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Question Number 50987 by naka3546 last updated on 23/Dec/18

$$\sqrt{x+y+9}+\sqrt{x-y+8}={33}^2$$ $$x>y$$ $$x,y\in {\mathbb{Z}}^{+}$$ $$(x^y+y^x)mod\left(1000\right)=?$$

Commented byRasheed.Sindhi last updated on 23/Dec/18

$$Nouniqueanswer.$$ $$Forexample:$$ $$x=320,y=247$$ $$({320}^{247}+{247}^{320})mod1000=401$$ $$AND$$ $$x=336,y=280$$ $$({336}^{280}+{280}^{336})mod1000=176$$

Commented bynaka3546 last updated on 23/Dec/18

$$sorry,Ihaveeditedmypost.$$

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