Difference between revisions of "2018 AMC 12A Problems/Problem 19"
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by a product of geometric series, from which the answer is <math>15+4=\boxed{\textbf{(C) } 19}.</math> | by a product of geometric series, from which the answer is <math>15+4=\boxed{\textbf{(C) } 19}.</math> | ||
− | ~athens2016 | + | ~athens2016 ~MRENTHUSIASM |
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− | ~MRENTHUSIASM | ||
== Solution 2 == | == Solution 2 == |
Revision as of 03:53, 21 October 2022
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 1
Note that the fractions of the form where
and
are nonnegative integers, span all terms of the infinite sum.
Therefore, the infinite sum becomes
by a product of geometric series, from which the answer is
~athens2016 ~MRENTHUSIASM
Solution 2
Separate into 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
. We can easily calculate these to be
respectively.
The fourth infinite series shall be all real numbers in the form , where
.
The fifth is all real numbers in the form , where
.
The sixth is all real numbers in the form , where
.
The seventh infinite series is all real numbers in the form , where
.
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
.
~~Edited by mprincess0229~~
Solution 3
This solution is Solution 1 but potentially clearer.
Clearly this is just summing over the reciprocals of the numbers of the form , where
. SO our desired sum is
. By the infinite geometric series formula,
is just
. Applying the infinite geometric series formula again gives that
. Applying the infinite geometric series formula again yields
. Hence our final answer is
.
-vsamc
Video Solution by LetsSolveMathProblems
https://www.youtube.com/watch?v=woXlEargLpI&ab_channel=LetsSolveMathProblems
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.