Difference between revisions of "1992 AIME Problems/Problem 1"
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===Solution 2=== | ===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>!</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. | 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>!</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=== | ||
| + | By the Gauss Summation Formula, there is a total sum of \frac{299*300}{2} for the numbers less than 10. The prime factorization of 30 is 2*3*5, therefore subtract 2\frac{149*150}{2} for the factors of 2, 3\frac{99*100}{2} for the factors of 3, and 5\frac{59*69}{2} for the factors of 5. Now add 6\frac{49*50}{2} for the factors of 6 subtracted twice, 10\frac{29*30}{2} for the factors of 10, and 15\frac{19*20}{2} for the factors of 15. Finally, subtract 30\frac{9*10}{2} to account for the intersection between the factors of 10 and 15. Dividing the final sum 12000 by 30 obtains the solution of 400. | ||
{{AIME box|year=1992|before=First question|num-a=2}} | {{AIME box|year=1992|before=First question|num-a=2}} | ||
Revision as of 10:02, 14 August 2009
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
By the Gauss Summation Formula, there is a total sum of \frac{299*300}{2} for the numbers less than 10. The prime factorization of 30 is 2*3*5, therefore subtract 2\frac{149*150}{2} for the factors of 2, 3\frac{99*100}{2} for the factors of 3, and 5\frac{59*69}{2} for the factors of 5. Now add 6\frac{49*50}{2} for the factors of 6 subtracted twice, 10\frac{29*30}{2} for the factors of 10, and 15\frac{19*20}{2} for the factors of 15. Finally, subtract 30\frac{9*10}{2} to account for the intersection between the factors of 10 and 15. Dividing the final sum 12000 by 30 obtains the solution of 400.
| 1992 AIME (Problems • Answer Key • Resources) | ||
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