NEW HERE? LEARN MORE
closeclose
Last visit was: 24 Apr 2024, 09:34 It is currently 24 Apr 2024, 09:34
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

What is the greatest value of m such that 4^m is a factor of 30! ?

SORT BY:
Date
Kudos
 1   2   
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
rxs0005
Senior Manager
Senior Manager
Joined: 07 Jun 2004
Posts: 438
Own Kudos [?]: 2829 [202]
Given Kudos: 22
Location: PA
Send PM
What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   03 Dec 2010, 18:12
18
Kudos
184
Bookmarks
00:00
A
B
C
D
E

Difficulty:

35% (medium)

Question Stats:

63% (01:17) correct 37% (01:38) wrong based on 2865 sessions
Hide Show timer Statistics
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
ShowHide Answer
Official Answer
A
Most Helpful Reply
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64895 [115]
Given Kudos: 426
Location: Pune, India
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   03 Dec 2010, 19:55
52
Kudos
63
Bookmarks
Expert Reply
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!?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92902
Own Kudos [?]: 618770 [80]
Given Kudos: 81587
Send PM
CAT Tests Forum Quiz Mode
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   04 Dec 2010, 02:42
39
Kudos
41
Bookmarks
Expert Reply
rxs0005 wrote:
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


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.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92902
Own Kudos [?]: 618770 [41]
Given Kudos: 81587
Send PM
CAT Tests Forum Quiz Mode
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   04 Dec 2010, 07:11
16
Kudos
24
Bookmarks
Expert Reply
Examples about the same concept from: everything-about-factorials-on-the-gmat-85592-20.html

Highest power of 12 in 18!:
Suppose we have the number \(18!\) and we are asked to to determine the power of \(12\) in this number. Which means to determine the highest value of \(x\) in \(18!=12^x*a\), where \(a\) is the product of other multiples of \(18!\).

\(12=2^2*3\), so we should calculate how many 2-s and 3-s are in \(18!\).

Calculating 2-s: \(\frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16\). So the power of \(2\) (the highest power) in prime factorization of \(18!\) is \(16\).

Calculating 3-s: \(\frac{18}{3}+\frac{18}{3^2}=6+2=8\). So the power of \(3\) (the highest power) in prime factorization of \(18!\) is \(8\).

Now as \(12=2^2*3\) we need twice as many 2-s as 3-s. \(18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a\). So \(18!=12^8*a\) --> \(x=8\).

The highest power of 900 in 50!:

\(50!=900^xa=(2^2*3^2*5^2)^x*a\), where \(x\) is the highest possible value of 900 and \(a\) is the product of other multiples of \(50!\).

Find the highest power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\) --> \(2^{47}\);

Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\) --> \(3^{22}\);

Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\) --> \(5^{12}\);

So, \(50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b\), where \(b\) is the product of other multiples of \(50!\). So \(x=6\).

Hope it helps.
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64895 [34]
Given Kudos: 426
Location: Pune, India
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   03 Dec 2010, 19:33
17
Kudos
17
Bookmarks
Expert Reply
rxs0005 wrote:
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


I think the question is
What is the greatest value of m such that \(4^m\) is a factor of 30! ?

The easiest way to do this is the following:
Divide 30 by 2. You get 15
Divide 15 by 2. You get 7. (Ignore remainder)
Divide 7 by 2. You get 3.
Divide 3 by 2. You get 1.
Add 15+7+3+1 = 26
Since greatest power of 2 in 30! is 26, greatest power of 4 in 30! will be 13.
I will explain the logic behind the process in a while. (It will take some effort to formulate.)
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64895 [21]
Given Kudos: 426
Location: Pune, India
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   04 Dec 2010, 06:49
11
Kudos
10
Bookmarks
Expert Reply
whiplash2411 wrote:
Would that also be 18? Since 12 is 6*2 and 6 would be the limiting factor? I feel like I might be missing something

Also kudos! Your explanation was fantastic.

Posted from my mobile device


Yes, greatest power of 12 in 40! will also be 18 because
12= 3*2^2
Total number of 3s = 13 + 4 + 1 = 18 (as shown above)
Total number of 2s in 40! is 20+10 + 5 + 2 + 1 = 38 (as shown above)
So you can make 19 4s. The limiting factor is still 3.
So rxs0005, bhushan288 and whiplash2411, you all had correct answers.

The interesting thing is the maximum power of 12 in 30!
30/2 = 15
15/2 = 7
7/2 = 3
3/2 = 1
Total 2s = 15 + 7 + 3 + 1 = 26 So you can make 13 4s
30/3 = 10
10/3 = 3
3/3 = 1
Total 3s = 10 + 3 + 1 = 14!

The maximum power of 12 is 13, not 14.
Here, the limiting factor is the number of 4s (i.e. half of the number of 2s). Of course the number of 2s is more but that number gets divided by 2 to make 4s. It becomes the limiting factor.
In such cases, you will need to check for both 2 and 3.
General Discussion
User avatar
whiplash2411
SVP
SVP
Joined: 09 Jun 2010
Status:Three Down.
Posts: 1764
Own Kudos [?]: 3468 [1]
Given Kudos: 210
Concentration: General Management, Nonprofit
Schools: HBS '17 (M) Stanford '17 (A)
Send PM
GMAT ToolKit User
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   03 Dec 2010, 20:05
1
Kudos
Would that also be 18? Since 12 is 6*2 and 6 would be the limiting factor? I feel like I might be missing something

Also kudos! Your explanation was fantastic.

Posted from my mobile device
User avatar
L
stne
SVP
SVP
Joined: 27 May 2012
Posts: 1680
Own Kudos [?]: 1422 [0]
Given Kudos: 632
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   19 Jun 2012, 06:45
VeritasPrepKarishma wrote:
whiplash2411 wrote:
Would that also be 18? Since 12 is 6*2 and 6 would be the limiting factor? I feel like I might be missing something

Also kudos! Your explanation was fantastic.

Posted from my mobile device


Yes, greatest power of 12 in 40! will also be 18 because
12= 3*2^2
Total number of 3s = 13 + 4 + 1 = 18 (as shown above)
Total number of 2s in 40! is 20+10 + 5 + 2 + 1 = 38 (as shown above)
So you can make 19 4s. The limiting factor is still 3.
So rxs0005, bhushan288 and whiplash2411, you all had correct answers.

The interesting thing is the maximum power of 12 in 30!
30/2 = 15
15/2 = 7
7/2 = 3
3/2 = 1
Total 2s = 15 + 7 + 3 + 1 = 26 So you can make 13 4s
30/3 = 10
10/3 = 3
3/3 = 1
Total 3s = 10 + 3 + 1 = 14!

The maximum power of 12 is 13, not 14.
Here, the limiting factor is the number of 4s (i.e. half of the number of 2s). Of course the number of 2s is more but that number gets divided by 2 to make 4s. It becomes the limiting factor.
In such cases, you will need to check for both 2 and 3.



Applauds to both karishma and Bunuel for such wonderful way to attack these problems

karishma's way seemed easier at first , until I encountered highest power of 12 in 30!

somehow I understood : that I found the highest power of 2's then halved it to get 13

no. of 3's is 14 , so as explained the answer is 13 and not 14 because in this case the highest prime is not the limiting factor
but rather 2^2 is the limiting factor. Till here it was clear.


but I got stuck when I came to bunuels example of highest power of 900 in 50!

Karishma how to do it by your method? ( want to grasp both methods and then decide , which i'd like to use )

Highest no of 2's as shown by bunuel is 47 , now I cannot half it to find the highest power of 4 , that will give me a non integer.

Also highest power of greatest prime ( 5) is 12 and answer to this question is 6

so my concern , what is the limiting factor here , or how to approach this problem ,lets say Karishma's way .
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64895 [4]
Given Kudos: 426
Location: Pune, India
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   19 Jun 2012, 07:03
2
Kudos
2
Bookmarks
Expert Reply
stne wrote:
but I got stuck when I came to bunuels example of highest power of 900 in 50!

Karishma how to do it by your method? ( want to grasp both methods and then decide , which i'd like to use )

Highest no of 2's as shown by bunuel is 47 , now I cannot half it to find the highest power of 4 , that will give me a non integer.

Also highest power of greatest prime ( 5) is 12 and answer to this question is 6

so my concern , what is the limiting factor here , or how to approach this problem ,lets say Karishma's way .


Let me ask you a question first:

What is the limiting factor in case you want to find the highest power of 6 in 50!
Would you say it is 3? Sure! You make a 6 using a 2 and a 3. You certainly will have fewer 3s as compared to number of 2s.

What is the limiting factor in case you want to find the highest power of 36 in 50!
Think! \(36 = 2^2 * 3^2\)
Whatever the number of 2s and number of 3s, you will halve both of them. So again, the limiting factor will be 3.

What about 900?
\(900 = 2^2 * 3^2 * 5^2\)
Again, 5 will be your limiting factor here. Whatever the number of 2, 3 and 5, each will be halved. So you will still have the fewest number of half 5s (so to say).
Number of 5s is 12. So you can make six 900s from 50!

The question mark arises only when you have different powers and the smaller number has a higher power.
What is the limiting factor in case of \(2^2*3\)? Not sure. We need to check.
What is the limiting factor in case of \(2^4*3^2*7\)? Not sure. We need to check.
What is the limiting factor in case of \(3*7^2\)? It has to be 7. We find the number of 7s (which is fewer than the number of 3s) and then further half it.
User avatar
kashishh
Manager
Manager
Joined: 02 Jun 2011
Posts: 95
Own Kudos [?]: 376 [0]
Given Kudos: 11
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   25 Jun 2012, 05:38
Bunuel wrote:
rxs0005 wrote:
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


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.


