Difference between revisions of "1992 AIME Problems/Problem 1"
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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, <math>\displaystyle 1+\frac{19}{30}=\frac{49}{30}.</math> Following this pattern, our answer is <math>4(10)+8(1+2+3+\cdots+9)=400.</math> | 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, <math>\displaystyle 1+\frac{19}{30}=\frac{49}{30}.</math> Following this pattern, our answer is <math>4(10)+8(1+2+3+\cdots+9)=400.</math> | ||
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Revision as of 21:54, 11 November 2007
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: 
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, 
Following this pattern, our answer is 
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