1991 AIME Problem 6

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

Write

r=n+f

with integer

n

and

0\leq f<1

After removing the common integer part, count how many of the

73

fractional terms cross

1

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.

Problem 6 in other years