GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 08 Dec 2019, 03:26 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest

Author Message
TAGS:

### Hide Tags

SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1727
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

$$10! = (5!)^2 * 2 * 3 * 7 * 6$$

So,

$$(5!)^2 * 2 * 3 * 7 * 6 - 2 * (5!)^2$$

$$= 2 * (5!)^2 (3 * 7 * 6 - 1)$$

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

$$= 2 * 5^2 * 4^2 * 3^2 * 2^2 * 5^3$$

$$= 20^2 * 10^3 * 3^2$$

Answer = 3+2 = 5 = E
Intern  Joined: 20 May 2014
Posts: 31
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

Dear Bunuel (or whoever sees this)

in this step : 250 * (5!)^2 = 250 x 120 x 120 = 5^2 x 12 x 12 x 1000, I don't get why you multiply by 1000 and also why is 250 turning to 5^2 , and the 120's turning to 12's?

Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 59590
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

sagnik2422 wrote:
Dear Bunuel (or whoever sees this)

in this step : 250 * (5!)^2 = 250 x 120 x 120 = 5^2 x 12 x 12 x 1000, I don't get why you multiply by 1000 and also why is 250 turning to 5^2 , and the 120's turning to 12's?

Thanks

$$250*(5!)^2=250*120^2=250*120*120=(25*10)*(12*10)*(12*10)=25*12*12*1000$$.
_________________
Intern  Joined: 14 Jul 2014
Posts: 1
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

How many questions of that difficulty should you expect to see to get a 750, for example?
Intern  Joined: 21 Jul 2014
Posts: 15
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

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.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59590
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

apsForGmat 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.

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.

The point is that 2*81*7 - 9 = 3^2*5^3. 2^5 provides enough 2's for 5^3, which will result in additional 3 0's.

Does this make sense?
_________________
Intern  Joined: 06 Jun 2013
Posts: 5
Location: United States
Concentration: Technology, Operations
GMAT 1: 600 Q42 V30 GPA: 3.55
WE: Engineering (Computer Software)
If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

1
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)

Hence n=5 which is option E.
Manager  B
Joined: 30 Mar 2013
Posts: 101
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

i did it this way and fail to see where I'm wrong:
10!/10^n - 2*(5!)^2/10^n

10! contains 2 trailing 0s....
5!=120 so we have 120*120*2. Thats two zeros.
So shouldn't the answer be 4?

We are allowed to split up fractions in this way, aren't we?
SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1727
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

usre123 wrote:
i did it this way and fail to see where I'm wrong:
10!/10^n - 2*(5!)^2/10^n

10! contains 2 trailing 0s....
5!=120 so we have 120*120*2. Thats two zeros.
So shouldn't the answer be 4?

We are allowed to split up fractions in this way, aren't we?

Yes, we are allowed to split the fractions.

As far as the computation of this problem is concerned, we require to get rid of the "subtraction" sign first to get the required answer
Manager  B
Joined: 30 Mar 2013
Posts: 101
If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

PareshGmat wrote:
usre123 wrote:
i did it this way and fail to see where I'm wrong:
10!/10^n - 2*(5!)^2/10^n

10! contains 2 trailing 0s....
5!=120 so we have 120*120*2. Thats two zeros.
So shouldn't the answer be 4?

We are allowed to split up fractions in this way, aren't we?

Yes, we are allowed to split the fractions.

As far as the computation of this problem is concerned, we require to get rid of the "subtraction" sign first to get the required answer

Oh haha. simple logic. Got it, thanks!! So this isn't the right way to solve.
Although I tried to do 4/2- 3/2 which gives me 1/2, and (4-3)/2 also gives the same answer. I wonder why that doesnt work here
SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1727
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

usre123 wrote:
PareshGmat wrote:
usre123 wrote:
i did it this way and fail to see where I'm wrong:
10!/10^n - 2*(5!)^2/10^n

10! contains 2 trailing 0s....
5!=120 so we have 120*120*2. Thats two zeros.
So shouldn't the answer be 4?

We are allowed to split up fractions in this way, aren't we?

Yes, we are allowed to split the fractions.

As far as the computation of this problem is concerned, we require to get rid of the "subtraction" sign first to get the required answer

Oh haha. simple logic. Got it, thanks!! So this isn't the right way to solve.
Although I tried to do 4/2- 3/2 which gives me 1/2, and (4-3)/2 also gives the same answer. I wonder why that doesnt work here

Actually in the problem in question, they are asking to count the zero's; that's why its better to calculate the numerator first.

