Difference between revisions of "1991 AIME Problems/Problem 6"
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== Problem == | == Problem == | ||
+ | Suppose <math>r^{}_{}</math> is a real number for which | ||
+ | <center><math> | ||
+ | \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. | ||
+ | </math></center> | ||
+ | Find <math>\lfloor 100r \rfloor</math>. (For real <math>x^{}_{}</math>, <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x^{}_{}</math>.) | ||
== Solution == | == Solution == | ||
+ | {{solution}} | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1991|num-b=5|num-a=7}} |
Revision as of 02:15, 2 March 2007
Problem
Suppose is a real number for which
Find . (For real , is the greatest integer less than or equal to .)
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
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 | ||
All AIME Problems and Solutions |