Difference between revisions of "1997 AIME Problems/Problem 5"
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The nearest fractions to <math>\frac 27</math> with numerator <math>1</math> are <math>\frac 13, \frac 14</math>; and with numerator <math>2</math> are <math>\frac 26, \frac 28 = \frac 13, \frac 14</math> anyway. For <math>\frac 27</math> to be the best approximation for <math>r</math>, the decimal must be closer to <math>\frac 27 \approx .28571</math> than to <math>\frac 13 \approx .33333</math> or <math>\frac 14 \approx .25</math>. | The nearest fractions to <math>\frac 27</math> with numerator <math>1</math> are <math>\frac 13, \frac 14</math>; and with numerator <math>2</math> are <math>\frac 26, \frac 28 = \frac 13, \frac 14</math> anyway. For <math>\frac 27</math> to be the best approximation for <math>r</math>, the decimal must be closer to <math>\frac 27 \approx .28571</math> than to <math>\frac 13 \approx .33333</math> or <math>\frac 14 \approx .25</math>. | ||
− | Thus <math>r</math> can range between <math>\frac{\frac 14 + \frac{2}{7}}{2} \approx .267857</math> and <math>\frac{\frac 13 + \frac{2}{7}}{2} \approx .309523</math>. At <math>r = . | + | Thus <math>r</math> can range between <math>\frac{\frac 14 + \frac{2}{7}}{2} \approx .267857</math> and <math>\frac{\frac 13 + \frac{2}{7}}{2} \approx .309523</math>. At <math>r = .2678, .3096</math>, it becomes closer to the other fractions, so <math>.2679 \le r \le .3095</math> and the number of values of <math>r</math> is <math>3095 - 2679 + 1 = \boxed{417}</math>. |
== See also == | == See also == | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:46, 4 June 2021
Problem
The number can be expressed as a four-place decimal where and represent digits, any of which could be zero. It is desired to approximate by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to is What is the number of possible values for ?
Solution
The nearest fractions to with numerator are ; and with numerator are anyway. For to be the best approximation for , the decimal must be closer to than to or .
Thus can range between and . At , it becomes closer to the other fractions, so and the number of values of is .
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.