sets permutation and combination

The answer is C - 10.

The procedure to solve this question is..

6 = 2 * 3 - Take the greater of the prime factors, here its 3

List the number of multiples of 3 in 25

3, 6 , 9, 12, 15, 18, 21, 24

Now count the number of 3's in the multiples of 3 in 25

3, 3 2, 3 3, 3 4, 3,5 3,3,2, 3,7 3,8

Total 10 - 3's

So 6 to power 10 divides 25! completely.


But be assured, this question would not appear on GMAT. GMAT does not test your knowledge of knowing a complex formula but your application of simple formulas.

Good question - I don't think I've ever seen one like this before. I'm not sure I am right but let me know what the OA is:

First thing you should realize is that you are going to have more 2s than 3s in the prime factorization of 25! This should be fairly obvious since 2 is a more commonly recurring factor of the numbers in the expansion of 25!

This in effect means that if you can count the number of 3s or the power of 3 in the prime factorization of 25! you have your answer for the maximum value of m since 6 = 2*3

The number of 2s and 3s you can pair to create a 6 which would help to divide into 25! would hence be limited by the number of 3s and not the number of 2s. I hope that makes sense.

25! contains the following multiples of 3:

3, 6, 9, 12, 15, 18, 21, 24

= 3 , 3*2, 3^2, 3*4, 3*5, 3^2*2, 3*7, 3*8

Count the number of 3s = 10 (8 + the two extra 3s in 9 and 18)

Pick C.

mohit11 wrote:
But be assured, this question would not appear on GMAT. GMAT does not test your knowledge of knowing a complex formula but your application of simple formulas.

I don't quite agree. There's no real complex formula involved here - it is rather intuitive to realize that counting 3s is going to give you your answer.

Based on my own GMAT experience in which I did see some unconventional questions, I would recommend that as long as the concepts being tested (in this case factorization and exponents) is withing the scope of the GMAT, get cracking.

Even so, what is the source of this question?

Ermm.. read your explanation. I had the formula memorized so never quite bothered to dwell too much into it. You're reasoning is rock solid. I concede !

mohit11 wrote:Ermm.. read your explanation. I had the formula memorized so never quite bothered to dwell too much into it. You're reasoning is rock solid. I concede !

Hah no worries

Actually this sort of Prime factorization related question is becoming increasingly common I've noticed. Good to get used to them

mohit11 wrote:The answer is C - 10.

The procedure to solve this question is..

6 = 2 * 3 - Take the greater of the prime factors, here its 3

List the number of multiples of 3 in 25

3, 6 , 9, 12, 15, 18, 21, 24

Now count the number of 3's in the multiples of 3 in 25

3, 3 2, 3 3, 3 4, 3,5 3,3,2, 3,7 3,8

Total 10 - 3's

So 6 to power 10 divides 25! completely.


But be assured, this question would not appear on GMAT. GMAT does not test your knowledge of knowing a complex formula but your application of simple formulas.

Excellent questions!!!!

hey guys... thanks a ton for the replies...

the OA for the question is C)10.

I got the trick in the question, but was not quite clear as to why the power of 3 has to be the answer and not the power of 2 ( which is 22 and also one of the answer choices).

Thanks again for the replies.

Regards,
Alay Naik

good quesiton

alaynaik wrote:hey guys... thanks a ton for the replies...

the OA for the question is C)10.

I got the trick in the question, but was not quite clear as to why the power of 3 has to be the answer and not the power of 2 ( which is 22 and also one of the answer choices).

Thanks again for the replies.

Regards,
Alay Naik

Lets take a simpler example, what is the highest power of 6 that divides 8!

6 = 2*3

8! = 8*7*6*5*4*3*2*1 = This is equal to 40320

Now we have to find the power of 6 that divides this number.

6 is = 2 * 3


In 40320 the first multiple of 2 is 2 and the last multiple is 40320 increment is 2 therefore, we have (40320-2)/2 + 1 multiples of 2 in 40320 which is equal to 20160 therefore we have 20160 multiples of 2 in 8!

Similarly, first multiple of 3 is 3 and last multiple is 40317, total number of multiples of 3 in 40320 are (40317-3)/3+ 1
= 13438

Note that 20160 (multiples of 2 ) is greater than 13438 (multiples of 3).

To get a 6 we need multiple 2 with 3

13438 multiples of 3 can multiple with 13438 multiples of 2 to create 13438 numbers of 6

So we are left with 20160 - 13438 = 6722 multiples of 2


The point i am trying to prove is that to find out what is the highest power of 6 that divides 8! completely we should merely focus on highest power of 3

Now back to original question,

Multiples of 3 in 8?

3, 3*2

Therefore, the power of 6 that divides 8! completely is 2


= 40320 / 3^2 = 4480

If you're still not convinced check how higher powers of 3 divide 8 factorial.

= 40320/ 3^3 = 1493.333

= 40320/3^4 = 497.777

= 40320/3^5 = 165.925

.. and so on...



Disclaimer: All the heavy calculation was done merely to prove a point.