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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 23"

(Created page with "To solve this problem, we can consider each unit of <math>1</math> to be separated into <math>n</math> equal increments. (For example, if each unit of <math>1</math> is separa...")
 
 
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Thus, our answer is <math>16+81=\boxed{97}</math>.
 
Thus, our answer is <math>16+81=\boxed{97}</math>.
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Note: This problem is probably way too hard to appear on an AMC 10
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Solution 2 (Fakesolve)
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Note that the answer probably looks like <math>\frac{\textbf{something}}{81}</math>, and based on the answer choices, we can guess that <math>\frac{16}{81}</math> is the desired probability, as <math>16</math> is a power of <math>2</math> and <math>\frac{16}{81}=\left(\frac{2}{3}\right)^{4}</math>

Latest revision as of 16:56, 30 September 2025

To solve this problem, we can consider each unit of $1$ to be separated into $n$ equal increments. (For example, if each unit of $1$ is separated into $2019$ equal increments, there would be $2019$ tiny fragments of $\frac{1}{2019}$ per unit of $1$.) Recall that $\text{probability} = \frac{\text{favorable}}{\text{total}}$, so we have to calculate (1) the number of ways to arrange $4$ “blocks” of $1$ in the whole “grid” of length $10$ given $n$ increments per unit (this is because the maximum least value for one of the sets is $9$) such that no two of them overlap and (2) the number of ways to arrange $4$ “blocks” of $1$ in the whole “grid” of length $10$ given $n$ increments per unit under no restrictions. We can easily see that the probability is

\[\frac{4! \times \dbinom{6x+4}{4}}{(9x+1)^4}\].

(The $4!$ is included due to the four blocks $S_1, S_2, S_3, S_4$ being able to be arranged order.)

Since there are infinite real numbers in a unit of $1$, there are infinite increments in a unit of $1$, so we should take the limit as $n$ approaches infinity:

\[\text{probability} = \lim_{x \to \infty} \frac{4! \times \dbinom{6x+4}{4}}{(9x+1)^4}\] Dividing leading coefficients, we get the probability to be $\frac{6^4}{9^4}$, which simplifies to $\frac{2^4}{3^4}$.

Thus, our answer is $16+81=\boxed{97}$.

Note: This problem is probably way too hard to appear on an AMC 10

Solution 2 (Fakesolve)

Note that the answer probably looks like $\frac{\textbf{something}}{81}$, and based on the answer choices, we can guess that $\frac{16}{81}$ is the desired probability, as $16$ is a power of $2$ and $\frac{16}{81}=\left(\frac{2}{3}\right)^{4}$

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