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> | ||
− | ==Solution== | + | ==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} + | + | 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} + \cdots</math> |
− | |||
− | + | Notice that this is equivalent to | |
+ | <cmath>2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )</cmath> | ||
− | -- | + | 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== | ||
+ | |||
+ | 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>, 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 <math>7</math> with the answer choices, or actually solve the quadratic, so that the answer is <math>\boxed{\textbf{(D) }16}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Just as in Solution 1, we arrive at the equation <math>\frac{2k+3}{k^2-1}=\frac{7}{51}</math>. | ||
+ | |||
+ | Therefore now, we can 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 number, 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 will lead us to the answer <math>\boxed{\textbf{(D) }16}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | 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^2-1},</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 by OmegaLearn == | ||
+ | https://youtu.be/SCGzEOOICr4?t=1110 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/3YhYGSneu70 | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/W5V1UF-vSDE | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Line 20: | Line 59: | ||
{{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 08:59, 17 July 2024
- 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 with the answer choices, or actually solve the quadratic, so that the answer is .
Solution 3
Just as in Solution 1, we arrive at the equation .
Therefore now, we can rewrite this as . Notice that . As is a prime number, we therefore must have that one of and is divisible by . Now, checking each of the answer choices, this will lead us to the answer .
Solution 4
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 5
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 by OmegaLearn
https://youtu.be/SCGzEOOICr4?t=1110
~ pi_is_3.14
Video Solution 1
Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
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.