Difference between revisions of "2018 AMC 12A Problems/Problem 19"
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== Solution 2== | == 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 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math> | + | 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 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math>a_{exp}</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}} | {{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:35, 17 September 2018
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.
(ayushk)
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 2, the second shall be reciprocals of the powers of 3, and the third is 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
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 |
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