HMMT-Feb 2022 guts, Problem 4

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 $[10(a+b)]$. (Here, any number equally between two integers or tenths of integers, respectively, is rounded up. For example, $[-4.5]=-4$ and $\langle 4.35\rangle=4.4$.)

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 $[10(a+b)]=988$. $\fbox{988}$.

Answer: 988