1991 AIME Problems/Problem 6

Problem

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

$\left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.$

Find $\lfloor 100r \rfloor$ . (For real $x^{}_{}$ , $\lfloor x \rfloor$ is the greatest integer less than or equal to $x^{}_{}$ .)

Solution

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 $\lfloor 100r \rfloor = 743$ .

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1991 AIME Problems/Problem 6

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