Difference between revisions of "1991 AIME Problems/Problem 6"
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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>.) | 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 (Hopefully Intuitive)== |
+ | THIS SOLUTION IS INCORRECT, PLEASE CORRECT IT IF YOU HAVE TIME! | ||
+ | ~Arcticturn | ||
+ | |||
+ | == Solution 1== | ||
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>. | 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>. | ||
== 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 \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100}\rfloor \ge ...\ge \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 | + | 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 \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100}\rfloor \ge ...\ge \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 r\rfloor = ... = \lfloor r+\frac{18}{100}\rfloor = 7</math>, and <math>\lfloor r+\frac{92}{100}\rfloor = ...= \lfloor r+1\rfloor = 8</math>. Since there are 19 terms in the former equation and 8 terms in the latter, our answer is <math>\lfloor nx\rfloor = 546+19\cdot 7+8\cdot 8=\boxed{743}</math> | ||
+ | |||
+ | === Note === | ||
+ | In the contest, you would just observe this mentally, and then calculate <math>546+19\cdot 7+8\cdot 8= 743</math>, hence the speed at which one can carry out this solution. | ||
== See also == | == See also == |
Latest revision as of 16:02, 24 November 2023
Contents
Problem
Suppose is a real number for which
Find . (For real , is the greatest integer less than or equal to .)
Solution (Hopefully Intuitive)
THIS SOLUTION IS INCORRECT, PLEASE CORRECT IT IF YOU HAVE TIME! ~Arcticturn
Solution 1
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, , and . Since there are 19 terms in the former equation and 8 terms in the latter, our answer is
Note
In the contest, you would just observe this mentally, and then calculate , hence the speed at which one can carry out this solution.
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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