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

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Manager
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What is the highest power of 12 that divides 54!? [#permalink]

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10 Apr 2018, 12:28
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What is the highest power of 12 that divides 54!?

(A) 25
(B) 26
(C) 19
(D) 50
(E) 31
[Reveal] Spoiler: OA

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Manager
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Location: India
Concentration: General Management, Marketing
WE: Information Technology (Computer Software)
What is the highest power of 12 that divides 54!? [#permalink]

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10 Apr 2018, 12:36
itisSheldon wrote:
What is the highest power of 12 that divides 54!?

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

Highest power of 12 depends on the multiple of 3.

Option A
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Re: What is the highest power of 12 that divides 54!? [#permalink]

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10 Apr 2018, 12:51
kunalcvrce wrote:
itisSheldon wrote:
What is the highest power of 12 that divides 54!?

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

Highest power of 12 depends on the multiple of 3.

Option A

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
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Re: What is the highest power of 12 that divides 54!? [#permalink]

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10 Apr 2018, 17:12
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(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!.
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What is the highest power of 12 that divides 54!? [#permalink]

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10 Apr 2018, 17:30
itisSheldon wrote:
kunalcvrce wrote:
itisSheldon wrote:
What is the highest power of 12 that divides 54!?

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

Highest power of 12 depends on the multiple of 3.

Option A

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

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

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Re: What is the highest power of 12 that divides 54!? [#permalink]

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11 Apr 2018, 01:38
Quote:
how do you get 31 powers of 3? I get 26, see below.

In this instance, the power of $$2^2$$ is the limiting factor. There are fewer 4s than 3s.

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!

How do you get 31 powers of 3 in 54!?

There are more threes either way (26 or 31).

Hence powers of 12 are limited by 2; there are 25 powers of $$2^2$$

Powers of 12 in 54! = 25

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
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What is the highest power of 12 that divides 54!? [#permalink]

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11 Apr 2018, 10:48
itisSheldon wrote:
Quote:
how do you get 31 powers of 3? I get 26, see below.

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
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Re: What is the highest power of 12 that divides 54!? [#permalink]

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12 Apr 2018, 16:22
itisSheldon wrote:
What is the highest power of 12 that divides 54!?

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

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!.

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Re: What is the highest power of 12 that divides 54!? [#permalink]

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19 Apr 2018, 22:11
54! Contains 2^50 and 3^26

Means 4^25

So for complete division 25 has to be answer

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Re: What is the highest power of 12 that divides 54!? [#permalink]

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19 Apr 2018, 22:27
itisSheldon wrote:
What is the highest power of 12 that divides 54!?

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

For more on this, check: https://www.veritasprep.com/blog/2011/0 ... actorials/
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Re: What is the highest power of 12 that divides 54!? [#permalink]

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19 Apr 2018, 23:18
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
Re: What is the highest power of 12 that divides 54!?   [#permalink] 19 Apr 2018, 23:18
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