# Find the floor value of a finite continued surd

Given $x=20062007$, and let $$A=\sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}.$$

Find the greatest integer not exceeding $A$.

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## 1 Answer

$$10x+1<\sqrt{100x^2+39x+\sqrt{3}}<10x+2$$ $$4x+1<\sqrt{16x^2+10x+1}<\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}<\sqrt{16x^2+10x+2}<4x+2$$ $$2x+1=\sqrt{4x^2+4x+1}<\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}<\sqrt{4x^2+4x+2}<2x+2$$ $$x+1=\sqrt{x^2+2x+1}<A<\sqrt{x^2+2x+2}<x+2$$

Thus $\lfloor A \rfloor=x+1=20062008$.

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Nice! my frend,Ivan Loh. –  math110 Mar 31 '13 at 9:12