Pretty simple, really. Answer is A.

If m = 13, then 4m = 52, which is 26x2, both of which are included in 30!

Since 13 is the largest number here, its the answer.

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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Karishma
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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..

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.
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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.
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Thanks for all the examples
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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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Trying to contribute in my small way.

Dear Bunuel,

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