1992 AIME Problems/Problem 1

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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

There are 8 fractions which fit the conditions between 0 and 1: $\displaystyle \frac{1}{30},\frac{7}{30},\frac{11}{30},\frac{13}{30},\frac{17}{30},\frac{19}{30},\frac{23}{30},\frac{29}{30}$

Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain them by adding 1 to each of our first 8 terms. For example, $\displaystyle 1+\frac{19}{30}=\frac{49}{30}.$ Following this pattern, our answer is $4(10)+8(1+2+3+\cdots+9)=400.$

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First question
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[[{{{year}}} AIME Problems/Problem 2|Problem 2]]
[[{{{year}}} AIME Problems/Problem 1|1]] [[{{{year}}} AIME Problems/Problem 2|2]] [[{{{year}}} AIME Problems/Problem 3|3]] [[{{{year}}} AIME Problems/Problem 4|4]] [[{{{year}}} AIME Problems/Problem 5|5]] [[{{{year}}} AIME Problems/Problem 6|6]] [[{{{year}}} AIME Problems/Problem 7|7]] [[{{{year}}} AIME Problems/Problem 8|8]] [[{{{year}}} AIME Problems/Problem 9|9]] [[{{{year}}} AIME Problems/Problem 10|10]] [[{{{year}}} AIME Problems/Problem 11|11]] [[{{{year}}} AIME Problems/Problem 12|12]] [[{{{year}}} AIME Problems/Problem 13|13]] [[{{{year}}} AIME Problems/Problem 14|14]] [[{{{year}}} AIME Problems/Problem 15|15]]
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