Difference between revisions of "1992 AIME Problems/Problem 1"
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By [[Euler's Totient Function]], there are <math>8</math> numbers that are relatively prime to <math>30</math>, less than <math>30</math>. Note that they come in pairs <math>(m,30-m)</math> which result in sums of <math>1</math>; thus the sum of the smallest <math>8</math> rational numbers satisfying this is <math>\frac12\cdot8\cdot1=4</math>. Now refer to solution 1. | By [[Euler's Totient Function]], there are <math>8</math> numbers that are relatively prime to <math>30</math>, less than <math>30</math>. Note that they come in pairs <math>(m,30-m)</math> which result in sums of <math>1</math>; thus the sum of the smallest <math>8</math> rational numbers satisfying this is <math>\frac12\cdot8\cdot1=4</math>. Now refer to solution 1. | ||
+ | === Solution 3=== | ||
+ | Note that if <math>x</math> is a solution, then <math>(300-x)</math> is a solution. We know that <math>\phi(300) = 80.</math> Therefore the answer is <math>\displaystyle\frac{80}{2} \cdot \displaystyle\frac{300}{30} = \boxed{400}.</math> | ||
{{AIME box|year=1992|before=First question|num-a=2}} | {{AIME box|year=1992|before=First question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:13, 16 September 2020
Problem
Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms.
Solution
Solution 1
There are 8 fractions which fit the conditions between 0 and 1:
Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain by adding 1 to each of our first 8 terms. For example, Following this pattern, our answer is
Solution 2
By Euler's Totient Function, there are numbers that are relatively prime to , less than . Note that they come in pairs which result in sums of ; thus the sum of the smallest rational numbers satisfying this is . Now refer to solution 1.
Solution 3
Note that if is a solution, then is a solution. We know that Therefore the answer is
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