Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Mar 2709:30 AM PDT
-10:30 AM PDT
Preparing for the GMAT? Avoid these 5 major mistakes that can lower your score and hurt your MBA dreams. In this expert panel discussion, 4 top GMAT coaches - GMAT Ninja, Jeff Miller from TTP, Rida Shafiq from eGMAT, and Chris Gentry from Manhattan Prep- Mar 27
10:00 AM PDT
-11:00 AM PDT
What happens after getting an MBA? Is it worth the investment? To find out, I interviewed 10 GMAT Club professionals who have completed their MBAs and built careers in consulting, finance, tech, and entrepreneurship. Their answers might surprise you! 
Mar 2911:30 AM EDT
-12:30 PM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Mar 3005:30 AM PDT
-07:30 AM PDT
We know Strengthen and Weaken questions account for more than 50% of the CR questions on the GMAT. With CR becoming even more important on GMAT Focus, it's time you strengthen your weaknesses with an approach that improves your solving time and accuracy!
Mar 3011:00 AM IST
-01:00 PM IST
Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions.
Mar 3110:00 AM EDT
-11:59 PM EDT
The Target Test Prep team has recently introduced TTP OnDemand GMAT, a 400-hour live masterclass video course created by Scott Woodbury-Stewart, founder and CEO of Target Test Prep.
Apr 0308:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
03 Dec 2010, 18:12
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
64% (01:17) correct
36%
(01:37)
wrong
based on 3122
sessions
History
Date
Time
Result
Not Attempted Yet
What is the greatest value of m such that 4^m is a factor of 30! ?
(A) 13
(B) 12
(C) 11
(D) 7
(E) 6
(A) 13
(B) 12
(C) 11
(D) 7
(E) 6
03 Dec 2010, 19:55
Kudos
Bookmarks
Ok.
So lets take a simple example first:
What is the greatest value of m such that 2^m is a factor of 10!
We need to find the number of 2s in 10!
Method:
Step 1: 10/2 = 5
Step 2: 5/2 = 2
Step 3: 2/2 = 1
Step 4: Add all: 5 + 2 + 1 = 8 (Answer)
Logic:
10! = 1*2*3*4*5*6*7*8*9*10
Every alternate number will have a 2. Out of 10 numbers, 5 numbers will have a 2. (Hence Step 1: 10/2 = 5)
These 5 numbers are 2, 4, 6, 8, 10
Now out of these 5 numbers, every alternate number will have another 2 since it will be a multiple of 4 (Hence Step 2: 5/2 = 2)
These 2 numbers will be 4 and 8.
Out of these 2 numbers, every alternate number will have yet another 2 because it will be a multiple of 8. (Hence Step 3: 2/2 = 1)
This single number is 8.
Now all 2s are accounted for. Just add them 5 + 2 + 1 = 8 (Hence Step 4)
These are the number of 2s in 10!.
Similarly, you can find maximum power of any prime number in any factorial.
If the question says 4^m, then just find the number of 2s and half it.
If the question says 6^m, then find the number of 3s and that will be your answer (because to make a 6, you need a 3 and a 2. You have definitely more 2s in 10! than 3s. So number of 3s is your limiting condition.)
Let's take this example: Maximum power of 6 in 40!.
40/3 = 13
13/3 = 4
4/3 = 1
Total number of 3s = 13 + 4 + 1 = 18
40/2 = 20
20/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1
Total number of 2s in 40! is 20+10 + 5 + 2 + 1 = 38
Definitely, number of 3s are less so we can make only 18 6s in spite of having many more 2s.
Usually, the greatest prime number will be the limiting condition.
And if you are still with me, then tell me, what happens if I ask for the greatest power of 12 in 40!?
So lets take a simple example first:
What is the greatest value of m such that 2^m is a factor of 10!
We need to find the number of 2s in 10!
Method:
Step 1: 10/2 = 5
Step 2: 5/2 = 2
Step 3: 2/2 = 1
Step 4: Add all: 5 + 2 + 1 = 8 (Answer)
Logic:
10! = 1*2*3*4*5*6*7*8*9*10
Every alternate number will have a 2. Out of 10 numbers, 5 numbers will have a 2. (Hence Step 1: 10/2 = 5)
These 5 numbers are 2, 4, 6, 8, 10
Now out of these 5 numbers, every alternate number will have another 2 since it will be a multiple of 4 (Hence Step 2: 5/2 = 2)
These 2 numbers will be 4 and 8.
Out of these 2 numbers, every alternate number will have yet another 2 because it will be a multiple of 8. (Hence Step 3: 2/2 = 1)
This single number is 8.
Now all 2s are accounted for. Just add them 5 + 2 + 1 = 8 (Hence Step 4)
These are the number of 2s in 10!.
Similarly, you can find maximum power of any prime number in any factorial.
If the question says 4^m, then just find the number of 2s and half it.
If the question says 6^m, then find the number of 3s and that will be your answer (because to make a 6, you need a 3 and a 2. You have definitely more 2s in 10! than 3s. So number of 3s is your limiting condition.)
Let's take this example: Maximum power of 6 in 40!.
40/3 = 13
13/3 = 4
4/3 = 1
Total number of 3s = 13 + 4 + 1 = 18
40/2 = 20
20/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1
Total number of 2s in 40! is 20+10 + 5 + 2 + 1 = 38
Definitely, number of 3s are less so we can make only 18 6s in spite of having many more 2s.
Usually, the greatest prime number will be the limiting condition.
And if you are still with me, then tell me, what happens if I ask for the greatest power of 12 in 40!?
04 Dec 2010, 02:42
Kudos
Bookmarks
rxs0005
What is the greatest value of m such that 4m is a factor of 30! ?
(A) 13
(B) 12
(C) 11
(D) 7
(E) 6
is there an easier way to do this other than brute force
(A) 13
(B) 12
(C) 11
(D) 7
(E) 6
is there an easier way to do this other than brute force
Finding the number of powers of a prime number k, in the n!.
The formula is:
\(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)
For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!)
\(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!.
Check for more: everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html
Back to the original question:
What is the greatest value of m such that 4^m is a factor of 30! ?
A. 13
B. 12
C. 11
D. 7
E. 6
First of all it should be 4^m instead of 4m.
Now, \(4^m=2^{2m}\), so we should check the highest power of 2 in 30!: \(\frac{30}{2}+\frac{30}{4}+\frac{30}{8}+\frac{30}{16}=15+7+3+1=26\). So the highest power of 2 in 30! is 26 --> \(2m=26\) --> \(m=13\).
Answer: A.
Hope it's clear.
