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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!?
_________________

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!.

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 --> \(2m=26\) --> \(m=13\).

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.)
_________________

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.
_________________

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 \(x=6\).

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!.

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.
_________________

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.

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 .
_________________

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.
_________________

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!.

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..

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