Difference between revisions of "2019 AMC 10A Problems/Problem 18"
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==Solution 3 (bash)== | ==Solution 3 (bash)== | ||
− | We can simply plug in all the answer choices as values of <math>k</math>, and see which one works. After | + | We can simply plug in all the answer choices as values of <math>k</math>, and see which one works. After legendary, amazingly, historically great calculations, this eventually gives us <math>\boxed{\textbf{(D) }16}</math> as the answer. |
+ | |||
+ | -ellpet | ||
==Solution 4== | ==Solution 4== | ||
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==Solution 5== | ==Solution 5== | ||
− | Assuming you are familiar with the rules for basic repeating decimals, <math>0.232323... | + | Assuming you are familiar with the rules for basic repeating decimals, <math>0.232323... = \frac{23}{99}</math>. Now we want our base, <math>k</math>, to conform to <math>23\equiv7\pmod k</math> and <math>99\equiv51\pmod k</math>, the reason being that we wish to convert the number from base <math>10</math> to base <math>k</math>. Given the first equation, we know that <math>k</math> must equal 9, 16, 23, or generally, <math>7n+2</math>. The only number in this set that is one of the multiple choices is <math>16</math>. When we test this on the second equation, <math>99\equiv51\pmod k</math>, it comes to be true. Therefore, our answer is <math>\boxed{\textbf{(D) }16}</math>. |
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+ | ==Solution 6== | ||
+ | |||
+ | Note that the LHS equals <cmath>\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2},</cmath> from which we see our equation becomes <cmath>\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \implies \ \ 51(2k+3) = 7(k^2-1).</cmath> | ||
+ | |||
+ | Note that <math>17</math> therefore divides <math>k^2 - 1,</math> but as <math>17</math> is prime this therefore implies <cmath>k \equiv \pm 1 \pmod{17}.</cmath> (Warning: This would not be necessarily true if <math>17</math> were composite.) Note that <math>\boxed{\textbf{(D)} 16 }</math> is the only answer choice congruent satisfying this modular congruence, thus completing the problem. <math>\square</math> | ||
+ | |||
+ | ~ Professor-Mom | ||
==Video Solution== | ==Video Solution== | ||
For those who want a video solution: https://www.youtube.com/watch?v=DFfRJolhwN0 | For those who want a video solution: https://www.youtube.com/watch?v=DFfRJolhwN0 | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/3YhYGSneu70 | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
+ | (Please put video solutions at the end in order of when they were edited in) | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/FFpBMkKnOk8?t=1263 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=A|num-b=17|num-a=19}} | {{AMC10 box|year=2019|ab=A|num-b=17|num-a=19}} | ||
{{AMC12 box|year=2019|ab=A|num-b=10|num-a=12}} | {{AMC12 box|year=2019|ab=A|num-b=10|num-a=12}} | ||
+ | [[Category:Introductory Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 07:04, 19 February 2021
- 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 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 .
Solution 6
Note that the LHS equals from which we see our equation becomes
Note that therefore divides but as is prime this therefore implies (Warning: This would not be necessarily true if were composite.) Note that is the only answer choice congruent satisfying this modular congruence, thus completing the problem.
~ Professor-Mom
Video Solution
For those who want a video solution: https://www.youtube.com/watch?v=DFfRJolhwN0
Video Solution
Education, the Study of Everything
(Please put video solutions at the end in order of when they were edited in)
Video Solution by TheBeautyofMath
https://youtu.be/FFpBMkKnOk8?t=1263
~IceMatrix
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.