1991 AIME Problems/Problem 6
Problem
Suppose is a real number for which
![$\left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.$](http://latex.artofproblemsolving.com/d/4/9/d49bf9d61cafb77fc2043857f815732ddef15f57.png)
Find . (For real
,
is the greatest integer less than or equal to
.)
Solution
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See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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All AIME Problems and Solutions |