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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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest   [#permalink] 18 Aug 2014, 11:26
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