Difference between revisions of "2019 AMC 10A Problems/Problem 18"
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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>. | 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>. | + | 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== | ==Solution 3== | ||
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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>. | 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>. | ||
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==Solution 4== | ==Solution 4== |
Latest revision as of 09:59, 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.