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FoolForMath
 3 years ago
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?
FoolForMath
 3 years ago
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?

This Question is Closed

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

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.3Obviously 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 \(2+2(7)+2(9)+2(9)+2(9)+11=81.\)
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