Take this example:

$$\frac{2333}{10} - \frac{333}{10}$$ ......... What is the power of 10 in this question... ??

Without solving, there is NO power of 10 in the numerator; however once you solve, its $$10^3$$ in the numerator.......

Does it makes sense?
Manager  B
Joined: 30 Mar 2013
Posts: 101
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

10! contains 2 trailing 0s....
5!=120 so we have 120*120*2. Thats two zeros.
So shouldn't the answer be 4?

We are allowed to split up fractions in this way, aren't we?[/quote]

Yes, we are allowed to split the fractions.

As far as the computation of this problem is concerned, we require to get rid of the "subtraction" sign first to get the required answer[/quote]

perfect sense, thank you!

Oh haha. simple logic. Got it, thanks!! So this isn't the right way to solve.
Although I tried to do 4/2- 3/2 which gives me 1/2, and (4-3)/2 also gives the same answer. I wonder why that doesnt work here[/quote]

Actually in the problem in question, they are asking to count the zero's; that's why its better to calculate the numerator first.

Take this example:

$$\frac{2333}{10} - \frac{333}{10}$$ ......... What is the power of 10 in this question... ??

Without solving, there is NO power of 10 in the numerator; however once you solve, its $$10^3$$ in the numerator.......

Does it makes sense?[/quote]

makes perfect sense, thank you. is this representative of a 750+ questions? or a 700+ questions
Intern  Joined: 02 Jul 2013
Posts: 1
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

For me, the easiest way to solve this question is as under:

7!*8*9*10 - 2(5!)^2
7! = 5040 & 5! = 120; Square of 120 = 12^2 *100 = 144*100=14400
Thus 5040*72*10 - 14400*2

As 504*72 = 36288;

36288*100 - 28800
3628800 - 28800 = 3600000 = 36*100000 = 36*10^5
Therefore n=5
Manager  Joined: 20 Jul 2013
Posts: 51
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

ninakath wrote:
For me, the easiest way to solve this question is as under:

7!*8*9*10 - 2(5!)^2
7! = 5040 & 5! = 120; Square of 120 = 12^2 *100 = 144*100=14400
Thus 5040*72*10 - 14400*2

As 504*72 = 36288;

36288*100 - 28800
3628800 - 28800 = 3600000 = 36*100000 = 36*10^5
Therefore n=5

Good ole' fashion brute force Mind as well simplify it bit:
10! - 2*(5!)^2
5!(6*7*8*9*10-2(5!))
5!(30240-240)
5!(30000) ----> With the 5 and 2 from 5!, we get a total of 5 zeros.
Intern  Joined: 07 Feb 2015
Posts: 45
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

Given expression : 10!-2(5!)(5!)
Since we have to find n in 10^n lets try and write it the given expression as a factor of 10
10!-2x5!x5!
5!(6.7.8.9.10-2x5!)=5!(120.7.4.9-2x5!)
since 5!=120
5!x5!(7.4.9-2)=5!x5!x250=14400x250=36x100000=36x10^5
hence greats value of n=5
E
Manager  Joined: 10 Jun 2015
Posts: 110
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

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! contains two 5's. Therefore, it ends with two 0's. The second part 2x5!x5! contains two 5's. Hence, it is definitely divisible by 100.
Math Expert V
Joined: 02 Sep 2009
Posts: 59590
If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

matvan 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

10! contains two 5's. Therefore, it ends with two 0's. The second part 2x5!x5! contains two 5's. Hence, it is definitely divisible by 100.

That's correct but that's not what the question is asking. Check here: if-10-2-5-2-is-divisible-by-10-n-what-is-the-greatest-106060.html#p830879

Hope it helps.
_________________
Intern  Joined: 04 Apr 2015
Posts: 16
Concentration: Human Resources, Healthcare
GMAT Date: 08-06-2015
GPA: 3.83
WE: Editorial and Writing (Journalism and Publishing)
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

The trick here is that it is not a product question. Addition and Subtraction can lead to formation of new numbers whose prime factorisation can give trailing zeroes.
Intern  Joined: 21 Mar 2014
Posts: 21
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

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
_________________
kinaare paaon phailane lage hian,
nadi se roz mitti kat rahi hai....
Manager  Joined: 29 Jul 2015
Posts: 152
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest  [#permalink]

### Show Tags

1
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*$$10^5$$

n=5 Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest   [#permalink] 19 Sep 2015, 13:02

Go to page   Previous    1   2   3    Next  [ 50 posts ]

Display posts from previous: Sort by

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