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

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18 Jun 2014, 10:23

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?

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

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18 Jun 2014, 10:35

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?

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

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18 Aug 2014, 11:47

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 :

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.

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

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18 Aug 2014, 12:26

Expert's post

1

This post was BOOKMARKED

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 :

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.

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

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18 Oct 2014, 01:01

1

This post received KUDOS

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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21 Oct 2014, 02:26

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 _________________

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

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21 Oct 2014, 02:50

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

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

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21 Oct 2014, 03:09

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.......

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

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21 Oct 2014, 14:05

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

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

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10 Mar 2015, 01:36

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 _________________

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

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02 Sep 2015, 02:56

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.

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

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19 Sep 2015, 08:37

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....

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

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19 Sep 2015, 13:02

1

This post received KUDOS

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

gmatclubot

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

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