Difference between revisions of "2019 AMC 10A Problems/Problem 18"
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==Solution 1== | ==Solution 1== | ||
− | We can expand the fraction <math>0.\overline{23}_k</math> as follows: <math>0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + ...</math> | + | We can expand the fraction <math>0.\overline{23}_k</math> as follows: <math>0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + ...</math> Notice that this is equivalent to |
<cmath>2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )</cmath> | <cmath>2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )</cmath> | ||
− | By summing the | + | By summing the geometric series and simplifying, we have <math>\frac{2k+3}{k^2-1} = \frac{7}{51}</math>. Solving this quadratic equation (or simply testing the answer choices) yields the answer <math>k = \boxed{\textbf{(D) }16}</math>. |
==Solution 2== | ==Solution 2== | ||
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Let <math>a = 0.2323\dots_k</math>. Therefore, <math>k^2a=23.2323\dots_k</math>. | Let <math>a = 0.2323\dots_k</math>. Therefore, <math>k^2a=23.2323\dots_k</math>. | ||
− | From this, we see that <math>k^2a-a=23_k</math>. | + | From this, we see that <math>k^2a-a=23_k</math>, so <math>a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}</math>. |
− | + | Now, similar to in Solution 1, we can either test if <math>2k+3</math> is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is <math>\boxed{\textbf{(D) }16}</math>. | |
− | <math> | + | ==Solution 3 (bash)== |
+ | We can simply plug in all the answer choices as values of <math>k</math>, and see which one works. After lengthy calculations, this eventually gives us <math>\boxed{\textbf{(D) }16}</math> as the answer. | ||
− | + | ==Solution 4== | |
+ | Just as in Solution 1, we arrive at the equation <math>\frac{2k+3}{k^2-1}=\frac{7}{51}</math>. | ||
− | + | We can now rewrite this as <math>\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}</math>. Notice that <math>2k+3=2(k+1)+1=2(k-1)+5</math>. As <math>17</math> is a prime, we therefore must have that one of <math>k-1</math> and <math>k+1</math> is divisible by <math>17</math>. Now, checking each of the answer choices, this gives <math>\boxed{\textbf{(D) }16}</math>. | |
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==Video Solution== | ==Video Solution== |
Revision as of 00:21, 27 February 2019
- The following problem is from both the 2019 AMC 10A #18 and 2019 AMC 12A #11, so both problems redirect to this page.
Contents
Problem
For some positive integer , the repeating base- representation of the (base-ten) fraction is . What is ?
Solution 1
We can expand the fraction as follows: Notice that this is equivalent to
By summing the geometric series and simplifying, we have . Solving this quadratic equation (or simply testing the answer choices) yields the answer .
Solution 2
Let . Therefore, .
From this, we see that , so .
Now, similar to in Solution 1, we can either test if is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is .
Solution 3 (bash)
We can simply plug in all the answer choices as values of , and see which one works. After lengthy calculations, this eventually gives us as the answer.
Solution 4
Just as in Solution 1, we arrive at the equation .
We can now rewrite this as . Notice that . As is a prime, we therefore must have that one of and is divisible by . Now, checking each of the answer choices, this gives .
Video Solution
For those who want a video solution : https://www.youtube.com/watch?v=DFfRJolhwN0
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 |
Followed by Problem 12 |
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.