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FoolForMath
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Another fiasco of wolframalpha:
How many integer solutions are there for the inequation \[ x^2 +y^2 \le 25 \]?
And this is what wolframalpha says: http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2+%3C%3D+25
But I got a different answer, Just for fun, can you find the right answer?
 2 years ago
 2 years ago
FoolForMath Group Title
Another fiasco of wolframalpha: How many integer solutions are there for the inequation \[ x^2 +y^2 \le 25 \]? And this is what wolframalpha says: http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2+%3C%3D+25 But I got a different answer, Just for fun, can you find the right answer?
 2 years ago
 2 years ago

This Question is Closed

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
@ash2326: I posted twice in the same group?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.3
Obviously the maximum value \(x\) or \(y\) can take is \(5\), and the minimum is \(5\). We can list all the choices as: At \(x=\pm 5\), \(y\) can take only the value \(0\). At \(x=\pm4\), \(y\) can take the values \(0, \pm1, \pm2\) and \( \pm3\). At \(x=\pm 3\), \(y\) can take the values \(0, \pm1, \pm2, \pm3\) and \(\pm4\). At \(x=\pm 2\), \(y\) can take the value \(0, \pm 1, \pm 2, \pm 3\) and \( \pm 4\). At \(x=\pm1\), \(y\) can take the values \(0, \pm1, \pm 2, \pm 3\) and \(\pm 4\). At \(x=0\) \(y\) can take the values \(0, \pm1, \pm2, \pm 3, \pm 4\) and \(\pm 5\). So the total number of integer solutions is \(2+2(7)+2(9)+2(9)+2(9)+11=81.\)
 2 years ago
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