Author 
Message 
TAGS:

Hide Tags

Manager
Status: Preparing Apps
Joined: 04 Mar 2009
Posts: 91
Concentration: Marketing, Strategy
GMAT 1: 650 Q48 V31 GMAT 2: 710 Q49 V38
WE: Information Technology (Consulting)

If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
10 Dec 2010, 07:20
Question Stats:
26% (01:30) correct 74% (01:28) wrong based on 3770 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 48110

Re: factorial one
[#permalink]
Show Tags
10 Dec 2010, 07:57
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. \(10!=5!*6*7*8*9*10=5!*(2*3)*7*(4*2)*9*(5*2)=5!*(2*3*4*5)*(7*2*9*2)*=252*(5!)^2\); \(252*(5!)^22*(5!)^2=250*(5!)^2=250*120*120=5^2*12*12*1,000\) > 5^2 with two 2s 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 \(252*(5!)^22*(5!)^2=250*(5!)^2=250*120*120=1,000*25*12*12=1,000*300*12=12*300,000=3,600,000\). Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Joined: 13 Aug 2012
Posts: 441
Concentration: Marketing, Finance
GPA: 3.23

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
21 Dec 2012, 01:07
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 \(=10*9*8*7*6*5!  2*5!*5*4*3*2*1\) \(=5!*10*3*8(125)\) \(=5*(2*2)*3*2*(5*2)*24*5*5*5\) \(=5*2*5*2*5*2*5*2*24*5*3\) \(=10^4*24*5*3\) \(=10^4*5*2*12*3\) \(=10^5*36\) Answer: E
_________________
Impossible is nothing to God.




Director
Joined: 23 Apr 2010
Posts: 553

Re: factorial one
[#permalink]
Show Tags
29 Dec 2010, 02:24
Bunuel, I don't understand: the number of trailing zeros for 10! is 2 according to the formula you explained there: gmatclubm12100599.htmlThe number of trailing zeros for \((5!)^2\) is 2 (one for 5!, again, according to the same formula and common sense). So the difference of two numbers should give 2 trailing zeros: xxx00  yyy00  zzz00 Could you please explain again? Thank you.



Math Expert
Joined: 02 Sep 2009
Posts: 48110

Re: factorial one
[#permalink]
Show Tags
29 Dec 2010, 02:34
nonameee wrote: Bunuel, I don't understand: the number of trailing zeros for 10! is 2 according to the formula you explained there: gmatclubm12100599.htmlThe number of trailing zeros for \((5!)^2\) is 2 (one for 5!, again, according to the same formula and common sense). So the difference of two numbers should give 2 trailing zeros: xxx00  yyy00  zzz00
Could you please explain again? Thank you. The red part is not correct. Consider this: 123,000 23,000  100,000 By the way: 10!2*(5!)^2=3,628,80028,800=3,600,000  five trailing zeros.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 23 Apr 2010
Posts: 553

Re: factorial one
[#permalink]
Show Tags
29 Dec 2010, 02:40
Yes, you are right. I will read your explanation above. Thank you.



Intern
Joined: 02 May 2013
Posts: 24
Concentration: International Business, Technology
WE: Engineering (Aerospace and Defense)

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
23 May 2013, 22:15
10! = 2^8*3^4*5^2*7= 2 zeros (10*5*2)
after rearranging 10! = 100*2^6*3^4*7
2*5!^2 = 2^7*3^2*5^2=2 zeroes(5*2*5*2)
after rearranging 100*2^5*3^2
now 10!2*5!^2 = 100(2^6*3^4*7  2^5*3^2) = 100*2^5*3^2(2*3^2*7  1) =100*2^5*3^2*125 =100 *2^5*3^2 *5^3 = 10^5 *2^2*3^2
OA:E



Retired Moderator
Joined: 27 Aug 2012
Posts: 1115

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
22 Jul 2013, 05:33
Hi Bunuel, I've a question based on the calculation given in the page 16 of GC Math book  20000 * (1.03)^8=25335.4How to carry out this sort of calculations ( (1.03)^8) during 2 minutetime constraint ?
_________________
UPDATED : eGMAT SC ResourcesConsolidated  ALL RC ResourcesConsolidated  ALL SC ResourcesConsolidated  UPDATED : AWA compilations109 Analysis of Argument Essays  GMAC's IR Prep Tool
Calling all Columbia (CBS) MBA Applicants: (2018 Intake) Class of 2020 !!! NEW !!!
GMAT Club guide  OG 111213  Veritas Blog  Manhattan GMAT Blog
KUDOS please, if you like the post or if it helps



