^n , 10! - 2*(5!)^2 and 10^n must have the same number of 10's. 10 in terms of prime factors can be written as 10 = 2*5 --> 10^n = (2^n)* (5^n). So the problem asks how many 2's equals to 5's the 10! - 2*(5!)^2 has.

Since we will loose valuable time by doing all these multiplications the easiest way is to find common terms. 10! - 2*(5!)^2 = 5!*6*7*8*9*10 - 2(5!)^2 = 5!*(2*3)*7*(2*4)*9*(2*5) - 2(5!)^2 = 5!* 5!* (7*9*2*2) - 2(5!)^2 = (5!)^2*252 -2(5!)^2 =(5!)^2 (252-2)=(5!)^2* 250

Since it is a product we can take its term separately. # of 2's and 5's in (5!) #of 2's in 5!: 5!/2 + 5!/4 = 2+1 = 3 #of 5's in 5!: 5!/5=1 BUT it is (5!)^2 --> There are 2 5's AND 6 2's in in (5!)^2 (5! has more factors of 2 than factors of 5 since every second number is multiple of 2. Thus, it would have been more effective to search only for factors of 5)

# of 2's and 5's in 250 With prime factorization 250 = 2*5^3. Thus there are 3 5's AND 1 2 in 250

In sum: (5!)^2* 250 = 2^6 * 5^2 * 2 * 5^3 * .... = 2^7 * 5^5. We want equal number of 5's and 2's so the greatest number of 5's and 2's in the (5!)^2* 250 is 5.

answer: E

gmatclubot

Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest
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09 Dec 2013, 06:22

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