[Math Revolution GMAT math practice question]
What is the smallest integer n such that 2^8 is a factor of n!?
A. 8
B. 10
C. 12
D. 14
E. 16
What is the smallest integer n such that 2^8 is a factor of
This topic has expert replies
- Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]
- fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
$$?\,\,\,:\,\,\,n \ge 0\,\,{\rm{int}}\,\,\,{\rm{min}}\,\,{\rm{such}}\,\,{\rm{that}}\,\,\,{{n!} \over {{2^8}}} = {\mathop{\rm int}} $$Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
What is the smallest integer n such that 2^8 is a factor of n!?
A. 8
B. 10
C. 12
D. 14
E. 16
$${{n!} \over {{2^8}}} = {\mathop{\rm int}} \,\,\,\,\, \Leftrightarrow \,\,\,\,n!\,\,{\rm{has}}\,\,\left( {{\rm{at}}\,\,{\rm{least}}} \right)\,\,{\rm{eight}}\,\,{\rm{2s}}$$
$$n = 8\,\,\,\,\, \Rightarrow \,\,\,\,\,8! = \left( {{2^3}} \right) \cdot 7 \cdot \left( {2 \cdot 3} \right) \cdot 5 \cdot \left( {{2^2}} \right) \cdot 3 \cdot 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{seven}}\,\,2{\rm{s}}$$
$${\rm{Conclusion:}}\,\,{\rm{10}}\,\,{\rm{is}}\,\,{\rm{the}}\,\,{\rm{answer}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{B}} \right)$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
-
- Legendary Member
- Posts: 2214
- Joined: Fri Mar 02, 2018 2:22 pm
- Followed by:5 members
n=a number from 0 to infinity i.e
$$0,1,2,3,4.................n$$
$$n!=The\ factorial\ of\ a\ number\ from\ 0\ to\ \inf inity$$
$$i.e\ 0!,1!,2!,3!,4!.......................n!$$
such that 2^8 is a factor of n!
when n=8
n!=8!
$$8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=720$$
$$Factors\ of\ 720=2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot5=2^4\cdot3^2\cdot5$$
$$;\ 2^8is\ not\ a\ factor\ of\ \ 8!$$
$$10!=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=3,628,800$$
$$factors\ of\ 3,628,800\ =2^8\cdot3^4\cdot5^2\cdot7\ \ $$
$$2^{8\ }is\ a\ factor\ of\ \ 10!$$
$$12!,\ 14!\ and\ 16!\ are\ greater\ than\ 10!\ hence\ 10\ is\ the\ smallest\ integer\ such\ that\ 2^8\ is\ a\ factor\ of\ 10!$$
$$answer\ =\ Option\ B$$
$$0,1,2,3,4.................n$$
$$n!=The\ factorial\ of\ a\ number\ from\ 0\ to\ \inf inity$$
$$i.e\ 0!,1!,2!,3!,4!.......................n!$$
such that 2^8 is a factor of n!
when n=8
n!=8!
$$8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=720$$
$$Factors\ of\ 720=2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot5=2^4\cdot3^2\cdot5$$
$$;\ 2^8is\ not\ a\ factor\ of\ \ 8!$$
$$10!=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=3,628,800$$
$$factors\ of\ 3,628,800\ =2^8\cdot3^4\cdot5^2\cdot7\ \ $$
$$2^{8\ }is\ a\ factor\ of\ \ 10!$$
$$12!,\ 14!\ and\ 16!\ are\ greater\ than\ 10!\ hence\ 10\ is\ the\ smallest\ integer\ such\ that\ 2^8\ is\ a\ factor\ of\ 10!$$
$$answer\ =\ Option\ B$$
- Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
=>
Since 2 = 2^1, 4 = 2^2, 6 = 2^1*3, 8 = 2^3, 10=2^1*5, the prime factorization of 10! = 1*2*3*...*10 has the form 28*m for some integer m, where m and 2 are relatively prime. Note that 9! = 2^7*k for some integer k, where k and 2 are relatively prime.
Therefore, the answer is B.
Answer: B
Since 2 = 2^1, 4 = 2^2, 6 = 2^1*3, 8 = 2^3, 10=2^1*5, the prime factorization of 10! = 1*2*3*...*10 has the form 28*m for some integer m, where m and 2 are relatively prime. Note that 9! = 2^7*k for some integer k, where k and 2 are relatively prime.
Therefore, the answer is B.
Answer: B
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7245
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We see that 10! has 8 factors of 2. Let's write out 10! as 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. Now let's factor out each factor of 2 from 10!Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
What is the smallest integer n such that 2^8 is a factor of n!?
A. 8
B. 10
C. 12
D. 14
E. 16
2 → 1 factor of 2
4 = 2^2 → 2 factors of 2
6 = 2 x 3 → 1 factor of 2
8 = 2^3 → 3 factors of 2
10 = 2 x 5 → 1 factor of 2
However, 8! (or 9!) only has 7 factors of 2, so the smallest value of n is 10.
Answer: B
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews