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# What is the greatest value of m such that 4^m is a factor of

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03 Dec 2010, 18:12
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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

is there an easier way to do this other than brute force

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03 Dec 2010, 19:55
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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!?
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03 Dec 2010, 19:33
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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.
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.)
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03 Dec 2010, 20:05
1
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.

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04 Dec 2010, 01:55
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Will it be 18 bcoz greatest prime number is 3..
40/3 13
13/3 4
4/3 1
so total 13+4+1 =18..
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04 Dec 2010, 02:42
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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 --> $$2m=26$$ --> $$m=13$$.

Hope it's clear.
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04 Dec 2010, 06:49
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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.

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.
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04 Dec 2010, 07:11
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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.
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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.

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 .
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19 Jun 2012, 07:03
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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.

$$900 = 2^2 * 3^2 * 5^2$$
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.
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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 --> $$2m=26$$ --> $$m=13$$.

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..
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25 Jun 2012, 05:42
1
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 --> $$2m=26$$ --> $$m=13$$.

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.
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Re: What is the greatest value of m such that 4^m is a factor of  [#permalink]

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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?
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30 Apr 2015, 23:00
1
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.

Here is a post with some more details on this concept: http://www.veritasprep.com/blog/2011/06 ... actorials/
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What is the greatest value of m such that 4^m is a factor of  [#permalink]

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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.
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01 Jun 2015, 20:17
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.
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What is the greatest value of m such that 4^m is a factor of  [#permalink]

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22 Jun 2015, 08:40
I would also do it using Bunuel's way. However, in this case we could actually count the 2's, by breaking each number to its primes, like this:

1
2
3
4 = 2,2
5
6 = 2,3
...
8 = 2,2,2
10 = 2,5
12 = 2,2,3
14 = 2,7
16 = 2,2,2,2
18 = 2,3,3
20 = 2,2,5
22 = 2,11
24 = 2,2,2,3
26 = 2, 13
28 = 2,2,7
30 = 2,5,3

We will then count 26 two's, which means that there are 13 four's. ANS A
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Re: What is the greatest value of m such that 4^m is a factor of  [#permalink]

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29 Dec 2016, 11:21
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

is there an easier way to do this other than brute force

Prime factorization of divisor i.e 4^m gives us 2^2m.

So we need to find the power of 2 in 30!.

30/2 + 30/4 + 30/8 + 30/16

15 + 7 + 3 + 1 = 26

So power of 2 = 2m = 26

So m = 13

(A)
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Re: What is the greatest value of m such that 4^m is a factor of  [#permalink]

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24 Aug 2017, 16:13
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!.

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Re: What is the greatest value of m such that 4^m is a factor of  [#permalink]

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03 Apr 2018, 08:43

4 has prime factorization = 2*2. Therefore 4 has a power of 2 in 4.
By following trailing zero concepts, we deduce that 2 has a total power of 26 in 30!. Therefore the total power of 4 in 30! equals 26 divided by 2 i.e. 13.
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