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

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Intern
Joined: 13 Sep 2015
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

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05 Nov 2015, 00:10
can this sort of question come up in exam, realisitically speaking, or is it out of scope? I get the basic trailing zero stuff, but I really doubt I could do this on the fly during an exam.
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

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28 Nov 2015, 02:47
Are there more questions similar to this one?
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

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28 Nov 2015, 08:25
srinjayc wrote:
Are there more questions similar to this one?

Check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

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

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24 Jan 2017, 18:59
1) We need to figure out how many 10 multipliers the numerator consists of.
2) 10!-2*5!*5!=5!(10*9*8*7*6-2*5*4*3*2)=5!*10*3*2*4(9*2*7-1)=5*4*3*2*10*3*2*4*125
3) Since there are plenty of 2's (7), we need to calculate the number of 5's to see how many of them will be in the numerator: 5 (5*1 - one 5), 10 (5*2 - one 5), 125 (5*5*5 - three 5's). This adds up to five 5's.

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

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12 Jul 2017, 07:52
10! = 10 * 9 * 8 * 7 * 6 * 5!
10! = 30240 * 5!

5!^2 = 120 * 5!

So now we can do the math

(30240 - 2*120) * 5! = 30,000 * 5! => we get 4 zeros from 30,000 + 1 zero from 5!, leading to 5 zeros, so the answer is E
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

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14 Oct 2018, 09:22
This is an excellent question, and one which requires understanding of the following topics:

1. Divisibility & trailing zeros
2. Simplification of complex expressions (e.g. factorials)
3. Most importantly - how to manage time, and use hints within questions

The most important piece of this is part (3), as this, like other questions, needs to be solved in 2 minutes. The GMAT will not give you a question like this that doesn’t have some sort of shortcut to it. In this case, the shortcut is how to actually perform the subtraction necessary to discover the amount of trailing zeros.

Peforming the subtraction:

10! - 2(5!)^2. Immediately think “how can i subtract away that squared factorial to view my expression more clearly.”

10! = 10*9*8*7*6*5*4*3*2*1
— 5! = the trail-end of the expression, so 10! = 10*9*8*7*6*5!
— However, that’s where the GMAT “tricks” you into getting stuck. But what if you could pull out another factorial hidden within the larger portion of 10!

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

Notice — you can pull out another 5! — your goal, again, is to get rid of the 5!^2 on the other side of the expression. Take some of your prime factors out and simply make another 5!.

10! = (5!^2) * (2*3*7*3*2)

Recall, your going to need to subtract the annoying expression “2(5!)^2” — so take out another two from your 10!

10! = 2(5!^2) * (3*7*3*2)

2(5!^2) * (3*7*3*2 - 1)

Simplify, and notice your 5s... 2(5!^2) *(126-1)

(2(5!^2)) * (125) = 2(5!^2)*(5^3)

Back to the original question - how to get those trailing zeros, so you can determine the max “n” value (or max power of 10).

5! = 5*4*3*2*1 == translate to one 5, 3 twos (take two from the number 4)
You have two expressions here, so a second 5, and a total of 3 more twos.

You know your expression also has 3 more 5s in it (From the 5^3), so you’ve got 5 5s, and >5 2s.

Since the BIGGEST answer choice is n=5, you can select that immediately, and you have your answer.

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

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16 Jan 2019, 09:58
Can someone help me with this question...
I did it in a very different way and can be totally wrong..really need to know if my approach is correct or not...

I found the trailing zeros for 10!
10/2= 5
5/2=2
2/2=1

5+2+1= 8

Then I found for 5!
5/2=2
2/2=1

2+1=3

8-3=5
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

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16 Jan 2019, 10:49
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

Inline is my approach.

10! - 2*(5!)^2 divided by 10^n

5! (10*9*8*7*6 - 2* 4*3*2*1)
5! * 60 * [504 - 4]
5! * 60 * 500

$$\frac{5! * 60 * 500}{10^n}$$

$$\frac{5*4*6 * 60 * 500}{10*10*10*10*10}$$

n = 5 is the greatest value.

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

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24 Jan 2019, 16:41
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

Hello Bunuel !

I was reading your answer but still do not get it, why does the minus is not affecting the whole term?

Kind regards!
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest   [#permalink] 24 Jan 2019, 16:41

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