Do you know number theory

If 21 x + 20 y = 420 21x+20y=420 for 0 < x < 20 0<x<20 and 0 < y < 21 0<y<21 , find the number of real ( x , y ) (x,y) pairs for which x 2 + y 2 x^2+y^2 is a perfect square.


The answer is 11.

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1 solution

We can represent,21x+20y=420 by a line which intersects y axis at (0,21)and x axis at (20,0) and the x and y have their limit between these numbers.We want the values of (x^2 +y^2)^1/2 which is nothing but distance of points of above line from origin.From the limits of x and y, we can conclude that,we will only take points of hypotenuse of triangle formed by x axis, y axis and the above line except end points.but as mentioned we only want integral solution.Now,from above facts the thing for which we are searching is nothing but integral value cevians to the hypotensue (29)whose other two sides are 21 and 20 and as we know perpendicular will be smallest cevian which is equal to 14.48(approx)..so your result came (20-15)+(21-15) will be your answer which is 11.... Sorry for rip english...if you required answer with actual figure....then telegram me (krity kamal)

Oh, I thought they were integers! I coded it for integers! :(

Vinayak Srivastava - 10 months, 3 weeks ago

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How do you code?

A Former Brilliant Member - 10 months, 3 weeks ago

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On Python, I know Number Theory easy codes.

Vinayak Srivastava - 10 months, 3 weeks ago

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@Vinayak Srivastava i even do not know easy codes.

A Former Brilliant Member - 10 months, 3 weeks ago

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