Difference between revisions of "2019 AMC 10A Problems/Problem 18"
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<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17</math> | <math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17</math> | ||
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+ | ==Video Solution 1== | ||
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+ | Education, the Study of Everything | ||
==Solution 1== | ==Solution 1== |
Revision as of 21:51, 22 November 2020
- 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 ?
Video Solution 1
Education, the Study of Everything
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 legendary, amazingly, historically great calculations, this eventually gives us as the answer.
-ellpet
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 .
Solution 5
Assuming you are familiar with the rules for basic repeating decimals, . Now we want our base, , to conform to and , the reason being that we wish to convert the number from base to base . Given the first equation, we know that must equal 9, 16, 23, or generally, . The only number in this set that is one of the multiple choices is . When we test this on the second equation, , it comes to be true. Therefore, our answer is .
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.