answer is A ...
from the argument : p = X^2 + Y^2 ..
From statement 1 & Y is odd, p may look like this:
13 = 4 + 9
13 = 12 + 1
21 = 20 + 1
21 = 12 + 9
hence, A satisfies the question..
From statement 2: X = 3 + Y ... as Y is odd, we get different answers which is divisible/not divisible by 4 ...
Remainder
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Source: Beat The GMAT — Data Sufficiency |
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ssy
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Tough question, let me give it a shot..
From the question stem - For x to be divisible by 4, it needs to be even. As 4=(2)(2), x also needs to have at least two 2s in its prime factorization to be divisible by 4.
1) Translated into an equation, this means:
[(x^2)+(y^2)]/8=int + 5
[(x^2)+(y^2)] =8(int)+5
As 8 is even, 8 multiplied by any integer is even. 5 is odd. Even+odd=odd So 8(int)+5 =odd
Hence. [(x^2)+(y^2)] = odd
We know that y^2 is odd.
So, x^2 is even as even+odd=odd.
The squre of all even integers is a multiple of 4, however, the square root of all even squares is not necessarily a multiple of 4.
Hence, Statement 1 is insufficient.
2) x-y=3
Plugging this into the equation given in the stem of the question:
P=(x^2)+(y^2)
X=3+y
P=[(3+y)^2}+(y^2)
P=9+6y+(y^2)+(y^2)
P=9+6y+2(y^2)
As p=x-y
X-y=9+6y+2(y^2)
X=9+6y+y^2
X=(y+3)(y+3)
Now, y is odd and odd+odd=even while (even)(even)=even. So, (y+3) is even, and (y+3)(y+3) is even.
So, x is even. Furthermore, as x is the product of 2 even numbers, it will have at least two 2s' in its factor. Hence, x is divisible by 4.
Statement 2 alone is sufficient.
Is the answer B?
From the question stem - For x to be divisible by 4, it needs to be even. As 4=(2)(2), x also needs to have at least two 2s in its prime factorization to be divisible by 4.
1) Translated into an equation, this means:
[(x^2)+(y^2)]/8=int + 5
[(x^2)+(y^2)] =8(int)+5
As 8 is even, 8 multiplied by any integer is even. 5 is odd. Even+odd=odd So 8(int)+5 =odd
Hence. [(x^2)+(y^2)] = odd
We know that y^2 is odd.
So, x^2 is even as even+odd=odd.
The squre of all even integers is a multiple of 4, however, the square root of all even squares is not necessarily a multiple of 4.
Hence, Statement 1 is insufficient.
2) x-y=3
Plugging this into the equation given in the stem of the question:
P=(x^2)+(y^2)
X=3+y
P=[(3+y)^2}+(y^2)
P=9+6y+(y^2)+(y^2)
P=9+6y+2(y^2)
As p=x-y
X-y=9+6y+2(y^2)
X=9+6y+y^2
X=(y+3)(y+3)
Now, y is odd and odd+odd=even while (even)(even)=even. So, (y+3) is even, and (y+3)(y+3) is even.
So, x is even. Furthermore, as x is the product of 2 even numbers, it will have at least two 2s' in its factor. Hence, x is divisible by 4.
Statement 2 alone is sufficient.
Is the answer B?
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mehravikas
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@Viju,
x and y are integers and 12 is not a square of any integer. Moreover, 36 + 9 = 45 that also leaves a remainder of 5.
in this case x is not divisible by 4
x and y are integers and 12 is not a square of any integer. Moreover, 36 + 9 = 45 that also leaves a remainder of 5.
in this case x is not divisible by 4
viju9162 wrote:answer is A ...
from the argument : p = X^2 + Y^2 ..
From statement 1 & Y is odd, p may look like this:
13 = 4 + 9
13 = 12 + 1
21 = 20 + 1
21 = 12 + 9
hence, A satisfies the question..
From statement 2: X = 3 + Y ... as Y is odd, we get different answers which is divisible/not divisible by 4 ...
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viju9162
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Hi mehravikas,
Thanks for pointing out my mistake. I will again solve the problem.. What is the OA ?
Regards,
Viju
Thanks for pointing out my mistake. I will again solve the problem.. What is the OA ?
Regards,
Viju
"Native of" is used for a individual while "Native to" is used for a large group
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mehravikas
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I have not solved it yet..i just saw your explanation. Will do it soon 
viju9162 wrote:Hi mehravikas,
Thanks for pointing out my mistake. I will again solve the problem.. What is the OA ?
Regards,
Viju
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heshamelaziry
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How did you each this p = x - yssy wrote:Tough question, let me give it a shot..
From the question stem - For x to be divisible by 4, it needs to be even. As 4=(2)(2), x also needs to have at least two 2s in its prime factorization to be divisible by 4.
1) Translated into an equation, this means:
[(x^2)+(y^2)]/8=int + 5
[(x^2)+(y^2)] =8(int)+5
As 8 is even, 8 multiplied by any integer is even. 5 is odd. Even+odd=odd So 8(int)+5 =odd
Hence. [(x^2)+(y^2)] = odd
We know that y^2 is odd.
