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

Question

(A) 25
(B) 26
(C) 19
(D) 50
(E) 31

FillFM · Accepted Answer

(A) 25.

12 = (2*2)*3

54/2 = 27
27/2 = 13
13/2 = 6
6/2 = 3
3/2 = 1
1/2 = 0 (We can stop)

> 27+13+6+3+1+0 = 50

So, 50 factors of 2 within 54! (or 25 factors of 4)

54/3 = 18
18/3 = 6
6/3 = 2
2/3 = 0

> 18+6+2+0 = 26 factors of 3 within 54!

25 factors of 4 and 26 factors of 3, so we can only make 25 factors of 12 within 54!.

kunalcvrce · Answer

Highest power of 12 depends on the multiple of 3.

Option A

itisSheldon · Answer

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

generis · Answer

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

itisSheldon · Answer

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

generis · Answer

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

ScottTargetTestPrep · Answer

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

push12345 · Answer

54! Contains 2^50 and 3^26

Means 4^25

So for complete division 25 has to be answer

Posted from my mobile device

KarishmaB · Answer

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

saraheja · Answer

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

Prasannathawait · Answer

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

Kinshook · Answer

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

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

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

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