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.
  
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=== Solution 3===
  
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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: $\frac{1}{30},\frac{7}{30},\frac{11}{30},\frac{13}{30},\frac{17}{30},\frac{19}{30},\frac{23}{30},\frac{29}{30}$

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, $1+\frac{19}{30}=\frac{49}{30}.$ Following this pattern, our answer is $4(10)+8(1+2+3+\cdots+9)=\boxed{400}.$

Solution 2

By Euler's Totient Function, there are $8$ numbers that are relatively prime to $30$, less than $30$. Note that they come in pairs $(m,30-m)$ which result in sums of $1$; thus the sum of the smallest $8$ rational numbers satisfying this is $\frac12\cdot8\cdot1=4$. Now refer to solution 1.

Solution 3

Note that if $x$ is a solution, then $(300-x)$ is a solution. We know that $\phi(300) = 80.$ Therefore the answer is $\displaystyle\frac{80}{2} \cdot \displaystyle\frac{300}{30} = \boxed{400}.$

1992 AIME (ProblemsAnswer KeyResources)
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Followed by
Problem 2
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