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

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If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 10 Dec 2010, 06:20
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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
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Re: factorial one [#permalink] New post 10 Dec 2010, 06:57
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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.

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Re: factorial one [#permalink] New post 29 Dec 2010, 01:24
Bunuel, I don't understand:

the number of trailing zeros for 10! is 2 according to the formula you explained there:
gmat-club-m12-100599.html

The number of trailing zeros for (5!)^2 is 2 (one for 5!, again, according to the same formula and common sense).

So the difference of two numbers should give 2 trailing zeros:
xxx00
-
yyy00
--------
zzz00

Could you please explain again? Thank you.
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Re: factorial one [#permalink] New post 29 Dec 2010, 01:34
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Bunuel, I don't understand:

the number of trailing zeros for 10! is 2 according to the formula you explained there:
gmat-club-m12-100599.html

The number of trailing zeros for (5!)^2 is 2 (one for 5!, again, according to the same formula and common sense).

So the difference of two numbers should give 2 trailing zeros:
xxx00
-
yyy00
--------
zzz00

Could you please explain again? Thank you.


The red part is not correct. Consider this:

123,000
-23,000
-------
100,000

By the way: 10!-2*(5!)^2=3,628,800-28,800=3,600,000 - five trailing zeros.

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Re: factorial one [#permalink] New post 29 Dec 2010, 01:40
Yes, you are right. I will read your explanation above. Thank you.
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 21 Dec 2012, 00:07
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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


=10*9*8*7*6*5! - 2*5!*5*4*3*2*1
=5!*10*3*8(125)
=5*(2*2)*3*2*(5*2)*24*5*5*5
=5*2*5*2*5*2*5*2*24*5*3
=10^4*24*5*3
=10^4*5*2*12*3
=10^5*36

Answer: E

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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 23 May 2013, 21:15
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
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 22 Jul 2013, 04:33
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Hi Bunuel,
I've a question based on the calculation given in the page 16 of GC Math book - 20000 * (1.03)^8=25335.4

How to carry out this sort of calculations ((1.03)^8) during 2 minute-time constraint ?

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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 26 Sep 2013, 09:31
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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


=10*9*8*7*6*5! - 2*5!*5*4*3*2*1
=5!*10*3*8(125)

=5*(2*2)*3*2*(5*2)*24*5*5*5
=5*2*5*2*5*2*5*2*24*5*3
=10^4*24*5*3
=10^4*5*2*12*3
=10^5*36

Answer: E


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.
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Re: factorial one [#permalink] New post 02 Nov 2013, 13:02
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,
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?
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Re: factorial one [#permalink] New post 03 Nov 2013, 10:45
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ronr34 wrote:
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,
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.

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Re: factorial one [#permalink] New post 17 Nov 2013, 13:44
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?
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Re: factorial one [#permalink] New post 17 Nov 2013, 13:56
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ronr34 wrote:
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?

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Re: factorial one [#permalink] New post 17 Nov 2013, 22:29
Bunuel wrote:
250 has one trailing zero,


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

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


How did you make this calculation?
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 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.

:lol:

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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 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
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 08 Jan 2014, 22:49
10! - 2*(5!)^2

=[2(5!)^2][126-1]
=[2(5!)^2][(5^3)]
five 5's all together and thus five 10's...(5^3) can borrow 2's from factorial.
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 01 Feb 2014, 03:12
10! - 2*(5!)^2 =5!-10-9-8-7-6-5!*5!*2=5!*5!*2(2*7*9-1)=5!*5!*5^3
we have five 5s

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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest   [#permalink] 01 Feb 2014, 03:12
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