Difference between revisions of "1992 AIME Problems/Problem 1"
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== Solution == | == Solution == | ||
| + | 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 18:08, 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: 
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, 
Following this pattern, our answer is 
