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

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

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28 Feb 2016, 08:13
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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

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Joined: 25 Dec 2012
Posts: 116
Re: If 12!/3^x is an integer, what is the greatest possible value of x?  [#permalink]

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28 Feb 2016, 12:19
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 - 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.
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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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28 Feb 2016, 15:18
2
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 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
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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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29 Oct 2016, 07:52
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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29 Oct 2016, 10:32
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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Joined: 26 Feb 2017
Posts: 3
Re: If 12!/3^x is an integer, what is the greatest possible value of x?  [#permalink]

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27 Jun 2018, 07:58
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.
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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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27 Jun 2018, 10:54
12=3*4
9=3*3
6=3*2
3

Answer C. count of 3 s
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Joined: 26 Feb 2017
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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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28 Jun 2018, 15:23
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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Joined: 22 May 2016
Posts: 2207
If 12!/3^x is an integer, what is the greatest possible value of x?  [#permalink]

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30 Jun 2018, 13:16
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 $$x=5$$

*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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30 Jun 2018, 13:16
1
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.

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]

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02 Jul 2018, 09:03
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.

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Re: If 12!/3^x is an integer, what is the greatest possible value of x? &nbs [#permalink] 02 Jul 2018, 09:03
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