What is the highest power of 12 that divides 54!?

Prime Factors of 12 are 2 and 3. Your analysis of highest power depends on the larger prime factor is absolutely correct, but in this problem the power of prime factor 2 is 2(i.e \(2^2\) *3=12).
So the highest power of 12 depends on the powers of prime factor 2
If the highest power of 12 depends on the power of 3, then the answer will be 31 though.
Its just a trap question, which differs slightly from these two questions
https://gmatclub.com/forum/how-many-val ... 63087.html
https://gmatclub.com/forum/how-many-val ... 63088.html

itisSheldon , how do you get 31 powers of 3? I get 26, see below.

The answer is A. There are 25 powers of \(2^2\) in 54!, and \(2^2\) turns out to be the limiting factor.

In other words, there are fewer 4s than 3s in 54!

Powers of 2 in 54!
\(\frac{54}{2}=27\)

\(\frac{54}{2^2}=\frac{54}{4}=13\)

\(\frac{54}{2^3}=\frac{54}{8}=6\)

\(\frac{54}{2^4}=\frac{54}{16}=3\)

\(\frac{54}{2^5}=\frac{54}{32}=1\)

\(27 + 13 + 6 + 3 + 1 = 50\)
But we need two 2s for every 12.

\(2^{50}=(2^2)^{25}\)
There are 25 powers of \(2^2\) in 54!

Powers of 3 in 54!

\(\frac{54}{3}=18\)

\(\frac{54}{3^2}=\frac{54}{9}=6\)

\(\frac{54}{3^3}=\frac{54}{27}=2\)

\(18 + 6 + 2 =\) 26 powers of 3 in 54!

There are more threes than fours. In 54!, powers of 12 are limited by 2; there are 25 powers of \(2^2\)

Powers of 12 in 54! = 25

ANSWER A

generis
You are right about the power of 3 as 26.
Its my bad that I did not give much thought before posting the solution and I agree that it is a silly mistake .
Thanks

itisSheldon
Not a silly mistake. An easy mistake to make. Eh, no big deal.

I hadn't yet seen FillFM 's answer. We posted almost simultaneously. I just saw it.

I was tired, and already boggled by
powers of 3 = 25;
then powers of 3 = 31;
and I got powers of 3 = 26

I was just checking.

It's a good question. +1

Since 12 = 2^2 x 3^1, and since the quantity of 2^2 is less than the quantity of 3^1 in 54!, we need to determine how many times 2^2 divides into 54!, so let’s begin by finding the number of 2’s in 54!.

To determine the number of factors of 2 within 54!, we can use the following shortcut in which we divide 54 by 2, and then divide the quotient of 54/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

54/2 = 27

27/2 = 13 (we can ignore the remainder)

13/2 = 6 (we can ignore the remainder)

6/2 = 3

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

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

The next step is to add up our quotients; that sum represents the number of factors of 2 within 54!.

Thus, there are 27 + 13 + 6 + 3 + 1 = 50 factors of 2 within 54!.

Since 2^50 = (2^2)^25, we see that there are 25 factors of 2^2 (and hence 25 factors of 12) in 54!.

Answer: A

54! Contains 2^50 and 3^26

Means 4^25

So for complete division 25 has to be answer

Posted from my mobile device

For more on this, check: https://anaprep.com/number-properties-h ... actorials/

This will depend on powers of 2 and not of 3 since 2^2 will exceed if we do for 3^26. hence 25 would work

trapped with B.
my bad.
A is correct.
We have to go for lowest power i. e. 25 for 2^2.
Nice question

Posted from my mobile device

Asked: What is the highest power of 12 that divides 54!?

Power of 2 in 54! = 27 + 13 + 6 + 3 + 1 = 50
Power of 3 in 54! = 18 + 6 + 2 = 26

54! = 2^50 * 3^26 k where k is any integer not multiple of 2 or 3
12 = 2^2 * 3
54! = (2^2*3)^25 * 3k
54! = 12^25 * 3k

The highest power of 12 that divides 54! = 25

IMO A

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.