So, x^2 is even as even+odd=odd.
The squre of all even integers is a multiple of 4, however, the square root of all even squares is not necessarily a multiple of 4.
Hence, Statement 1 is insufficient.
2) x-y=3
Plugging this into the equation given in the stem of the question:
P=(x^2)+(y^2)
X=3+y
P=[(3+y)^2}+(y^2)
P=9+6y+(y^2)+(y^2)
P=9+6y+2(y^2)
As p=x-y
X-y=9+6y+2(y^2)
X=9+6y+y^2
X=(y+3)(y+3)
Now, y is odd and odd+odd=even while (even)(even)=even. So, (y+3) is even, and (y+3)(y+3) is even.
So, x is even. Furthermore, as x is the product of 2 even numbers, it will have at least two 2s' in its factor. Hence, x is divisible by 4.
Statement 2 alone is sufficient.
Is the answer B?
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mehravikas
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Hi Viju,
The answer to this problem is indeed 'A' You got it man...but there was a flaw in your approach. X^2 cannot be 20 or 12 because X has to be an integer.
Now the main part - If x is divisible by 4 then the remainder of x^2 + y^2 will always be 1 because x^2 will always be divisible by 8.
The remainder is 5 so we can say that x is not divisible by 4.
Answer 'A'
Regards,
Vikas
The answer to this problem is indeed 'A'
You got it man...but there was a flaw in your approach. X^2 cannot be 20 or 12 because X has to be an integer.Now the main part - If x is divisible by 4 then the remainder of x^2 + y^2 will always be 1 because x^2 will always be divisible by 8.
The remainder is 5 so we can say that x is not divisible by 4.
Answer 'A'
Regards,
Vikas
viju9162 wrote:Hi mehravikas,
Thanks for pointing out my mistake. I will again solve the problem.. What is the OA ?
Regards,
Viju
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viju9162
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Hi Vikas,
Thank you. But I am getting confused. Please help me here..
From statement (1) and argument, my understanding is as follows:
->When p is divided by 8, the remainder is 5.
-> P = X^2 + Y^2 ( where Y is odd)
Hence, depending on the above conditions, I select the below numbers:
13=4+9 : where 13/8 gives a reminder of 5, Sq rt of 9 is 3 ( which is odd) and Sq rt of 4 is 2 ( which is not divisible by 4)
29 = 4+25 : where 29/8 gives a reminder of 5, Sq rt of 25 is 5 ( which is odd) and Sq rt of 4 is 2 ( which is not divisible by 4)
45=36+9 : where 45/8 gives a reminder of 5, Sq rt of 9 is 3 ( which is odd) and Sq rt of 36 is 6 ( which is not divisible by 4)
This also implies for numbers 53, 61 and so on .. hence A is the answer...
Am I correct here? Please let me know.
Thank you,
Viju
Thank you. But I am getting confused. Please help me here..
From statement (1) and argument, my understanding is as follows:
->When p is divided by 8, the remainder is 5.
-> P = X^2 + Y^2 ( where Y is odd)
Hence, depending on the above conditions, I select the below numbers:
13=4+9 : where 13/8 gives a reminder of 5, Sq rt of 9 is 3 ( which is odd) and Sq rt of 4 is 2 ( which is not divisible by 4)
29 = 4+25 : where 29/8 gives a reminder of 5, Sq rt of 25 is 5 ( which is odd) and Sq rt of 4 is 2 ( which is not divisible by 4)
45=36+9 : where 45/8 gives a reminder of 5, Sq rt of 9 is 3 ( which is odd) and Sq rt of 36 is 6 ( which is not divisible by 4)
This also implies for numbers 53, 61 and so on .. hence A is the answer...
Am I correct here? Please let me know.
Thank you,
Viju
"Native of" is used for a individual while "Native to" is used for a large group
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mehravikas
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yes...absolutely correct !!
viju9162 wrote:Hi Vikas,
Thank you. But I am getting confused. Please help me here..
From statement (1) and argument, my understanding is as follows:
->When p is divided by 8, the remainder is 5.
-> P = X^2 + Y^2 ( where Y is odd)
Hence, depending on the above conditions, I select the below numbers:
13=4+9 : where 13/8 gives a reminder of 5, Sq rt of 9 is 3 ( which is odd) and Sq rt of 4 is 2 ( which is not divisible by 4)
29 = 4+25 : where 29/8 gives a reminder of 5, Sq rt of 25 is 5 ( which is odd) and Sq rt of 4 is 2 ( which is not divisible by 4)
45=36+9 : where 45/8 gives a reminder of 5, Sq rt of 9 is 3 ( which is odd) and Sq rt of 36 is 6 ( which is not divisible by 4)
This also implies for numbers 53, 61 and so on .. hence A is the answer...
Am I correct here? Please let me know.
Thank you,
Viju
