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! = 25ANSWER A

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