Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer United States

Problem:

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 $\langle a\rangle+**=98.6$ and $[a]+\langle b\rangle=99.3$ . Compute the minimum possible value of $[10(a+b)]$ .

(Here, any number equally between two integers or tenths of integers, respectively, is rounded up. For example, $[-4.5]=-4$ and $\langle 4.35\rangle=4.4$ .)

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.