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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
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Suppose <math>r^{}_{}</math> is a [[real number]] for which
 
<div style="text-align:center"><math>
 
<div style="text-align: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.
 
\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></div>
 
</math></div>
−
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>.)
+
Find <math>\lfloor 100r \rfloor</math>. (For real <math>x^{}_{}</math>, <math>\lfloor x \rfloor</math> is the [[floor function|greatest integer]] less than or equal to <math>x^{}_{}</math>.)
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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There are <math>91 - 19 + 1 = 73</math> numbers in the [[sequence]]. Since <math>\lfloor r + \frac{91}{100} \rfloor</math> can be at most <math>1</math> apart, all of the numbers in the sequence can take one of two possible values. Since <math>\frac{546}{73} = 7 R 35</math>, the numbers must be either <math>7</math> or <math>8</math>. As the remainder is <math>35</math>, <math>8</math> must take on <math>35</math> of the values, with <math>7</math> being the value of the remaining <math>73 - 35 = 38</math> numbers. The 39th number is <math>19 + 39 - 1= 57</math>, and so <math>8 \le \lfloor r + \frac{57}{100} < 8.01</math>. Solving shows that <math>\frac{743}{100} \le r < \frac{744}{100}</math>, so <math>\lfloor r \rfloor = 743</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=5|num-a=7}}
 
{{AIME box|year=1991|num-b=5|num-a=7}}
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 +
[[Category:Intermediate Number Theory Problems]]

Revision as of 18:20, 11 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

There are $91 - 19 + 1 = 73$ numbers in the sequence. Since $\lfloor r + \frac{91}{100} \rfloor$ 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 $8 \le \lfloor r + \frac{57}{100} < 8.01$. Solving shows that $\frac{743}{100} \le r < \frac{744}{100}$, so $\lfloor r \rfloor = 743$.

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
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