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# Factors questions

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Manager
Joined: 01 Jun 2006
Posts: 77

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21 Sep 2006, 11:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi all,

Got stuck with the following questions...need your help guys

Q1: Is "m" a multiple of 6?
1) More than 2 of the first 5 positive integer multiples of m are multiples of 3

2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12

Couldn't get this one.....how should i go about such questions?

Q2: What is the sum of positive integers x and y?
1) x^2 + 2xy + y^2 = 16
2) x^2 - y^2 = 8

I got it D but dont know how number 2 statement can be proved...can anybody explain?

Q3: If x is an integer, is x even?
1) x^2 - y^2 = 0
2) x^2 + y^2 = 18

usman

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SVP
Joined: 01 May 2006
Posts: 1794

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21 Sep 2006, 12:02
Q1) E

Statment 1 & 2 do not provide enough of information to conclude.

Q2) B

Statment 1:
x^2 + 2xy + y^2 = 16
<=> (x+y)^2= (4)^2 = (-4)^2
thus, x+y = 4 or x+y = -4

INSUFF

Statment 2:
x^2-y^2=8
<=> (x+y)(x-y) = 8 = 4*2 (=8*1 not working)
Since x & y are integers, we can say than:
o x+y = 4 (1)
and
o x-y = 2 (2)
Thus (1)+(2) <=> 2*x= 6 <=> x=3
=> y=1

SUFF

Q3) C

Statment 1: brings nothing alone
x^2 - y^2 = 0 (1)

INSUFF

Statment 2: brings nothing alone because is a real number (not the integer 3)
x^2 + y^2 = 18 (2)

INSUFF

Statments 1 & 2 : we can eliminate y
(1) + (2) <=> 2*x^2 = 18
<=> x^2 = 3^2 = (-3)^2
X is not an even integer.

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Director
Joined: 06 May 2006
Posts: 791

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Re: Factors questions [#permalink]

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21 Sep 2006, 12:23
Q1: Is "m" a multiple of 6?
1) More than 2 of the first 5 positive integer multiples of m are multiples of 3

2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12

Plugging in numbers should give it to you.
1) If more than 2 of the first 5 positive integer multiples of m are multiples of 3, then the number itself must be a multiple of 3. However, this does not mean it is a multiple of 6. Insufficient.
e.g. m = 9, then first 5 integer multiples are 9, 18, 27, 36, 45.
or m = 6, then we have 6, 12, 18, 24, 30. Both satisfy the criteria, but 9 is not a multiple of 6.

2) If you take m=6, then 2 of the first 5 integer multiples are multiples of 12. Not so if you take any other number. Sufficient.

Hence B.

Q2: What is the sum of positive integers x and y?
1) x^2 + 2xy + y^2 = 16
2) x^2 - y^2 = 8

Exactly as Fig has explained.

Q3: If x is an integer, is x even?
1) x^2 - y^2 = 0
2) x^2 + y^2 = 18

Also exactly as Fig has explained.
_________________

Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

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VP
Joined: 02 Jun 2006
Posts: 1258

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21 Sep 2006, 13:35
1. Is m = 6k?

S1: > 2 multiples of 3, in first 5 multiples of m.

m, 2m, 3m, 4m, 5m.

As only 3m is multiple of 3, m has to be a multiple of 3. i.e. 3,6,9 etc.

If m = 6, then > 2 multiples of 3, and 6 = 6x1.
If m = 9, then > 2 multiples of 3, but 9 != 6xk
Not sufficient.

S2: < 2 multiples of 3 ,in first 5 multiples of m.
m, 2m,3m, 4m, 5m

Again as only 3 is a multiple and less than 2 multiples then m is not a multiple of 3.

If m is not a multiple of 3 then enough to say that m is not divisible by 6 (to be divisble m to be divisible by both 3 and 2).

Sufficient.

-------------------------------------------------------------
2. Q: x+y? If x,y integers.

1) x^2 + 2xy + y^2 = 16

(x+y)^2 = 16
(x+y) = +4

Not sufficient

2) x^2 - y^2 = 8
(x+y)(x-y) = 8
Sufficient.
-------------------------------------------------------------
3. x integer, Q: x = even?
S1: Not sufficient.

S2: x^2+y^2 = 18, Not sufficient as x can be even or odd ( if y is even or odd).

S1 & S2: Sufficient.

------------------------------------------------------------

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Manager
Joined: 03 Jul 2006
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21 Sep 2006, 14:36
I think (2) is 'D'

if X & Y are +ve integers(given), why should we consider x+y = -4 ?

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SVP
Joined: 01 May 2006
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21 Sep 2006, 14:44
rkatl wrote:
I think (2) is 'D'

if X & Y are +ve integers(given), why should we consider x+y = -4 ?

Yes, u are right Of course a sum of positive integer cannot give a negative integer... Answer D (a long day, skipping not reading well)

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Manager
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22 Sep 2006, 00:41
//Yezz'z reply from the other post

---------------------------------------------------------------------------------------
Q1: Is "m" a multiple of 6?
1) More than 2 of the first 5 positive integer multiples of m are multiples of 3

2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12

M to be a multiple of 6 it takes the form = 3*2*k where k is positive intiger

from one

first five multiples are M ,m*2 , M*3 , m*4 , m*5

worst case scenario m*5,m*3,m are multiples of 3 , and m is thus a multiple of three but we need a factor of 2 at least to make sure m is multiple of 6........insuff

from 2

clearly insuff

both together still insuff

my answer is E

Q2: What is the sum of positive integers x and y?
1) x^2 + 2xy + y^2 = 16
2) x^2 - y^2 = 8
from one

(x+y)^2 = +or -ve sqrt 16 ie (x+y) might be +ve or -ve 4 but stem say they are positive thus =4.....suff

from two

(x-y)(x+y) = 8( 8 = 8*1 or 4*2 )

it is obvious that there is no two intigers that when added together =8 and when subtracted = 1

thus( x-y)(x+y) must = 4*2

x+y > x-y (positive intigers) thus x+y = 4 ...suff

Q3: If x is an integer, is x even?
1) x^2 - y^2 = 0
2) x^2 + y^2 = 18

from one

x^2 = y^2 we have no clue about values of y....... and zero is even

even - even = even or odd - odd = even .insuff

from two

x^2+y^2 = 18

18 is even ie even+even or odd +odd.......insuff

one and two together

2x^2 = 18

ie x^2 = 9 thus x = odd.....suff

most probably i should go to sleep
_________________
Nothing is impossible and nothing is as bad as it seems
------------------------------------------------------------------------------------

Guys im still not clear....will shoot my questions on the in the evening....

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Manager
Joined: 01 Jun 2006
Posts: 77

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22 Sep 2006, 08:09
Haas_mba07 wrote

[/quote]S2: < 2 multiples of 3 ,in first 5 multiples of m.
m, 2m,3m, 4m, 5m

Again as only 3 is a multiple and less than 2 multiples then m is not a multiple of 3.

If m is not a multiple of 3 then enough to say that m is not divisible by 6 (to be divisble m to be divisible by both 3 and 2).

Sufficient.

---------------------------
The second statement was
2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12 (not of 3)

Hass can you explain the second point with respect to 12...
Are there any tricks to get to these questions.......

Thanks guys I got the other two questions.....as per your explanations. Answer for Q2 is D and for Q3 it is C.......

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22 Sep 2006, 08:09
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