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If a is divisible by 6!, then a/5 must be

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Bunuel
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If a is divisible by 6!, then a/5 must be [#permalink] New post   06 Dec 2016, 07:31
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If a is divisible by 6!, then a/5 must be

I. a prime number
II. divisible by 3
III. a multiple of 5

A. None
B. II only
C. III only
D. I and II only
E. I, II, and III
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B
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Abhishek009
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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   06 Dec 2016, 07:55
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Bunuel wrote:
If a is divisible by 6!, then a/5 must be

I. a prime number
II. divisible by 3
III. a multiple of 5

A. None
B. II only
C. III only
D. I and II only
E. I, II, and III


Least value of a = 720

Now, a/5 = 144 ( Divisible by 3 )

Let value of a = 1440

Now, a/5 = 288 ( Divisible by 3 )

Answer will be (B) Only II...
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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   18 Apr 2017, 05:23
Abhishek009 wrote:
Bunuel wrote:
If a is divisible by 6!, then a/5 must be

I. a prime number
II. divisible by 3
III. a multiple of 5

A. None
B. II only
C. III only
D. I and II only
E. I, II, and III


Least value of a = 720

Now, a/5 = 144 ( Divisible by 3 )

Let value of a = 1440

Now, a/5 = 288 ( Divisible by 3 )

Answer will be (B) Only II...


Hi tell me one thing since least value of a is 720 so it should be multiple of 5 as well
Please explain Abhishek009
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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   18 Apr 2017, 05:27
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ehsan090 wrote:
Abhishek009 wrote:
Bunuel wrote:
If a is divisible by 6!, then a/5 must be

I. a prime number
II. divisible by 3
III. a multiple of 5

A. None
B. II only
C. III only
D. I and II only
E. I, II, and III


Least value of a = 720

Now, a/5 = 144 ( Divisible by 3 )

Let value of a = 1440

Now, a/5 = 288 ( Divisible by 3 )

Answer will be (B) Only II...


Hi tell me one thing since least value of a is 720 so it should be multiple of 5 as well
Please explain Abhishek009


The least value of a is 6! = 720, but we are asked about a/5, not a. The least value of a/5 is 6!/5 = 144.

Hope it's clear.
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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   18 Apr 2017, 05:42
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Hi ehsan090
It is not necessary to compute this value 720 at all here.
Its good to remember factorials upto 6.
But i did not depend on my knowledge to remember that 6! is 720.


Simply put let a=6!(smallest value)

So we are asked about a/5

Statement 1=>
a/5 will never be a prime.There would be a lot of other integers in the numerator--> 2,3,4 etc.
Hence this is false
Statement 2 =>
a/5 will take aways the five from 6!
But three would always be there in the numerator.
Hence this is always true.

Statement 2 =>
a/5 may or may not contain 5.
Example if a=6! then a/5 will be 1*2*3*4*6.
We don't have any five here.
Hence this statement may or may be true.

((This statement can be true if a was say 6!*5 ))



Smash that B.
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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   18 Apr 2017, 07:13
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a is divisible by 6!
i.e. a is divisible by 1*2*3*4*5*6


So least value of a is 6!= 1*2*3*4*5*6

a/5= 1*2*3*4*6

a has at least 2 prime nos. (2 and 3). So a is not a prime no.

As a has 3 as its prime factor, it is divisible by 3.

We just eliminated 5 and we are not sure if there is any other 5 or a multiple of 5 in the factors of a. So choice 3 is also not true as this is a MUST BE question.
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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   24 Apr 2017, 16:54
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Bunuel wrote:
If a is divisible by 6!, then a/5 must be

I. a prime number
II. divisible by 3
III. a multiple of 5

A. None
B. II only
C. III only
D. I and II only
E. I, II, and III


Since 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720, we see that a = 720k for some positive integer k, and hence a/5 = 720k/5 = 144k.

Furthermore, since 144 is a multiple of 3, 144k will be always divisible by 3 for all values of k. However, since 144 is not a multiple of 5, 144k will only be a multiple of 5 for some values of k. Last but not least, since 144 is not a prime number, 144k will never be a prime number for any values of k.

Answer: B
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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   25 Apr 2017, 13:09
If a is divisible by 6!, the lowest value of is 6!. a/5 removes a factor of 5.
So, if a=6x5x4x3x2, then a/5 is not divisible by 5. In addition, it's clear that a/5 is not a prime.
Lastly, dividing by 5 does not remove the factor of 3. So we know that II is true.
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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   03 Dec 2017, 08:22
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Bunuel wrote:
If a is divisible by 6!, then a/5 must be

I. a prime number
II. divisible by 3
III. a multiple of 5

A. None
B. II only
C. III only
D. I and II only
E. I, II, and III


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Re: If a is divisible by 6!, then a/5 must be [#permalink] New post   12 Nov 2020, 00:59
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Bunuel wrote:
If a is divisible by 6!, then a/5 must be

I. a prime number
II. divisible by 3
III. a multiple of 5

A. None
B. II only
C. III only
D. I and II only
E. I, II, and III


a is multiple of 6!
6!=6*5*4*3*2*1/5 IS NOT A PRIME NUMBER BECAUSE THE RESULT IS MULTIPLIED BY 2, SO IT'S A EVEN NUMBER (I) IS OUT.

II. 6*5*4*3*2*1/5 = 6*4*3*2*1 IS DIVIDED BY 3; TRUE

III. 6*5*4*3*2*1/5 = 6*4*3*2*1=144 IS NOT MULTIPLE OF 5

THE ANSWER IS B.
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Re: If a is divisible by 6!, then a/5 must be [#permalink]
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