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Divisibility

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Source: — Data Sufficiency |
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by fibbonnaci » Sun Mar 21, 2010 7:42 am
statement 1:
x^3+x is divisible by 4
x(x^2 + 1) = 4* integer.
any number multiplied by an even number is even.

case 1:
x- even and x^2+1 is odd,
Then we know that x is even and therefore divisble by 2

Case 2:
x- odd and x^2+1 is even.
We know when x= odd, x is not divisible by 2.

case 3:
x- even and x^2+1 is even
so here again we have x is divisible by 2.

Not sufficient.

Statement 2:
5x+4 is divisible by 6

in effect it means, 5x+4 = 6* integer.
ie. 5x+4 = even
5x= even - 4 => which is again even.
5x = even
5-> odd number. so x has to be an even number to make the multiple 5x even.
since we know x is even, we know x is divisible by 2.
Sufficient.

My answer is B

Hope this helps!
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by harshavardhanc » Mon Mar 22, 2010 2:36 am
beat_gmat_09 wrote:If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
Statement 1 :

x^3+x = X ( X^2 + 1) is divisible by 4.

Try to find out whether for any X, is X^2 + 1 divisible by 4? X in that case would have to be odd. try for 5, 7, 9, 13 ....

you'll find that for X=13, (X^2 + 1) becomes a multiple. Hence, we cannot definitely say, based on st1, that it is even.

Statement 2 :

5x+4 is divisible by 6. I.e 5x+4 is divisible by 3 & 2. I.e 5x+4 is EVEN.

therefore, 5X is even. Hence, X is even.

That's enough for us to conclude that X is divisible by 2.

Hence, B.
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by sanju09 » Mon Mar 22, 2010 3:26 am
beat_gmat_09 wrote:If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
(1) If x^3 + x is divisible by 4, then x^3 + x is even, x could be either even or odd. Insufficient

(2) If 5 x + 4 is divisible by 6, then 5 x is even; and x may or may not be even here (what if x is 8/5?). Insufficient

Taken together, (1) confirms x is an integer and (2) confirms that 5 x is even, therefore x is even. [spoiler]Sufficient

C[/spoiler]
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by harshavardhanc » Mon Mar 22, 2010 3:36 am
sanju09 wrote:
beat_gmat_09 wrote:If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
(1) If x^3 + x is divisible by 4, then x^3 + x is even, x could be either even or odd. Insufficient

(2) If 5 x + 4 is divisible by 6, then 5 x is even; and x may or may not be even here (what if x is 8/5?). Insufficient

Taken together, (1) confirms x is an integer and (2) confirms that 5 x is even, therefore x is even. [spoiler]Sufficient

C[/spoiler]
sooooper!!!

didn't pay attention to this fact. :) thanks mate!
Regards,
Harsha
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by outreach » Mon Mar 22, 2010 10:14 am
but (8/5) is divisible by 2. The Q asks if x is divisible by 2

i think the ans should be B
harshavardhanc wrote:
sanju09 wrote:
beat_gmat_09 wrote:If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
(1) If x^3 + x is divisible by 4, then x^3 + x is even, x could be either even or odd. Insufficient

(2) If 5 x + 4 is divisible by 6, then 5 x is even; and x may or may not be even here (what if x is 8/5?). Insufficient

Taken together, (1) confirms x is an integer and (2) confirms that 5 x is even, therefore x is even. [spoiler]Sufficient

C[/spoiler]
sooooper!!!

didn't pay attention to this fact. :) thanks mate!
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by harshavardhanc » Mon Mar 22, 2010 10:38 am
outreach wrote:but (8/5) is divisible by 2. The Q asks if x is divisible by 2

i think the ans should be B
harshavardhanc wrote:
sanju09 wrote:
beat_gmat_09 wrote:If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
(1) If x^3 + x is divisible by 4, then x^3 + x is even, x could be either even or odd. Insufficient

(2) If 5 x + 4 is divisible by 6, then 5 x is even; and x may or may not be even here (what if x is 8/5?). Insufficient

Taken together, (1) confirms x is an integer and (2) confirms that 5 x is even, therefore x is even. [spoiler]Sufficient

C[/spoiler]



sooooper!!!

didn't pay attention to this fact. :) thanks mate!

So what about 5/3? is it divisible by 2? No? How about 10/6? Now is it divisible? ;)

divisibility rules don't apply to fractions. Hence, Sanju's approach is correct. The answer should be C.
Regards,
Harsha
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by papumba2011 » Mon Mar 22, 2010 11:48 am
In case of stmt 1.

It is given x(x^2+1) is multiple of 4.

In all the discussion above, i think when considering whether x is even or odd we forgot to maintain x(x^2 + 1) to be divisible by 4.

If x is even and not a multiple of 4 then x(x^2 + 1) cannot be a multiple of 4.
for example x = 2 , then x(x^2 + 1) becomes 2 * 5 which is not didvidible by 4. Similar is the case for 6, 10 , etc.

So x must be a multiple of 4 itself.

If x is odd, x^2 +1 need to be multiple of 4 which again is not possible. Consider values of x to be 3, 5, 7, 9 , etc.

Hence from stmt 1 , x must always be multiple of 4 for x(x^2 + 1) to be multiple of 4.

Hence stmt 1 is sufficient.

In case of stmt well as pointed out in earlier post it is insufficient.

Hence answer is A.

