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

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== Solution ==
 
== Solution ==
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There are 8 fractions which fit the conditions from 0 to 1: <math>\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}</math>
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Their sum is 4. Note that there are also 8 terms from 1 to 2, and 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>
  
 
== See also ==
 
== See also ==
 
* [[1992 AIME Problems]]
 
* [[1992 AIME Problems]]

Revision as of 19:08, 26 July 2006

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 from 0 to 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 from 1 to 2, and 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 $4(10)+8(1+2+3+\cdots+9)=400.$

See also

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