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:

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, Following this pattern, our answer is

Solution 2

By Euler's Totient Function, there are numbers that are relatively prime to , less than . Note that they come in pairs which result in sums of ; thus the sum of the smallest rational numbers satisfying this is . Now refer to solution 1.

Solution 3

By the Gauss Summation Formula, there is a total sum of \frac{299*300}{2} for the numbers less than 10. The prime factorization of 30 is 2*3*5, therefore subtract 2\frac{149*150}{2} for the factors of 2, 3\frac{99*100}{2} for the factors of 3, and 5\frac{59*69}{2} for the factors of 5. Now add 6\frac{49*50}{2} for the factors of 6 subtracted twice, 10\frac{29*30}{2} for the factors of 10, and 15\frac{19*20}{2} for the factors of 15. Finally, subtract 30\frac{9*10}{2} to account for the intersection between the factors of 10 and 15. Dividing the final sum 12000 by 30 obtains the solution of 400.