Manager
Joined: 26 Sep 2013
Posts: 200
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
26 Sep 2013, 10:31
mbaiseasy 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 \(=10*9*8*7*6*5!  2*5!*5*4*3*2*1\) \(=5!*10*3*8(125)\)\(=5*(2*2)*3*2*(5*2)*24*5*5*5\) \(=5*2*5*2*5*2*5*2*24*5*3\) \(=10^4*24*5*3\) \(=10^4*5*2*12*3\) \(=10^5*36\) Answer: E Can someone explain the bolded part to me? Maybe I'm missing something, but going from that first step to the 2nd step makes no sense to me. 10*9*8*7*6*52*5!*5*4*3*2*1, shouldn't that equal 10*9*8*7*62*5*4*3*2*1, since the second part of that equation contains  2*5!, and there's also another 5! in the 5*4*3*2*1 portion? Can someone link me to a page explaining the logic here, because this factorial stuff is making no sense to me. I read the factorial guide post, but it doesn't seem to have anything that applies here. Desperate for help here. I have GMATs in about 6 weeks, and these factorial problems keep beating my head in! I can't find anything that explains or helps to solve these problems at all, and the explanations make no sense to me.



Senior Manager
Joined: 08 Apr 2012
Posts: 395

Re: factorial one
[#permalink]
Show Tags
02 Nov 2013, 14:02
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. \(10!=5!*6*7*8*9*10=5!*(2*3)*7*(4*2)*9*(5*2)=5!*(2*3*4*5)*(7*2*9*2)*=252*(5!)^2\); \(252*(5!)^22*(5!)^2=250*(5!)^2=250*120*120=5^2*12*12*1,000\) > 5^2 with two 2s 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 \(252*(5!)^22*(5!)^2=250*(5!)^2=250*120*120=1,000*25*12*12=1,000*300*12=12*300,000=3,600,000\). Hope it's clear. Hi Bunuel, When I see these kinds of questions I immediatly go to the regular formula, where 10! has 2 trailing "0". A. is there a way to do it with the formula above? B. if not, can you try and specify how you come to solve this kind of question? how did you know not to use the formula above?



Math Expert
Joined: 02 Sep 2009
Posts: 48110

Re: factorial one
[#permalink]
Show Tags
03 Nov 2013, 11:45
ronr34 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. \(10!=5!*6*7*8*9*10=5!*(2*3)*7*(4*2)*9*(5*2)=5!*(2*3*4*5)*(7*2*9*2)*=252*(5!)^2\); \(252*(5!)^22*(5!)^2=250*(5!)^2=250*120*120=5^2*12*12*1,000\) > 5^2 with two 2s 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 \(252*(5!)^22*(5!)^2=250*(5!)^2=250*120*120=1,000*25*12*12=1,000*300*12=12*300,000=3,600,000\). Hope it's clear. Hi Bunuel, When I see these kinds of questions I immediatly go to the regular formula, where 10! has 2 trailing "0". A. is there a way to do it with the formula above? B. if not, can you try and specify how you come to solve this kind of question? how did you know not to use the formula above? We need to find the number of trailing zeros of 10!  2*(5!)^2, not 10!. You cannot find the number of trailing zeros of 10!  2*(5!)^2 directly with that formula. To do that you should use algebraic manipulations as shown.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 08 Apr 2012
Posts: 395

Re: factorial one
[#permalink]
Show Tags
17 Nov 2013, 14:44
Thanks. But when I try to find the number of "10" in 250 I tried doing factorization of 250 (2*5*5*5) and only got one "10" out of it.... Is there another way it can be done if I don't see what you see the numbers?



Math Expert
Joined: 02 Sep 2009
Posts: 48110

Re: factorial one
[#permalink]
Show Tags
17 Nov 2013, 14:56



Senior Manager
Joined: 08 Apr 2012
Posts: 395

Re: factorial one
[#permalink]
Show Tags
17 Nov 2013, 23:29
Bunuel wrote: 250 has one trailing zero, So if I follow your logic, in 250*120*120 there should only be 3 trailing zeros. one in 250, one in 120 and one in 120 again.... This rises our count to 3, yet somewhere in there are hiding an additional 2 trailing zeros. If I were to arrive at this equation on test day, I would have selected "3". I am trying to see where in this form of the equation the additional zeros are hiding...



