Difference between revisions of "1991 AIME Problems/Problem 6"
(→Solution) |
Jackshi2006 (talk | contribs) (→Solution) |
||
Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
− | There are <math>91 - 19 + 1 = 73</math> numbers in the [[sequence]]. Since the terms of the sequence 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 values of each of the terms of the sequence 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>\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor</math>, which is also the first term of this sequence with a value of <math>8</math>, so <math>8 \le r + \frac{57}{100} < 8.01</math>. Solving shows that <math>\frac{743}{100} \le r < \frac{744}{100}</math>, so <math>743\le 100r < 744</math>, and <math>\lfloor 100r \rfloor = 743</math>. | + | There are <math>91 - 19 + 1 = 73</math> numbers in the [[sequence]]. Since the terms of the sequence 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 values of each of the terms of the sequence 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>\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor</math>, which is also the first term of this sequence with a value of <math>8</math>, so <math>8 \le r + \frac{57}{100} < 8.01</math>. Solving shows that <math>\frac{743}{100} \le r < \frac{744}{100}</math>, so <math>743\le 100r < 744</math>, and <math>\lfloor 100r \rfloor = \boxed{743}</math>. |
== See also == | == See also == |
Revision as of 17:51, 27 August 2020
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 the terms of the sequence can be at most apart, all of the numbers in the sequence can take one of two possible values. Since , the values of each of the terms of the sequence 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 , which is also the first term of this sequence with a value of , so . Solving shows that , so , and .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.