1991 AIME Problems/Problem 6

Problem

Suppose $r^{}_{}$ is a real number for which

$\left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{10...$

There are $91 - 19 + 1 = 73$ numbers in the sequence. Since $\left\lfloor r + \frac{91}{100} \right\rfloor$ can be at most $1$ apart, all of the numbers in the sequence can take one of two possible values. Since $\frac{546}{73} = 7 R 35$ , the numbers must be either $7$ or $8$ . As the remainder is $35$ , $8$ must take on $35$ of the values, with $7$ being the value of the remaining $73 - 35 = 38$ numbers. The 39th number is $19 + 39 - 1= 57$ , and so $8 \le \left\lfloor r + \frac{57}{100}\right\rfloor < 8.01$ . Solving shows that $\frac{743}{100} \le r < \frac{744}{100}$ , so $\lfloor 100r \rfloor = 743$ .