2013 AIME II Problems/Problem 6
Contents
[hide]Problem 6
Find the least positive integer such that the set of consecutive integers beginning with contains no square of an integer.
Solutions
Solution 1
The difference between consecutive integral squares must be greater than 1000. , so \implies x\geq500x=500x>500n=x-500nn^21000\Nn^2=1000n=10\sqrt{10}\sqrt{10}3^2=910=(x+3)^2=x^2+6x+9x^210=6x+9x=1/6\implies \sqrt{10}\approx 19/6$. This is 3.16.
Then,$ (Error compiling LaTeX. Unknown error_msg)n\approx 31.6n^2<1000n31531^2532^2531^2=281961532^2=283024282,000NN=282000\implies\boxed{282}x^2(x+1)^2x\ge 0(x+1)^2-x^2=2x+1x\ge 02x+1x2x+1\ge 1000x\ge \frac{999}{2}x\ge 500\overline{N000}\rightarrow \overline{N999}xx=500x^2=250000(x+1)^2=251001250000\overline{N000}310009619742x+1531^2=281961532^2=283024282000\boxed{282}$.
===Solution 3=== Let$ (Error compiling LaTeX. Unknown error_msg)xN250x500kx-500x^2(500+k)^2250000+1000k+k^22500001000k\overline{N000}\rightarrow \overline{N999}k^210001000k=31k^2=961(k+1)^2=32^2=10241000k1000x(500+31)^2=281961(500+32)^2=283024282000\boxed{282}$.
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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