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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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12/3 + 12/9 = 4 + 1 = 5

3^5
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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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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) 5
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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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) 5


i didnt understand why you divided 4 by 3.
please explain
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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
12=3*4
9=3*3
6=3*2
3

Answer C. count of 3 s
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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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) 5


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.
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If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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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

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.
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If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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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) 5

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.

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.
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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
Expert Reply
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


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


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

Max power of 3 in 12! = [12/3] + [12/3^2] = 4 + 1 = 5

IMO C
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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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?
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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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

So, Highest power of x is 4 + 1 = 5 , Answer will be (C)
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Re: If 12!/3^x is an integer, what is the greatest possible value of x? [#permalink]
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