Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 11 Mar 2015
Posts: 18
GPA: 3.4

Re: Math Revolution and GMAT Club Contest! If the greatest common divisor
[#permalink]
Show Tags
19 Dec 2015, 01:29
GCD =120
120 = 5! So (n2)=5!
Hence n=7
Answer choice is D !!



Intern
Joined: 05 Jun 2013
Posts: 35

Re: Math Revolution and GMAT Club Contest! If the greatest common divisor
[#permalink]
Show Tags
19 Dec 2015, 03:18
(n+2)!, (n2)!, and (n+4)! is 120 GCD of two numbers would be the lowest powers of prime numbers from each of the two numbers!
We can try out one of answer choices say 3 n=3, which is GCD of 5!, 3!,9! 5! = 1*2*3*2*2*5 , 3! = 1*2*3, 9!= 1*2*3*2*2*5*2*3*7*2*2*2*3*3 Therefore of GCD when n =3 is 6.
Since all 3 numbers are factorials , obviously the middle number is factor of other two. Therefore to get GCD of 120, (n2)! = 120 which is (n2)! = 5! and therefore n =7.
Hence answer choice D



Moderator
Joined: 22 Jun 2014
Posts: 1053
Location: India
Concentration: General Management, Technology
GPA: 2.49
WE: Information Technology (Computer Software)

Re: Math Revolution and GMAT Club Contest! If the greatest common divisor
[#permalink]
Show Tags
19 Dec 2015, 03:18
We can rewrite the factors to see what is there in common: (n+4)! = (n+4)!*(n+3)!*(n+2)!*(n+1)!*n*(n1)!*(n2)! (n+2)! = (n+2)!*(n+1)!*n*(n1)!*(n2)! (n2)! = (n2)! greatest common divisor of (n+2)!, (n2)!, and (n+4)! is (n2)! because this common in both. it can divide all the three numbers. its value is given as 120. hence (n2)! = 120 = 5*4*3*2*1 = 5! = (72)! n=7. Option D is the correct answer.
_________________
 Target  720740 http://gmatclub.com/forum/informationonnewgmatesrreportbeta221111.html http://gmatclub.com/forum/listofoneyearfulltimembaprograms222103.html



Intern
Joined: 08 Aug 2015
Posts: 4
Location: India
WE: Manufacturing and Production (Aerospace and Defense)

Re: Math Revolution and GMAT Club Contest! If the greatest common divisor
[#permalink]
Show Tags
20 Dec 2015, 03:19
GCD depends on least of (n+2)!, (n2)! and (n+2)!, (n2)! being the least of the three it shall be a multiple of 120 n=7 gives 5!=120 Answer is D



Math Expert
Joined: 02 Sep 2009
Posts: 47206

Re: Math Revolution and GMAT Club Contest! If the greatest common divisor
[#permalink]
Show Tags
20 Dec 2015, 10:47
Bunuel wrote: Math Revolution and GMAT Club Contest Starts! QUESTION #11:If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120, what is the value of n? A. 4 B. 5 C. 6 D. 7 E. 3 Check conditions below: Math Revolution and GMAT Club ContestThe Contest Starts November 28th in Quant Forum We are happy to announce a Math Revolution and GMAT Club Contest For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday). To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize: PS + DS course with 502 videos that is worth $299! All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
Thank you! MATH REVOLUTION OFFICIAL SOLUTION:\((n+2)!=(n+2)(n+1)n(n1)(n2)!\) and \((n+4)!=(n+4)(n+3)(n+2)(n+1)n(n1)(n2)!\). From this, we can observe that (n+4)!, (n+2)! and (n2)! include (n2)!, which means the greatest common divisor for them is (n2)!. So, since we get n2=5 and n=7 from \((n2)!=120=5!\), the correct answer is D.
_________________
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



Intern
Joined: 09 Mar 2018
Posts: 8

Re: Math Revolution and GMAT Club Contest! If the greatest common divisor
[#permalink]
Show Tags
25 Apr 2018, 11:37
Prime factorize 120 = 2*2*2*3*5 Pick up the answer choices and fit those in the question.
For e.g., picking 7 would give you  9!, 5!, 11! All these three numbers will definitely contain at least 3 2's, one 3 and one 5.
Hence D is the answer



Manager
Joined: 31 Jan 2018
Posts: 72

Re: Math Revolution and GMAT Club Contest! If the greatest common divisor
[#permalink]
Show Tags
25 Apr 2018, 13:04
Bunuel wrote: Math Revolution and GMAT Club Contest Starts! QUESTION #11:If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120, what is the value of n? A. 4 B. 5 C. 6 D. 7 E. 3 Check conditions below: Math Revolution and GMAT Club ContestThe Contest Starts November 28th in Quant Forum We are happy to announce a Math Revolution and GMAT Club Contest For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday). To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize: PS + DS course with 502 videos that is worth $299! All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
Thank you! The answer must be 7 (D) 120 is basically a 5! so all the numbers (n+2)!, (n2) and (n+4)! should be divided completely by 5! Looking into the options provided, only for n = 7 all the numbers are divided by 5!




Re: Math Revolution and GMAT Club Contest! If the greatest common divisor &nbs
[#permalink]
25 Apr 2018, 13:04



Go to page
Previous
1 2
[ 27 posts ]



