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What is the remainder of x^2 + y^2, when it is divided by 4?

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What is the remainder of x^2 + y^2, when it is divided by 4?  [#permalink]

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New post 08 Nov 2017, 23:39
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A
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C
D
E

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  75% (hard)

Question Stats:

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[GMAT math practice question]

What is the remainder of x^2 + y^2, when it is divided by 4?

1) x and y are different prime numbers.
2) x – y = 2

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Re: What is the remainder of x^2 + y^2, when it is divided by 4?  [#permalink]

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New post 08 Nov 2017, 23:58
=>

Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution.

Since we have 2 variables and 0 equation, C is most likely to be the answer.

Conditions 1) & 2)

Since x and y are different prime numbers and x – y = 2, both x and y are odd integers. x = 2a + 1 and y = 2b + 1 for some integers a and b. x^2 + y^2 = (2a+1)^2 + (2b+1)^2 = 4a^2 + 4a + 1 + 4b^2 + 4b + 1 = 4(a^2 + a + b^2 + b ) + 2.
Thus, its remainder is 2, when it is divided by 4.
Both condition 1) and 2) are sufficient.
The answer is C.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously, there may be cases where the answer is A, B, D or E.
Answer: C
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Re: What is the remainder of x^2 + y^2, when it is divided by 4?  [#permalink]

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New post 09 Nov 2017, 00:11
X, Y are prime numbers :--

1, 3, 5,7,11

(A) x=1, y=3

Remainder is 2

X=3,Y=5

R= 2

X= 5, Y =11

R =2

We get R =2 in all cases

Why A can't win.


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What is the remainder of x^2 + y^2, when it is divided by 4?  [#permalink]

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New post 09 Nov 2017, 09:06
1
sun01 wrote:
X, Y are prime numbers :--

1, 3, 5,7,11

(A) x=1, y=3

Remainder is 2

X=3,Y=5

R= 2

X= 5, Y =11

R =2

We get R =2 in all cases

Why A can't win.


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Hi sun01

highlighted part is wrong. \(1\) is not a prime number but \(2\) is. so if \(x=2\) & \(y=3\) then the remainder will be \(1\). and for other primes the remainder you have mentioned is \(2\). so no unique value. Hence Insufficient

Statement 1: \(x\) & \(y\) can be any prime nos. hence Insufficient

Statement 2: \(x-y=2\). square both sides to get

\(x^2+y^2-2xy=4 =>x^2+y^2=4+2xy\). Divide this by \(4\) to get

\(\frac{{x^2+y^2}}{4} = 1+\frac{xy}{2}\). But we don't have the value of \(x\) & \(y\) to get the remainder. Insufficient

Combining 1 & 2, we know that both \(x\) & \(y\) are different prime numbers and both are odd. so the remainder \(\frac{xy}{2}\) will be \(1\) because any odd number divided by \(2\) will leave \(1\) as remainder. Sufficient

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Re: What is the remainder of x^2 + y^2, when it is divided by 4?  [#permalink]

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New post 10 Nov 2017, 07:57
sun01 wrote:
X, Y are prime numbers :--

1, 3, 5,7,11

(A) x=1, y=3

Remainder is 2

X=3,Y=5

R= 2

X= 5, Y =11

R =2

We get R =2 in all cases

Why A can't win.


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We have two following cases.

\(x = 2\), \(y = 3\) : \(x^2 + y^2 = 4 + 9 = 13\).
It has the remainder 1 when it is divided by 4.

\(x = 3\), \(y = 5\) : \(x^2 + y^2 = 9 + 25 = 34\).
It has the remainder 2 when it is divided by 4.

From your solution, 1 is not a prime number.
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Re: What is the remainder of x^2 + y^2, when it is divided by 4? &nbs [#permalink] 10 Nov 2017, 07:57
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