ankurkshl wrote:

Abhishek009 wrote:

Bunuel wrote:

If 12!/3^x is an integer, what is the greatest possible value of x?

A. 3

B. 4

C. 5

D. 6

E. 7

12/3 = 4

4/3 = 1

4 + 1 = 5

Hence correct answer will be (C) 5Kindly specify dividing 4 by 3 by the same shortcut method that you have applied in the tagged answer.

i have got the factorisation concept.

thanks.

regards.

ankurkshl , the method is one way to find the number of powers of a prime number in \(n!\).

Divide \(n\) (without the !) by the prime number (here, 3). \(n=12\)

Use the

resulting quotient as your new dividend, and divide again by \(3\) until the resulting number is too small to divide by \(3\)

1) 12 divided by 3: \(\frac{12}{3^1}=4\)

2) Now use \(4\), and divide by 3. Do not worry about remainders.* \(\frac{4}{3}=1\)

3) Sum the number of 3s from all stages:

(4 + 1) = 5

There are five factors of 3 in 12!, so the greatest possible value for x is 5.

Answer C

Hope that helps.

*

We just need to know whether \(4\) can be divided by \(3\) such that the result \(\geq1\)

(How many times does 3 go into 4? One time, with a remainder about which we do not care). This method is a variation on the method I posted above. Both are VERY handy when n! is huge.
_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"