Difference between revisions of "1992 AIME Problems/Problem 1"
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== Solution == | == Solution == | ||
| + | ===Solution 1=== | ||
| + | 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{400}.</math> | ||
| − | + | ===Solution 2=== | |
| + | By [[Euler's Totient Function]], there are <math>8</math> numbers that are relatively prime to <math>30</math>, less than <math>30</math>. Note that they come in pairs <math>(m,30-m)</math> which result in sums of <math>1</math>; thus the sum of the smallest <math>8</math> rational numbers satisfying this is <math>\frac12\cdot8\cdot1=4</math>. Now refer to solution 1. | ||
| + | |||
| + | === Solution 3=== | ||
| + | |||
| + | Note that if <math>x</math> is a solution, then <math>(300-x)</math> is a solution. We know that <math>\phi(300) = 80.</math> Therefore the answer is <math>\frac{80}{2} \cdot\frac{300}{30} = \boxed{400}.</math> | ||
| + | {{AIME box|year=1992|before=First question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:02, 23 February 2022
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
Note that if 
is a solution, then 
is a solution. We know that 
Therefore the answer is 
| 1992 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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