What is the smallest integer n such that 2^8 is a factor of

This topic has expert replies
User avatar
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 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

User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pm
Thanked: 59 times
Followed by:33 members

by fskilnik@GMATH » Sat Dec 22, 2018 4:03 pm
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
$$?\,\,\,:\,\,\,n \ge 0\,\,{\rm{int}}\,\,\,{\rm{min}}\,\,{\rm{such}}\,\,{\rm{that}}\,\,\,{{n!} \over {{2^8}}} = {\mathop{\rm int}} $$
$${{n!} \over {{2^8}}} = {\mathop{\rm int}} \,\,\,\,\, \Leftrightarrow \,\,\,\,n!\,\,{\rm{has}}\,\,\left( {{\rm{at}}\,\,{\rm{least}}} \right)\,\,{\rm{eight}}\,\,{\rm{2s}}$$
$$n = 8\,\,\,\,\, \Rightarrow \,\,\,\,\,8! = \left( {{2^3}} \right) \cdot 7 \cdot \left( {2 \cdot 3} \right) \cdot 5 \cdot \left( {{2^2}} \right) \cdot 3 \cdot 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{seven}}\,\,2{\rm{s}}$$
$${\rm{Conclusion:}}\,\,{\rm{10}}\,\,{\rm{is}}\,\,{\rm{the}}\,\,{\rm{answer}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{B}} \right)$$

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

Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pm
Followed by:5 members

by deloitte247 » Sun Dec 23, 2018 1:18 pm
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!
$$8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=720$$
$$Factors\ of\ 720=2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot5=2^4\cdot3^2\cdot5$$
$$;\ 2^8is\ not\ a\ factor\ of\ \ 8!$$
$$10!=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=3,628,800$$
$$factors\ of\ 3,628,800\ =2^8\cdot3^4\cdot5^2\cdot7\ \ $$
$$2^{8\ }is\ a\ factor\ of\ \ 10!$$
$$12!,\ 14!\ and\ 16!\ are\ greater\ than\ 10!\ hence\ 10\ is\ the\ smallest\ integer\ such\ that\ 2^8\ is\ a\ factor\ of\ 10!$$
$$answer\ =\ Option\ B$$

User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

by Max@Math Revolution » Sun Dec 23, 2018 6:08 pm
=>
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

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 7243
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Fri Feb 15, 2019 5:21 pm
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
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!

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]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage