Difference between revisions of "1993 AHSME Problems/Problem 30"
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We are going to look at this problem in binary. | We are going to look at this problem in binary. | ||
− | + | <math>x_0 = (0.a_1 a_2 \cdots )_2</math> | |
+ | <math>2x_0 = (a_1.a_2 a_3 \cdots)_2</math> | ||
− | + | If <math>2x_0 < 1</math>, then <math>x_0 < \frac{1}{2}</math> which means that <math>a_1 = 0</math> and so <math>x_1 = (.a_2 a_3 a_4 \cdots)_2</math> | |
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− | If <math>2x_0 < 1</math>, then <math>x_0 < \frac{1}{2}</math> which means that <math> | ||
If <math>2x_0 \geq 1</math> then <math>x \geq \frac{1}{2}</math> which means that <math>x_1 = 2x_0 - 1 = (.a_2 a_3 a_4 \cdots)_2</math>. | If <math>2x_0 \geq 1</math> then <math>x \geq \frac{1}{2}</math> which means that <math>x_1 = 2x_0 - 1 = (.a_2 a_3 a_4 \cdots)_2</math>. | ||
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Using the same logic, we notice that this sequence cycles and that since <math>x_0 = x_5</math> we notice that <math>a_n = a_{n+5}</math>. | Using the same logic, we notice that this sequence cycles and that since <math>x_0 = x_5</math> we notice that <math>a_n = a_{n+5}</math>. | ||
− | We have <math> | + | We have <math>2</math> possibilities for each of <math>a_1</math> to <math>a_5</math> but we can't have <math>a_1 = a_2 = a_3 = a_4 = a_5 = 1</math> so we have <math>2^5 - 1 = \boxed{(D)31}</math> |
-mathman523 | -mathman523 | ||
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== See also == | == See also == |
Latest revision as of 21:22, 1 June 2020
Problem
Given , let for all integers . For how many is it true that ?
Solution
We are going to look at this problem in binary.
If , then which means that and so
If then which means that .
Using the same logic, we notice that this sequence cycles and that since we notice that .
We have possibilities for each of to but we can't have so we have
-mathman523
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Problem 30 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.