Please let me know whether I am wrong.
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by thephoenix » Mon Mar 22, 2010 8:29 pm
harshavardhanc wrote:
outreach wrote:but (8/5) is divisible by 2. The Q asks if x is divisible by 2

i think the ans should be B
harshavardhanc wrote:
sanju09 wrote:
beat_gmat_09 wrote:If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
(1) If x^3 + x is divisible by 4, then x^3 + x is even, x could be either even or odd. Insufficient

(2) If 5 x + 4 is divisible by 6, then 5 x is even; and x may or may not be even here (what if x is 8/5?). Insufficient

Taken together, (1) confirms x is an integer and (2) confirms that 5 x is even, therefore x is even. [spoiler]Sufficient

C[/spoiler]



sooooper!!!

didn't pay attention to this fact. :) thanks mate!

So what about 5/3? is it divisible by 2? No? How about 10/6? Now is it divisible? ;)

divisibility rules don't apply to fractions. Hence, Sanju's approach is correct. The answer should be C.
IMO Ans is D
harsh 5/3 is not div by 2 but even the exp is not div by 6
we need to chose such numbers where the given statments hold true ........
for s1) when x=4,8,12,16....(2x+2)
the exp is div by 4
and x is div by 2

for s2)for x=4,10,16,22,28,34,.........
the expis div by 6
and x is div by 2
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by sanju09 » Tue Mar 23, 2010 1:08 am
papumba2011 wrote:In case of stmt 1.

It is given x(x^2+1) is multiple of 4.

In all the discussion above, i think when considering whether x is even or odd we forgot to maintain x(x^2 + 1) to be divisible by 4.

If x is even and not a multiple of 4 then x(x^2 + 1) cannot be a multiple of 4.
for example x = 2 , then x(x^2 + 1) becomes 2 * 5 which is not didvidible by 4. Similar is the case for 6, 10 , etc.

So x must be a multiple of 4 itself.

If x is odd, x^2 +1 need to be multiple of 4 which again is not possible. Consider values of x to be 3, 5, 7, 9 , etc.

Hence from stmt 1 , x must always be multiple of 4 for x(x^2 + 1) to be multiple of 4.

Hence stmt 1 is sufficient.

In case of stmt well as pointed out in earlier post it is insufficient.

Hence answer is A.

Please let me know whether I am wrong.
It's very hard to disagree with you here, really!!

[spoiler]heart changed to A[/spoiler]
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by harshavardhanc » Tue Mar 23, 2010 2:03 am
sanju09 wrote:
It's very hard to disagree with you here, really!!

[spoiler]heart changed to A[/spoiler]
yes, it is very correct! thanks you ppl!


And I hate myself more after analyzing my mistake :
harshavardhanc wrote: Try to find out whether for any X, is X^2 + 1 divisible by 4? X in that case would have to be odd. try for 5, 7, 9, 13 ....

you'll find that for X=13, (X^2 + 1) becomes a multiple.
How stupid am I? :shock: :roll: can't I distinguish between + and - ????

booo hooo!!!!
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by kstv » Tue Mar 23, 2010 3:02 am
Seems party here is over , so one for the road.
If x is positive is x divisible by 2 ?
(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
(2) 5x+ 4 is divisible by 6 so 5x-2 is divisible by 6 (5x+4 - 6) = 5x - 2
5x - 2 if it is divisible by 6 it has to be divisible by 2 & 3
if 5x-2 is divisible by 2 then , 5x (5x-2+2) is divisible by 2
so 5x +4 is divisible by 2 Sufficient.

(1) x(x²+1) is divisible by 4 so either x is divisible by 4 or (x²+1) is divisible by 4
It is not possible that x is divisible just by 2 then x is even and (x²+1) has to be odd
what if x is odd of the form 2a+1 then x²+1 = 4a²+4a+1 + 1 this is divisible by 2 not 4
so x is a multiple of 4 Sufficient
IMO D
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by sanju09 » Tue Mar 23, 2010 3:37 am
kstv wrote:Seems party here is over , so one for the road.
If x is positive is x divisible by 2 ?
(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
(2) 5x+ 4 is divisible by 6 so 5x-2 is divisible by 6 (5x+4 - 6) = 5x - 2
5x - 2 if it is divisible by 6 it has to be divisible by 2 & 3
if 5x-2 is divisible by 2 then , 5x (5x-2+2) is divisible by 2
so 5x +4 is divisible by 2 Sufficient.

(1) x(x²+1) is divisible by 4 so either x is divisible by 4 or (x²+1) is divisible by 4
It is not possible that x is divisible just by 2 then x is even and (x²+1) has to be odd
what if x is odd of the form 2a+1 then x²+1 = 4a²+4a+1 + 1 this is divisible by 2 not 4
so x is a multiple of 4 Sufficient
IMO D
The one for the road had been the heaviest one perhaps, hence it resulted in few accidents (see in bold)

5 x + 4 is given divisible by 6 thus by 2, but that again is not sufficient to call x as even; isn't so?
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by kstv » Tue Mar 23, 2010 3:49 am
@ sanju09 Hi ! Hic !
if 5x + 4 is divisible by 2,
do you agree with this part
If so what does it say about x
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by sanju09 » Tue Mar 23, 2010 4:02 am
kstv wrote:@ sanju09 Hi ! Hic !
if 5x + 4 is divisible by 2,
do you agree with this part
If so what does it say about x
It says nothing concrete about x, I suppose.

Have some lime to avoid those hiccups, not good for the road
The mind is everything. What you think you become. -Lord Buddha



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