Difference between revisions of "2019 AMC 10A Problems/Problem 15"
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Noticing that there is clearly a pattern, but the formula for it is hidious, we first find the first few terms of the sequence to see if there is any pattern: <math>1, \frac{1}{7}, \frac{3}{11}, \frac{1}{5}, \frac{3}{19}, \frac{3}{23} ...</math> | Noticing that there is clearly a pattern, but the formula for it is hidious, we first find the first few terms of the sequence to see if there is any pattern: <math>1, \frac{1}{7}, \frac{3}{11}, \frac{1}{5}, \frac{3}{19}, \frac{3}{23} ...</math> | ||
− | Now, we notice that the denominator seems to be in a pattern: <math>1, 3, 3, 1, 3, 3, 1, 3, 3...</math> | + | Now, we notice that the denominator seems to be in a pattern: <math>1, 3, 3, 1, 3, 3, 1, 3, 3...</math> Then, using the positions of each term, we can come up with a simple formula for the denominator with n as the position or index: <math>3n + (n - 1)</math>. |
Plugging <math>n</math> in for <math>2019</math>, we get <math>\boxed{(E) 8078}</math> | Plugging <math>n</math> in for <math>2019</math>, we get <math>\boxed{(E) 8078}</math> |
Revision as of 15:35, 12 February 2020
- The following problem is from both the 2019 AMC 10A #15 and 2019 AMC 12A #9, so both problems redirect to this page.
Contents
Problem
A sequence of numbers is defined recursively by ,
, and
for all
Then
can be written as
, where
and
are relatively prime positive integers. What is
Solution 1 (Induction)
Using the recursive formula, we find ,
, and so on. It appears that
, for all
. Setting
, we find
, so the answer is
.
To prove this formula, we use induction. We are given that and
, which satisfy our formula. Now assume the formula holds true for all
for some positive integer
. By our assumption,
and
. Using the recursive formula,
so our induction is complete.
Solution 2
Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by .
We have , so in other words,
.
By recursively following this pattern, we can see that .
By plugging in 2019, we thus find . Since the numerator and the denominator are relatively prime, the answer is
.
-eric2020
Solution 3
It seems reasonable to transform the equation into something else. Let ,
, and
. Therefore, we have
Thus,
is the harmonic mean of
and
. This implies
is a harmonic sequence or equivalently
is arithmetic. Now, we have
,
,
, and so on. Since the common difference is
, we can express
explicitly as
. This gives
which implies
.
~jakeg314
Solution 4 (Not rigorous at all)
Noticing that there is clearly a pattern, but the formula for it is hidious, we first find the first few terms of the sequence to see if there is any pattern:
Now, we notice that the denominator seems to be in a pattern: Then, using the positions of each term, we can come up with a simple formula for the denominator with n as the position or index:
.
Plugging in for
, we get
~EricShi1685
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 10 Problems and Solutions |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.