Difference between revisions of "2018 AMC 12A Problems/Problem 19"
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Knowing <math>a_{1}</math> and <math>a_{2}</math>, we can substitute and solve for <math>a_{4}</math>, and we get <math>\frac{1}{2}</math>. If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them <math>\frac{1}{4}</math> and <math>\frac{1}{8}</math>. | Knowing <math>a_{1}</math> and <math>a_{2}</math>, we can substitute and solve for <math>a_{4}</math>, and we get <math>\frac{1}{2}</math>. If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them <math>\frac{1}{4}</math> and <math>\frac{1}{8}</math>. | ||
− | Finally, for the seventh sequence, we see <math>a_{7}=\frac{a_{8}}{30}</math>, where <math>a_{8}</math> is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get <math>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30} | + | Finally, for the seventh sequence, we see <math>a_{7}=\frac{a_{8}}{30}</math>, where <math>a_{8}</math> is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get <math>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}</math>, but when we separated the sequence into its parts, we ignored the <math>1/1</math>, so adding in the <math>1</math>, we get <math>1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}</math>, which when we solve for, we get <math>\frac{29}{8}=\frac{29a_{8}}{30}</math>, <math>\frac{1}{8}=\frac{a_{8}}{30}</math>, <math>\frac{30}{8}=(a_{8})</math>, <math>\frac{15}{4}=(a_{8})</math>. So our answer is <math>\frac{15}{4}</math>, but we are asked to add the numerator and denominator, which sums up to <math>19</math>, which is the answer. |
~~Edited by mprincess0229~~ | ~~Edited by mprincess0229~~ |
Revision as of 02:07, 25 January 2020
Contents
Problem
Let be the set of positive integers that have no prime factors other than , , or . The infinite sum of the reciprocals of the elements of can be expressed as , where and are relatively prime positive integers. What is ?
Solution
It's just since this represents all the numbers in the denominator. (athens2016)
Solution 2
Separate into 7 separate infinite series's so we can calculate each and find the original sum:
The first infinite sequence shall be all the reciprocals of the powers of , the second shall be reciprocals of the powers of , and the third will consist of reciprocals of the powers of 5. We can easily calculate these to be respectively.
The fourth infinite series shall be all real numbers in the form , where and are greater than or equal to 1.
The fifth is all real numbers in the form , where and are greater than or equal to 1.
The sixth is all real numbers in the form , where and are greater than or equal to 1.
The seventh infinite series is all real numbers in the form , where and and are greater than or equal to 1.
Let us denote the first sequence as , the second as , etc. We know , , , let us find . factoring out from the terms in this subsequence, we would get .
Knowing and , we can substitute and solve for , and we get . If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them and .
Finally, for the seventh sequence, we see , where is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get , but when we separated the sequence into its parts, we ignored the , so adding in the , we get , which when we solve for, we get , , , . So our answer is , but we are asked to add the numerator and denominator, which sums up to , which is the answer.
~~Edited by mprincess0229~~
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.