Difference between revisions of "1991 AIME Problems/Problem 6"

 
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== Problem ==
 
== Problem ==
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Suppose <math>r^{}_{}</math> is a real number for which
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<center><math>
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\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.
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</math></center>
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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 ==
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{{solution}}
  
 
== See also ==
 
== See also ==
* [[1991 AIME Problems]]
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{{AIME box|year=1991|num-b=5|num-a=7}}

Revision as of 02:15, 2 March 2007

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

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See also

1991 AIME (ProblemsAnswer KeyResources)
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
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