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

Question

A. 3
B. 4
C. 5
D. 6
E. 7

sameerspice · Accepted Answer

This question is asking how many power of 3 exists in 12!
easiest way is to identify factor of three
12 -> 4*3 => 1 power of 3
9 -> 3*3=> 2 power of 3
6 -> 2*3=> 1 power of 3
3 -> 1*3=> 1 power of 3
Total equal 5 hence C

sowragu · Answer

12 - 4*3
9 - 3*3
6 - 2*3
3 - 1*3

Hence max of 3^5 is allowed. IMO C.

This is not the right way of doing. Forgot the easy approach. Will update.

law258 · Answer

12/3 + 12/9 = 4 + 1 = 5

3^5

Abhishek009 · Answer

12/3 = 4
4/3 = 1

4 + 1 = 5
 
Hence correct answer will be (C) 5

ankurkshl · Answer

i didnt understand why you divided 4 by 3.
please explain

gmatzpractice · Answer

12=3*4
9=3*3
6=3*2
3

Answer C. count of 3 s

ankurkshl · Answer

Kindly 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.

generis · Answer

This question is easy, and a good way to demonstrate one way to find the number of "powers of a prime in  n! ." (See footnote.*)

This  n!  is small. Most will not be.

The method: Divide  n=12  (without !) by increasing powers of  3 . Don't worry about remainders.

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

(2) 12 divided by  3^2: (\frac{12}{3^2})=(\frac{12}{9})=1

(3) 12 divided by  3^3=27 : 
Will not work.  \frac{12}{27}<1

(4) Add up the results of division by each power of 3 that "worked":  (4 + 1) = 5

There are 5 powers of 3 in 12!, so the greatest possible value of  x=5

Answer C

*The theory is described by  Bunuel  in Everything About Factorials,  Finding the Number of Powers of a Prime, p, in the n!, here . Important! The post also has an example.

generis · Answer

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.

JeffTargetTestPrep · Answer

o determine the number of factors of 3 within 12!, we can use the following shortcut in which we divide 12 by 3, and then divide the quotient of 12/3 by 3 and continue this process until we can no longer get a nonzero integer as the quotient.

12/3 = 4

4/3 = 1 (we can ignore the remainder)

Since 1/3 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 3 within 12!.

Thus, there are 4 + 1 = 5 factors of 3 within 12!. Thus, the greatest value of x is 5.

Answer: C

Kinshook · Answer

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

Max power of 3 in 12! =   +   = 4 + 1 = 5

IMO C

QQRamanQQ · Answer

Why didn't we include 3^0 in the option? question doesn't specify there cant be zero. Can anyone help me with this one? if we include 3^0, we get D as the correct answer. Can anyone help me with this?

Abhishek009 · Answer

12/3 = 4
4/3 = 1

So, Highest power of x is 4 + 1 = 5 , Answer will be (C)

gautam25gs · Answer

this could be solved by applying legendre's formula 
 n/p+n/p^2+n/p^3+n/p^4...... where p^k>n

here 12/3=4----- eq.1
12/3^2 i.e 9
12/9=1 
only the quotient is added 4+1=5
hence 3^5 answer
solution is option c

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

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GMAT Problem Solving (PS) Questions

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

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