Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

For a real number $x$ , let $[x]$ be $x$ rounded to the nearest integer and $\langle x\rangle$ be $x$ rounded to the nearest tenth. Real numbers $a$ and $b$ satisfy $\langle a\rangle+[b]=98.6$ and $[a]+\langle b\rangle=99.3$ . Compute the minimum possible value of $[10(a+b)]$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Without loss of generality, let $a$ and $b$ have the same integer part or integer parts that differ by at most 1, as we can always repeatedly subtract 1 from the larger number and add 1 to the smaller to get another solution. Next, we note that the decimal part of $a$ must round to .6 and the decimal part of $b$ must round to .3. We note that $(a, b)=(49.55,49.25)$ is a solution and is clearly minimal in fractional parts, giving us $[10(a+b)]=988$ .

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