Dear Bunuel,

is this formulae is applied with primes ( 2, 3 ..) only?
the doubt is because you reduced 4 into 2..
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92902
Own Kudos [?]: 618770 [1]
Given Kudos: 81587
Send PM
CAT Tests Forum Quiz Mode
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   25 Jun 2012, 05:42
1
Kudos
Expert Reply
kashishh wrote:
Bunuel wrote:
rxs0005 wrote:
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


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.


Dear Bunuel,

is this formulae is applied with primes ( 2, 3 ..) only?
the doubt is because you reduced 4 into 2..


You can apply this formula to ANY prime. I used the base of 2 instead of 4 since 4 is not a prime.
avatar
MeliMeds
Intern
Intern
Joined: 17 Apr 2015
Posts: 5
Own Kudos [?]: [0]
Given Kudos: 53
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   30 Apr 2015, 06:16
VeritasPrepKarishma wrote:
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!?


Wow thanks for the A-1 explanation and breakdown! To answer your final question, would it be 9? (applying the same principle we used in attacking the original question, where we found the number of 2's in 30 then divided it by 2 to derive the number of 4's, I subsequently found the number of 3's (as the larger of the two primes, 2 and 3 found in 12) in 40 which is 18, and because that would correspond with the number of 6's in 40, I proceeded to divide this by 2 to get 9 as a final choice... Hope I'm not joining the party too late, please advise if this is correct? :)
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64895 [2]
Given Kudos: 426
Location: Pune, India
Send PM
What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   Updated on: 17 Oct 2022, 02:04
1
Kudos
1
Bookmarks
Expert Reply
MeliMeds wrote:
VeritasPrepKarishma wrote:
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!?


Wow thanks for the A-1 explanation and breakdown! To answer your final question, would it be 9? (applying the same principle we used in attacking the original question, where we found the number of 2's in 30 then divided it by 2 to derive the number of 4's, I subsequently found the number of 3's (as the larger of the two primes, 2 and 3 found in 12) in 40 which is 18, and because that would correspond with the number of 6's in 40, I proceeded to divide this by 2 to get 9 as a final choice... Hope I'm not joining the party too late, please advise if this is correct? :)


Power of 12 in 40!
12 = 2*2*3
So to make a 12, you need two 2s and a 3.

How many 2s are there in 40!?
40/2 = 20
20/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1
Total = 38
For every 12, we need two 2s so we can make 38/2 = 19 12s.

How many 3s are there in 40!?
40/3 = 13
13/3 = 4
4/3 = 1
Total = 18
Every 12 needs only one 3 so out of 18 3s, we can make 18 12s.

We have enough 2s to make 19 12s and enough 3s to make 18 12s. So 3 is the limiting condition so you can make 18 12s.
So highest power of 12 in 40! is 18.

Originally posted by KarishmaB on 30 Apr 2015, 23:00.
Last edited by KarishmaB on 17 Oct 2022, 02:04, edited 1 time in total.
User avatar
B
ankushbagwale
Current Student
Joined: 21 Jul 2014
Posts: 72
Own Kudos [?]: 34 [0]
Given Kudos: 59
Location: India
Schools: Kelley '21 (D) Kenan-Flagler '21 (II) ISB '20 (A) Johnson '21 (WL) Booth '21 (D) Ross '21 (D) Goizueta '21 (D) LBS '21 (D) Tuck '21 (D) Insead Sept19 Intake (D) Mendoza '21 (I) Fisher '21 (I) Carlson '21 (I)
GPA: 4
WE:Project Management (Energy and Utilities)
Send PM
GMAT ToolKit User Reviews Badge
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   01 Jun 2015, 17:20
VeritasPrepKarishma wrote:
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!?



Till now I was using the formula to calculate the number of primes in any factorial.

Now, I think I could understand what is the logic behind the formule. It feels awesome to know the reasoning behind a set formula & yes I think Ican answer your question posed above.

so we have 38 # of 2's & 18 # of 3's in 40!.
Now to make one 12 = 2^2 x 3 that means we need two 2's & one 3.

So pairs of such 2's = 19.
& # of 3's =18. Here the limiting factor is 3 as it is less.

Thus we will have 18 such 12's in 40!.

Pl. let me know if I have comprehended your approach.
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64895 [0]
Given Kudos: 426
Location: Pune, India
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   01 Jun 2015, 20:17
Expert Reply
ankushbagwale wrote:
Till now I was using the formula to calculate the number of primes in any factorial.

