Difference between revisions of "1992 AIME Problems/Problem 1"
(→See also: AIME box) |
(year) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | There are 8 [[fraction]]s which fit the conditions between 0 and 1: <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 them by adding 1 to each of our first 8 terms. For example, <math> | + | Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain them 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)=400.</math> |
| − | {{AIME box|before=First question|num-a=2}} | + | {{AIME box|year=1992|before=First question|num-a=2}} |
Revision as of 21:55, 11 November 2007
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 between 0 and 1: 
Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain them by adding 1 to each of our first 8 terms. For example, 
Following this pattern, our 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 | ||
