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If x is an even integer, is x divisible by 6?

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chetan2u
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If x is an even integer, is x divisible by 6? [#permalink] New post   14 Jan 2018, 06:04
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If x is an even integer, is x divisible by 6?

(1) \(x=27!-17!\)
(2) \(x=3y\)


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chetan2u
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   18 Jan 2018, 06:15
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poul_249 wrote:
Could anyone please explain why OA is marked A. I think both the statements r sufficient . Ans should be D


Hi..

ans should be A

chetan2u wrote:
Is x divisible by 6, if x is an even integer?

(1) \(x=27!-17!\)
(2) \(x=3y\)


New Question


INFO given-
x is even integer..
(1) \(x=27!-17!\)
x=27!-17!=17!(18*19*..27-1)
17! is a MULTIPLE of 6, so ans is YES
suff

(2) \(x=3y\)
this is a TRAP...
it is nowhere given that y is an integer...
x=3y....
say y is 20....ans is YES
BUT if y is \(\frac{20}{3}\)...x becomes 20 and ans is NO..
Insuff

A..

IMPORTANT... 80% answering it wrong shows us to be careful in reading the info provided
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AshutoshB
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   14 Jan 2018, 06:21
D

As 1, can be written as 17!(18*19*--27. -1)

As 17! contains 6

And for 2)

No. to be divisble by 2 and 3

Its even and equals to 3y condition satisfied

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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   14 Jan 2018, 06:21
Statement 1 : 17!(27x26x25...18-1). 17! is divisible by 6 so this statement is sufficient
Statement 2: x=3y and x is even integer so y should be even integer so is also sufficient


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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   18 Jan 2018, 05:26
Could anyone please explain why OA is marked A. I think both the statements r sufficient . Ans should be D
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poul_249
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   18 Jan 2018, 06:20
chetan2u wrote:
poul_249 wrote:
Could anyone please explain why OA is marked A. I think both the statements r sufficient . Ans should be D


Hi..

ans should be A

chetan2u wrote:
Is x divisible by 6, if x is an even integer?

(1) \(x=27!-17!\)
(2) \(x=3y\)


New Question


INFO given-
x is even integer..
(1) \(x=27!-17!\)
x=27!-17!=17!(18*19*..27-1)
17! is a MULTIPLE of 6, so ans is YES
suff

(2) \(x=3y\)
this is a TRAP...
it is nowhere given that y is an integer...
x=3y....
say y is 20....ans is YES
BUT if y is \(\frac{20}{3}\)...x becomes 20 and ans is NO..
Insuff

A..

IMPORTANT... 80% answering it wrong shows us to be careful in reading the info provided



Thanks for the explanation.
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   18 Jan 2018, 06:48
poul_249 wrote:
Could anyone please explain why OA is marked A. I think both the statements r sufficient . Ans should be D


It should be A.
Because there is no mention that x must be an integer.
(1) We can find x value and it would be interger.
-> possible to judge
(2) x
If x is 6.6? It also devided by 6.
We cannot assure the value of x. interger or not. So (2) is wrong.

I hope this can help you.
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chetan2u
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   18 Jan 2018, 06:57
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HG0815 wrote:
poul_249 wrote:
Could anyone please explain why OA is marked A. I think both the statements r sufficient . Ans should be D


It should be A.
Because there is no mention that x must be an integer.
(1) We can find x value and it would be interger.
-> possible to judge
(2) x
If x is 6.6? It also devided by 6.
We cannot assure the value of x. interger or not. So (2) is wrong.

I hope this can help you.


coloured portion is wrong..

x is given as even integer, so it cannot be 6.6
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   18 Jan 2018, 07:21
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Excellent Question.

Here is what I did on this one ->


Given data => x is even. So x is divisible by 2.
To be checked => If x is divisible by 6 or not.

For a number to be divisible by 6 it must be divisible by both 2 and 3.

So basically the question can be rephrased as -> Is x divisible by 3?

