ankurkshl
Abhishek009
Bunuel
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.