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

## 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 $\displaystyle\frac{80}{2} \cdot \displaystyle\frac{300}{30} = \boxed{400}.$

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