Question

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 $x.$ )

Answer

Answer: 743

Concepts:floor and ceiling functions counting integers in a range summation

Difficulty rating: 1980

Small Hint:

Write $r=n+f$ with integer $n$ and $0\leq f<1$

Big Hint:

After removing the common integer part, count how many of the $73$ fractional terms cross $1$

Solution:

Write $r=n+f,$ where $n$ is an integer and $0\leq f<1.$ There are $73$ summands. Since $546=73\cdot7+35,$ we must have $n=7,$ and exactly $35$ of the numbers $f+\frac{19}{100},$ $\ldots,$ $f+\frac{91}{100}$ have floor $1.$ These must be the terms with numerators $57,$ $58,$ $\ldots,$ $91.$ Hence $f+\frac{56}{100}<1\leq f+\frac{57}{100},$ so $43\leq100f<44.$ Therefore $\lfloor100r\rfloor=700+43=743.$

1991 AIME Problem 6

Answer: 743

Difficulty rating: 1980

Small Hint:

Big Hint:

Solution:

Problem 6 in Other Years