, 2013

Algebra Difficulty 5.4 AIME, harder Prove it Slovenia

For every real number $a$ , let $[a]$ be the greatest integer that is not greater than $a$ . Find all integers $y$ for which there exists a real number $x$ such that $\left[\frac{x+23}{8}\right] = [\sqrt{x}] = y$ .

Solution

Let $y$ be such a number. Then $\sqrt{x} \ge |\sqrt{x}| = y$ . Since $\sqrt{x} \ge 0$ we have $y = [\sqrt{x}] \ge 0$ , so we may square the inequality to get $x \ge y^2$ . Also, $\frac{x+23}{8} < [\frac{x+23}{8}]+1 = y+1$ , or $x < 8y-15$ . This implies $y^2 < 8y-15$ , or $(y-3)(y-5) < 0$ , which means that $3 < y < 5$ . But $y$ is an integer, so $y=4$ . In this case there indeed exists a real number $x$ which satisfies the condition, for example $x=16$ .

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