Difference between revisions of "1992 AIME Problems/Problem 1"
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=== Solution 3=== | === 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> | + | 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}} | {{AIME box|year=1992|before=First question|num-a=2}} | ||
{{MAA Notice}} | {{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 | |
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