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1991 AIME Question 6

AIME · 1991 AIME

Question 6

Suppose $r$ is a real number for which $$
\begin{aligned}&\left\lfloor r+\frac{19}{100}\right\rfloor+\left\lfloor r+\frac{20}{100}\right\rfloor\\&\quad+\left\lfloor r+\frac{21}{100}\right\rfloor+\cdots\\&\quad+\left\lfloor r+\frac{91}{100}\right\rfloor=546.\end{aligned}
$$ Find $\lfloor100r\rfloor.$ (For real $x,$ $\lfloor x\rfloor$ is the greatest integer less than or equal to $x.$)