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

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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 27 Mar 2014, 22:21
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
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 18 Jun 2014, 09:23
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
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 18 Jun 2014, 09:35
Expert's post
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.
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 14 Jul 2014, 01:23
How many questions of that difficulty should you expect to see to get a 750, for example?
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 18 Aug 2014, 10:47
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.
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 18 Aug 2014, 11:26
Expert's post
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.

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.


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

Hence n=5 which is option E.
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 21 Oct 2014, 01:13
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?
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 21 Oct 2014, 01:26
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
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If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 21 Oct 2014, 01:50
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
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 21 Oct 2014, 02:09
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?
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 21 Oct 2014, 13:05
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
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 11 Nov 2014, 09:18
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
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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink] New post 29 Jan 2015, 16:41
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.
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest   [#permalink] 29 Jan 2015, 16:41

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