Difference between revisions of "1992 AIME Problems/Problem 1"
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There are 8 [[fraction]]s which fit the conditions between 0 and 1: <math>\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> | There are 8 [[fraction]]s which fit the conditions between 0 and 1: <math>\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> | ||
| − | 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, <math>1+\frac{19}{30}=\frac{49}{30}.</math> Following this pattern, our answer is <math>4(10)+8(1+2+3+\cdots+9)=\boxed{ | + | 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, <math>1+\frac{19}{30}=\frac{49}{30}.</math> Following this pattern, our answer is <math>4(10)+8(1+2+3+\cdots+9)=\boxed{400}.</math> |
===Solution 2=== | ===Solution 2=== | ||
Revision as of 18:27, 2 September 2017
Contents
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.
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