Difference between revisions of "1991 AIME Problems/Problem 6"
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== Solution 2 (Faster) == | == Solution 2 (Faster) == | ||
− | Recall by Hermite's Identity that <math>\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor</math> for positive integers <math>n</math>, and real <math>x</math>. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, <math>\lfloor r\rfloor,...,\lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100},...,\lfloor r+1\rfloor \ge 8</math>. We can see that <math>\lfloor r\rfloor +1=\lfloor r+1\rfloor</math>. Because <math>\lfloor r\rfloor</math> is at most 7, and <math>\lfloor r+1\rfloor</math> is at least 8, we can clearly see their values are <math>7</math> and <math>8</math> respectively. So, <math>lfloor nx\rfloor</math> is <math>546+19\cdot 7+8\cdot 8=\boxed{743}</math> | + | Recall by Hermite's Identity that <math>\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor</math> for positive integers <math>n</math>, and real <math>x</math>. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, <math>\lfloor r\rfloor,...,\lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100}\rfloor,...,\lfloor r+1\rfloor \ge 8</math>. We can see that <math>\lfloor r\rfloor +1=\lfloor r+1\rfloor</math>. Because <math>\lfloor r\rfloor</math> is at most 7, and <math>\lfloor r+1\rfloor</math> is at least 8, we can clearly see their values are <math>7</math> and <math>8</math> respectively. So, <math>lfloor nx\rfloor</math> is <math>546+19\cdot 7+8\cdot 8=\boxed{743}</math> |
== See also == | == See also == |
Revision as of 11:18, 1 October 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 .
Solution 2 (Faster)
Recall by Hermite's Identity that for positive integers , and real . Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, and . We can see that . Because is at most 7, and is at least 8, we can clearly see their values are and respectively. So, is
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.