Difference between revisions of "1997 AIME Problems/Problem 5"

Revision as of 01:14, 5 October 2008

Problem

The number $r$ can be expressed as a four-place decimal $0.abcd,$ where $a, b, c,$ and $d$ represent digits, any of which could be zero. It is desired to approximate $r$ by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to $r$ is $\frac 27.$ What is the number of possible values for $r$ ?

Solution

The nearest fractions to $\frac 27$ with numerator $1$ are $\frac 13, \frac 14$ ; and with numerator $2$ are $\frac 26, \frac 28 = \frac 13, \frac 14$ anyway. For $\frac 27$ to be the best approximation for $r$ , the decimal must be closer to $\frac 27 \approx .28571$ than to $\frac 13 \approx .33333$ or $\frac 14 \approx .25$ .

Thus $r$ can range between $\frac{\frac 14 + \frac{2}{7}}{2} \approx .267857$ and $\frac{\frac 13 + \frac{2}{7}}{2} \approx .309523$ . At $r = .2678, .3096$ , it becomes closer to the other fractions, so $.2679 \le r \le .3095$ and the number of values of $r$ is $3096 - 2678 + 1 = \boxed{417}$ .

Solution 2

1/3=.3333 and 1/4=.2500, This range contains two times the amount in the desired set

3333-2500=833

(833+1)/2=417

1997 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1997 AIME Problems/Problem 5"

Revision as of 01:14, 5 October 2008

Contents

Problem

Solution

Solution 2

See also

@@ Line 10: / Line 10: @@
 /3=.3333 and 1/4=.2500,
 This range contains two times the amount in the desired set
 -2500=833
 (833+1)/2=417