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

 
m
Line 1: Line 1:
 
== 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 ==
* [[1991 AIME Problems]]
+
{{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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

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