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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    Find the number of integers solutions of \(\large \color{black}{\begin{align} |x|+|y|=200\hspace{1.5em}\\~\\ \end{align}}\)

  2. skullpatrol
    • one year ago
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    any ideas?

  3. anonymous
    • one year ago
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    x must be between -200 and 200. For each x we have a corresponding value for |y|, so 2 y solutions per x, unless y = 0. Thus there are 2 (for |x| = 200, y = 0), 2 (for x=0, |y| = 200), and 199*2 = 398 for other cases. So we have 402 solutions.

  4. anonymous
    • one year ago
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    Whoops, I meant 199*2*2, so 796 for the other cases. So there are 800 :P

  5. mathmath333
    • one year ago
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    didnt understand ur soln

  6. anonymous
    • one year ago
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    So first if x is not in the range -200 <= x <= 200, then that means |y| is negative which is impossible. We then consider the special cases where x or y is 0: If x is 0 then |y| = 200. So y = -200, 200; 2 solutions. Similarly if y is 0, x = -200, 200; 2 solutions. So now we consider any x in the range -200 <= x <= 200 where x is not 0, -200, or 200. There are 199 positive x and 199 negative x, so 398 in total. For each x, we get |y| = some number that is not 0, so y = + or - this number; 2 solutions per value of x. Hence this gives an additional 398*2 = 796 solutions. Altogether there are 2 + 2 + 796 = 800 solutions.

  7. ybarrap
    • one year ago
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    Yes. I confirm that result. x can be positive or negative: Two solutions $$ x=200-|y|,|y|\le200\text{ here x is positive}\\ x=|y|-200,|y|\le 199\text{ here x is negative} $$ Now count number of solutions. One solution for every value of y, which can be positive or negative for each |y| above: $$ 200\times2+199\times 2+1\times 2\text{(for when |y|=0)}\\ =800 $$

  8. ybarrap
    • one year ago
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    Does that make sense?

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