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If x and y are positive integers and x^2 – y^2 = 101, what is the valu

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Bunuel
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If x and y are positive integers and x^2 – y^2 = 101, what is the valu [#permalink] New post   21 Feb 2020, 05:47
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If x and y are positive integers and x^2 – y^2 = 101, what is the value of x^2 + y^2 ?


A. 5050
B. 5101
C. 5150
D. 5201
E. 5301


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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu [#permalink] New post   21 Feb 2020, 05:48
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Bunuel wrote:
If x and y are positive integers and x^2 – y^2 = 101, what is the value of x^2 + y^2 ?


A. 5050
B. 5101
C. 5150
D. 5201
E. 5301


\(x^2 - y^2 = (x-y)(x+y)=101\). 101 is a prime number, so it can be broken into the product of two positive multiples only in one way 101=1*101. Now, since x and y are positive integers, then x - y < x + y, thus x - y = 1 and x + y = 101. Solving for x and y gives x = 51 and y = 50. Thus \(x^2 + y^2=51^2+50^2=5,101\).

Answer: B.
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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu [#permalink] New post   21 Feb 2020, 10:36
Expert Reply

Solution



Given
In this question, we are given that
    • The numbers x and y are positive integers
    • Also, \(x^2 – y^2 = 101\)

To find
We need to determine
    • The value of \(x^2 + y^2\)

Approach and Working out
The given equation can be written as \(x^2 – y^2 = 101\)
    • Or, (x + y) (x – y) = 101

Now, 101 is a prime number and can be written as a product of two integers only as 101 * 1
Hence, we can say x + y = 101 and x – y = 1

Solving, we get x = 51 and y = 50
    • Therefore, \(x^2 + y^2 = 51^2 + 50^2 = 2601 + 2500 = 5101\)

Thus, option B is the correct answer.

Correct Answer: Option B
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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu [#permalink] New post   21 Feb 2020, 15:27
Expert Reply
Bunuel wrote:
If x and y are positive integers and x^2 – y^2 = 101, what is the value of x^2 + y^2 ?


A. 5050
B. 5101
C. 5150
D. 5201
E. 5301


We can PLUG IN THE ANSWERS, which represent the value of \(x^2 + y^2\).
Since \((x^2 + y^2) + (x^2 - y^2) = 2x^2\), the correct answer must yield an EVEN INTEGER when added to \(x^2 – y^2 = 101\).
Implication:
The correct answer must be ODD.
Eliminate A and C.

B: \(x^2 + y^2 = 5101\)
Adding together \(x^2 + y^2 = 5101\) and \(x^2 – y^2 = 101\), we get:
\(2x^2 = 5202\)
\(x^2 = 2601\)

Since \(50^2 = 2500\) and \(60^2 = 3600\), \(x\) in this case must be between 50 and 60.
For its square to yield a units digit of 1, \(x\) must have a units digit of 1 or 9.
Test \(x=51\):
\(51^2 = (50+1)^2 = 50^2 + 1^2 + 2*50*1 = 2500 + 1 + 100 = 2601\)
Success!

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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu [#permalink] New post   16 Apr 2021, 00:11
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\(x^2 - y^2 = 101\)

=> (x + y) (x - y) = 101 * 1

=> x + y = 101 and x - y = 1

Solving linear equations we get, x = 51 and y = 50.

=> \(x^2 + y^2 = 51^2 + 50^2 = 2601+ 2500 = 5101\)

Answer B
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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu [#permalink]
16 Apr 2021, 00:11
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