Manager
Joined: 28 Apr 2013
Posts: 143
Location: India
GPA: 4
WE: Medicine and Health (Health Care)

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
18 Nov 2013, 19:12
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 If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest value of n? =10!2(5!*5!)/10^n =10*9*8*7*6*5!2(5!*5!)/10^n =5!*10*8*3(9*7*21)/10^n 2400*5!/10^n 2400*5*4*3*2/10^n 2400*10*12/10^n max value of n for the above can be 5
_________________
Thanks for Posting
LEARN TO ANALYSE
+1 kudos if you like



Senior Manager
Joined: 08 Apr 2012
Posts: 395

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
19 Nov 2013, 10:01
rango 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 If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest value of n? =10!2(5!*5!)/10^n =10*9*8*7*6*5!2(5!*5!)/10^n =5!*10*8*3(9*7*21)/10^n 2400*5!/10^n 2400*5*4*3*2/10^n 2400*10*12/10^n max value of n for the above can be 5 How did you make this calculation?



Manager
Joined: 28 Apr 2013
Posts: 143
Location: India
GPA: 4
WE: Medicine and Health (Health Care)

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
19 Nov 2013, 16:37
ronr34 wrote: rango 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 If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest value of n? =10!2(5!*5!)/10^n =10*9*8*7*6*5!2(5!*5!)/10^n =5!*10*8*3(9*7*21)/10^n 2400*5!/10^n 2400*5*4*3*2/10^n 2400*10*12/10^n max value of n for the above can be 5 How did you make this calculation? in the above calculation; i have deliberately eliminated (9*7*21) after step 3 for the ease of calculation and understanding. Try to find how many 10 multiples can be required to divide the numerator. 2400 req 10^2; 10*12 req 10^2; also 9*7*21 req 10^2; so after adding it comes to be 10^6 ; but we have in options max n0. 5 ; so right answer.
_________________
Thanks for Posting
LEARN TO ANALYSE
+1 kudos if you like



Intern
Joined: 24 Nov 2013
Posts: 3

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
09 Dec 2013, 07:22
LOGIC: In fact this is a divisibility problem. In order 10!  2*(5!)^2 to be divisible by 10^n , 10!  2*(5!)^2 and 10^n must have the same number of 10's. 10 in terms of prime factors can be written as 10 = 2*5 > 10^n = (2^n)* (5^n). So the problem asks how many 2's equals to 5's the 10!  2*(5!)^2 has.
Since we will loose valuable time by doing all these multiplications the easiest way is to find common terms. 10!  2*(5!)^2 = 5!*6*7*8*9*10  2(5!)^2 = 5!*(2*3)*7*(2*4)*9*(2*5)  2(5!)^2 = 5!* 5!* (7*9*2*2)  2(5!)^2 = (5!)^2*252 2(5!)^2 =(5!)^2 (2522)=(5!)^2* 250
Since it is a product we can take its term separately. # of 2's and 5's in (5!) #of 2's in 5!: 5!/2 + 5!/4 = 2+1 = 3 #of 5's in 5!: 5!/5=1 BUT it is (5!)^2 > There are 2 5's AND 6 2's in in (5!)^2 (5! has more factors of 2 than factors of 5 since every second number is multiple of 2. Thus, it would have been more effective to search only for factors of 5)
# of 2's and 5's in 250 With prime factorization 250 = 2*5^3. Thus there are 3 5's AND 1 2 in 250
In sum: (5!)^2* 250 = 2^6 * 5^2 * 2 * 5^3 * .... = 2^7 * 5^5. We want equal number of 5's and 2's so the greatest number of 5's and 2's in the (5!)^2* 250 is 5.
answer: E



Retired Moderator
Joined: 20 Dec 2013
Posts: 180
Location: United States (NY)
GMAT 1: 640 Q44 V34 GMAT 2: 710 Q48 V40 GMAT 3: 720 Q49 V40
GPA: 3.16
WE: Consulting (Venture Capital)

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
08 Jan 2014, 23:49
10!  2*(5!)^2 =[2(5!)^2][1261] =[2(5!)^2][(5^3)] five 5's all together and thus five 10's...(5^3) can borrow 2's from factorial.
_________________
MY GMAT BLOG  ADVICE  OPINIONS  ANALYSIS



Senior Manager
Joined: 23 Oct 2010
Posts: 361
Location: Azerbaijan
Concentration: Finance

Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
Show Tags
01 Feb 2014, 04:12
10!  2*(5!)^2 =5!1098765!*5!*2=5!*5!*2(2*7*91)=5!*5!*5^3 we have five 5s
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth




Re: If 10!  2*(5!)^2 is divisible by 10^n, what is the greatest &nbs
[#permalink]
01 Feb 2014, 04:12



Go to page
1 2 3
Next
[ 46 posts ]



