Difference between revisions of "1992 AIME Problems/Problem 1"
(→Problem) |
MRENTHUSIASM (talk | contribs) m (Undo vandalism) |
||
(19 intermediate revisions by 15 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Find the sum of all positive rational | + | Find the sum of all [[positive number |positive]] [[rational number]]s that are less than 10 and that have [[denominator]] 30 when written in [[reduced fraction | lowest terms]]. |
== 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 20:02, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.