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FoolForMath

  • 2 years ago

Another fiasco of wolframalpha: How many integer solutions are there for the in-equation \[ x^2 +y^2 \le 25 \]? 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?

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  1. FoolForMath
    • 2 years ago
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    @ash2326: I posted twice in the same group?

  2. Mr.Math
    • 2 years ago
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    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 \(2+2(7)+2(9)+2(9)+2(9)+11=81.\)

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