Now, I think I could understand what is the logic behind the formule. It feels awesome to know the reasoning behind a set formula & yes I think Ican answer your question posed above.

so we have 38 # of 2's & 18 # of 3's in 40!.
Now to make one 12 = 2^2 x 3 that means we need two 2's & one 3.

So pairs of such 2's = 19.
& # of 3's =18. Here the limiting factor is 3 as it is less.

Thus we will have 18 such 12's in 40!.

Pl. let me know if I have comprehended your approach.


Yes, that's correct. And before using any formula, ensure that you fully understand why the formula is what it is, how you get it and the conditions in which you can or cannot use it.
User avatar
B
munnabhaimbbs
Intern
Intern
Joined: 17 Jul 2016
Posts: 24
Own Kudos [?]: 11 [2]
Given Kudos: 141
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   29 Dec 2016, 06:53
2
Kudos
Bunuel wrote:
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

I would go with A.
Step 1- Prime factorize the 4^m.That is equal to 2^2m.
Step2-Let's divide 30! successively by 2 and keep adding the quotient till you don't have anything left to divide
30/2=15
15/2=7
7/2=3
3/2=1
1/2=0
Thus ,15+7+3+1+0=26.
Step-3: Calculate how many such numbers can be created using the highest power of each of its prime factors
30!=2^26*k=4^13*k
User avatar
L
Abhishek009
Board of Directors
Joined: 11 Jun 2011
Status:QA & VA Forum Moderator
Posts: 6072
Own Kudos [?]: 4689 [3]
Given Kudos: 463
Location: India
GPA: 3.5
WE:Business Development (Commercial Banking)
Send PM
GMAT ToolKit User
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   29 Dec 2016, 08:22
2
Kudos
1
Bookmarks
Bunuel wrote:
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

\(4 = 2^2\)

Thus the highest power of 2 in 30! is - 26

30/2 = 15
15/2 = 7
7/2 = 3
3/2 = 1

So, the highest power of 4 in 30! will be 26/2 = 13

Thus, answer will be (A) 13
User avatar
L
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18754
Own Kudos [?]: 22044 [1]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   24 Aug 2017, 16:13
1
Bookmarks
Expert Reply
rxs0005 wrote:
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


Since 4 = 2^2, we are actually trying to determine the greatest value of m such that 2^2m is a factor 30!.

Let’s determine the number of factors of 2 within 30!. To do that, we can use the following shortcut in which we divide 30 by 2, and then divide that quotient by 2 again, continuing this process until we no longer get a nonzero integer as the quotient.

30/2 = 15

15/2 = 7 (we can ignore the remainder)

7/2 = 3 (we can ignore the remainder)

3/2 = 1 (we can ignore the remainder)

Since 1/2 does not produce a nonzero quotient, we can stop.

The next step is to add up our quotients; that sum represents the number of factors of 2 within 30!.

Thus, there are 15 + 7 + 3 + 1 = 26 factors of 2 within 30!, and thus 26/2 = 13 factors of 4 within 30!.

Answer: A
User avatar
L
Kinshook
GMAT Club Legend
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5343
Own Kudos [?]: 3964 [0]
Given Kudos: 160
Location: India
Schools: HBS '22 CBS '22 Wharton '22
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Send PM
Reviews Badge
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   28 Mar 2020, 05:41
rxs0005 wrote:
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


Asked: What is the greatest value of m such that 4^m is a factor of 30! ?

4 = 2^2

Power of 2 in 30! = 15 + 7 + 3 + 1 = 26
Power of 4 in 30! = 13

IMO A
avatar
B
rohitgupta2362
Intern
Intern
Joined: 02 Dec 2017
Posts: 8
Own Kudos [?]: 2 [0]
Given Kudos: 3
Send PM
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink] New post   28 Mar 2020, 19:21
VeritasKarishma wrote:
rxs0005 wrote:
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


I think the question is
What is the greatest value of m such that \(4^m\) is a factor of 30! ?

The easiest way to do this is the following:
Divide 30 by 2. You get 15
Divide 15 by 2. You get 7. (Ignore remainder)
Divide 7 by 2. You get 3.
Divide 3 by 2. You get 1.
Add 15+7+3+1 = 26
Since greatest power of 2 in 30! is 26, greatest power of 4 in 30! will be 13.
I will explain the logic behind the process in a while. (It will take some effort to formulate.)


+1 Kudos for the explanation.

Also, I found that these problems follow the Legendre's formula:

So, if I were to solve this problem, I would do:

[40/2] + [40/2^2] + [40/2^3] .............
Total = 15+7+3+ 1 = 26/2 = 13
GMAT Club Bot
gmatclubot
Re: What is the greatest value of m such that 4^m is a factor of 30! ? [#permalink]
28 Mar 2020, 19:21
 1   2   
Moderators:
Bunuel
Math Expert
92902 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions