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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0Find the number of integers solutions of \(\large \color{black}{\begin{align} x+y=200\hspace{1.5em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, I meant 199*2*2, so 796 for the other cases. So there are 800 :P

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0didnt understand ur soln

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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.

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Yes. I confirm that result. x can be positive or negative: Two solutions $$ x=200y,y\le200\text{ here x is positive}\\ x=y200,y\le 199\text{ here x is negative} $$ Now count number of solutions. One solution for every value of y, which can be positive or negative for each y above: $$ 200\times2+199\times 2+1\times 2\text{(for when y=0)}\\ =800 $$
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