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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