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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"

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(→‎Solution 2 (Bash))
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==Solution 2 (Bash)==
 
==Solution 2 (Bash)==
  
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WORKING IN PROGRESS
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Consider all the cases where <math>a+b=15</math>, and construct the following table:
 +
 
 +
<cmath>\begin{array}{|c|c|c|}
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& & \\ [-2ex]
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\textbf{a} & \textbf{b} & \textbf{a/b} \\ [0.5ex]
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\hline
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& & \\ [-2ex]
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\ \ \ \ 1 \ \ \ \ & \ \ \ \ 14 \ \ \ \ & \ \ \ \ 1/14 \ \ \ \ \\ [1ex]
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2 & 13 & 2/13 \\ [1ex]
 +
3 & 12 & 1/4 \\ [1ex]
 +
4 & 11 & 4/11 \\ [1ex]
 +
5 & 10 & 1/2 \\ [1ex]
 +
6 & 9 & 2/3 \\ [1ex]
 +
7 & 8 & 7/8 \\ [1ex]
 +
8 & 7 & 8/7 \\ [1ex]
 +
9 & 6 & 3/2 \\ [1ex]
 +
10 & 5 & 2 \\ [1ex]
 +
11 & 4 & 11/4 \\ [1ex]
 +
12 & 3 & 4 \\ [1ex]
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13 & 2 & 13/2 \\ [1ex]
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14 & 1 & 14 \\ [1ex]
 +
\end{array}</cmath>
 +
 
 +
Let <math>\frac{a}{b}=n</math>.
 +
Now, we list all the possible integers obtained from an addition of two values of <math>n</math>:
 +
 
 +
<cmath>\begin{array}{|c|c|c|c|}
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& & &\\ [-2ex]
 +
\textbf{n1} & \textbf{n2} & \textbf{sum} & \textbf{condition} \\ [0.5ex]
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\hline
 +
& & &\\ [-2ex]
 +
\ \ \ \ 2 \ \ \ \ & \ \ \ \ 2 \ \ \ \ & \ \ \ \ 4 \ \ \ \ & \ \\ [1ex]
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& 4 & 6 & \\ [1ex]
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& 14 & 16 & \\ [1ex]
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4 & 4 & 8 & \\ [1ex]
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& 14 & 18 & \\ [1ex]
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14 & 14 & 28 & \\ [1ex]
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1/2 & 1/2 & 1 & \\ [1ex]
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& 3/2 & 2 & \\ [1ex]
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& 13/2 & 7 & \\ [1ex]
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3/2 & 3/2 & 3 & \\ [1ex]
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& 13/2 & 8 & \ \ Rep.\ \ \\ [1ex]
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13/2 & 13/2 & 13 & \\ [1ex]
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1/4 & 11/4 & 3 & Rep. \\ [1ex]
 +
\end{array}</cmath>
 +
 
 +
Although <math>13</math> terms are found in total, <math>8</math> and <math>3</math> have appeared twice respectively. Taken repetition into account, we have a total of <math>\boxed{\textbf{(C)}\ 11}</math> terms.
  
 
~Wilhelm Z
 
~Wilhelm Z

Revision as of 09:55, 24 November 2021

Problem 5

Call a fraction $\frac{a}{b}$, not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?

$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$

Solution 1

$a=15-b$ so the fraction is $\frac{15-b}{b}$ which is $\frac{15}{b}-1$. We can just ignore the $-1$ part and only care about $\frac{15}{b}$. Now we just group $\frac{15}{1}, \frac{15}{3}, \frac{15}{5}$ as the integers and $\frac{15}{2}, \frac{15}{6}, \frac{15}{10}$ as the halves. We get $30, 20, 18, 10, 8, 6$ from the integers group and $15, 10, 9, 5, 4, 3$ from the halves group. These are both $6$ integers and we see that $10$ overlaps, so the answer is $\boxed{\textbf{(C)}\ 11}$.

~lopkiloinm

Solution 2 (Bash)

Consider all the cases where $a+b=15$, and construct the following table:

\[\begin{array}{|c|c|c|} & & \\ [-2ex] \textbf{a} & \textbf{b} & \textbf{a/b} \\ [0.5ex] \hline & & \\ [-2ex] \ \ \ \ 1 \ \ \ \ & \ \ \ \ 14 \ \ \ \ & \ \ \ \ 1/14 \ \ \ \ \\ [1ex] 2 & 13 & 2/13 \\ [1ex] 3 & 12 & 1/4 \\ [1ex] 4 & 11 & 4/11 \\ [1ex] 5 & 10 & 1/2 \\ [1ex] 6 & 9 & 2/3 \\ [1ex] 7 & 8 & 7/8 \\ [1ex] 8 & 7 & 8/7 \\ [1ex] 9 & 6 & 3/2 \\ [1ex] 10 & 5 & 2 \\ [1ex] 11 & 4 & 11/4 \\ [1ex] 12 & 3 & 4 \\ [1ex] 13 & 2 & 13/2 \\ [1ex] 14 & 1 & 14 \\ [1ex] \end{array}\]

Let $\frac{a}{b}=n$. Now, we list all the possible integers obtained from an addition of two values of $n$:

\[\begin{array}{|c|c|c|c|} & & &\\ [-2ex] \textbf{n1} & \textbf{n2} & \textbf{sum} & \textbf{condition} \\ [0.5ex] \hline & & &\\ [-2ex] \ \ \ \ 2 \ \ \ \ & \ \ \ \ 2 \ \ \ \ & \ \ \ \ 4 \ \ \ \ & \ \\ [1ex] & 4 & 6 & \\ [1ex] & 14 & 16 & \\ [1ex] 4 & 4 & 8 & \\ [1ex] & 14 & 18 & \\ [1ex] 14 & 14 & 28 & \\ [1ex] 1/2 & 1/2 & 1 & \\ [1ex] & 3/2 & 2 & \\ [1ex] & 13/2 & 7 & \\ [1ex] 3/2 & 3/2 & 3 & \\ [1ex] & 13/2 & 8 & \ \ Rep.\ \ \\ [1ex] 13/2 & 13/2 & 13 & \\ [1ex] 1/4 & 11/4 & 3 & Rep. \\ [1ex] \end{array}\]

Although $13$ terms are found in total, $8$ and $3$ have appeared twice respectively. Taken repetition into account, we have a total of $\boxed{\textbf{(C)}\ 11}$ terms.

~Wilhelm Z

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