Difference between revisions of "1992 AIME Problems/Problem 1"

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== Solution ==
 
== Solution ==
===Solution 1===
 
There are 8 [[fraction]]s which fit the conditions between 0 and 1: <math>\frac{1}{30},\frac{7}{30},\frac{11}{30},\frac{13}{30},\frac{17}{30},\frac{19}{30},\frac{23}{30},\frac{29}{30}</math>
 
  
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, <math>1+\frac{19}{30}=\frac{49}{30}.</math> Following this pattern, our answer is <math>4(10)+8(1+2+3+\cdots+9)=\boxed{400}.</math>
 
  
===Solution 2===
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This is cool. Thanks Bye
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}}
 
{{MAA Notice}}
 

Revision as of 18:59, 23 February 2022

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

This is cool. Thanks Bye