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 | + | 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 == | ||
− | {{ | + | 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}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 19:20, 11 March 2007
Problem
Suppose is a real number for which
Find . (For real , is the greatest integer less than or equal to .)
Solution
There are numbers in the sequence. Since can be at most apart, all of the numbers in the sequence can take one of two possible values. Since , the numbers must be either or . As the remainder is , must take on of the values, with being the value of the remaining numbers. The 39th number is , and so . Solving shows that , so .
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 |