Lets jump into statements ->


Statment 1 ->

x= 27! - 17!
Notice that x is divisible by 3 as both 27! and 17! are divisible by 3 (Remember => Multiple of 3+ Multiple of 3 = Multiple of 3)
Hence since x is divisible by 3 -> It is divisible by 6.


But wait wait. wait.....

Why do we even bother to this. Lets think about this for a second.

Forget calculating. We have the value of x.
And this is DS question.

So we would be able to answer the question if x is divisible by 6 or not.No matter what the value of x be.


So it has to be sufficient.

Damn, I like this second approach.
TAKEAWAY -> If the value is given, DO NOT CALCULATE ANYTHING.

Sufficient.


Statment 2 ->
we don't have any clue of what y is

y = 2/3 => x = 2 => x is not divisible by 6.
y= 6 => x = 18 => x is divisible by 6.

Not sufficient.

Smash that A.
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   18 Jan 2018, 08:44
Hi AshutoshB
Could you please describe the property below?

17!(18*19*--27. -1)

Thank you!


AshutoshB wrote:
D

As 1, can be written as 17!(18*19*--27. -1)

As 17! contains 6

And for 2)

No. to be divisble by 2 and 3

Its even and equals to 3y condition satisfied

Hit kudos

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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   18 Jan 2018, 23:47
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[quote="robertops"]Hi AshutoshB
Could you please describe the property below?

17!(18*19*--27. -1)

Thank you!


Hi

Factorial of a non-negative integer N is denoted as N! and its formula is:
N! = N*(N-1)*(N-2)*(N-3)*...3*2*1
So basically its the product of all integers from 1 to N (important - both 0! and 1! are = 1 only)

So 6! = 6*5*4*3*2*1, which can also be written as 6! = 6*5*4! (because 4! = 4*3*2*1)
Similarly, 27! can be written as = 27*26*25....19*18*17! (we have written product of all integers from 17*16.. to 1 as 17! only)
Thus 27! - 17! = 27*26*25....19*18*17! - 17!

we can take 17! common to get
17!*(27*26*...19*18 - 1)

Hope this is clear.
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   19 Jan 2018, 03:06
You are awesome.
Thanks alot.

amanvermagmat wrote:
robertops wrote:
Hi AshutoshB
Could you please describe the property below?

17!(18*19*--27. -1)

Thank you!


Hi

Factorial of a non-negative integer N is denoted as N! and its formula is:
N! = N*(N-1)*(N-2)*(N-3)*...3*2*1
So basically its the product of all integers from 1 to N (important - both 0! and 1! are = 1 only)

So 6! = 6*5*4*3*2*1, which can also be written as 6! = 6*5*4! (because 4! = 4*3*2*1)
Similarly, 27! can be written as = 27*26*25....19*18*17! (we have written product of all integers from 17*16.. to 1 as 17! only)
Thus 27! - 17! = 27*26*25....19*18*17! - 17!

we can take 17! common to get
17!*(27*26*...19*18 - 1)

Hope this is clear.
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   05 Nov 2020, 13:59
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chetan2u wrote:
Is x divisible by 6, if x is an even integer?

(1) \(x=27!-17!\)
(2) \(x=3y\)


New Question


(1) 17!(27*26*25*24*......18) Green part is divisibleby by 6 (24 or 18) so the x is divisible by 6. Sufficient

(2) If y is 2 then x is divisible by 6 but if y is 3 then is not divisible by 6. Insufficient.

The answer is A.
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Re: If x is an even integer, is x divisible by 6? [#permalink] New post   05 Nov 2020, 16:27
Is x divisible by 6, if x is an even integer?

(1) \(x=27!-17!\)

x = 17! (27*26*25... - 1)
17! is a multiple of 6; x is a multiple of 6. SUFFICIENT.

(2) \(x=3y\)

We know x = even. If y = 2, x = a multiple of 6. If y = 2/3, then x is not a multiple of 6. INSUFFICIENT.

Answer is A.
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Re: If x is an even integer, is x divisible by 6? [#permalink]
05 Nov 2020, 16:27
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