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

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~Wilhelm Z
 
~Wilhelm Z
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== Solution 3 ==
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All special fractions are: <math>\frac{1}{14}</math>, <math>\frac{2}{13}</math>, <math>\frac{3}{12}</math>, <math>\frac{4}{11}</math>, <math>\frac{5}{10}</math>, <math>\frac{6}{9}</math>, <math>\frac{7}{8}</math>, <math>\frac{8}{7}</math>, <math>\frac{9}{6}</math>, <math>\frac{10}{5}</math>, <math>\frac{11}{4}</math>, <math>\frac{12}{3}</math>, <math>\frac{13}{2}</math>, <math>\frac{14}{1}</math>.
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Hence, the following numbers are integers: <math>\frac{3}{12} + \frac{11}{4}</math>, <math>\frac{5}{10} + \frac{5}{10}</math>, <math>\frac{5}{10} + \frac{9}{6}</math>, <math>\frac{5}{10} + \frac{13}{2}</math>, <math>\frac{9}{6} + \frac{9}{6}</math>, <math>\frac{9}{6} + \frac{13}{2}</math>, <math>\frac{10}{5} + \frac{10}{5}</math>, <math>\frac{10}{5} + \frac{12}{3}</math>, <math>\frac{10}{5} + \frac{14}{1}</math>, <math>\frac{12}{3} + \frac{12}{3}</math>, <math>\frac{12}{3} + \frac{14}{1}</math>, <math>\frac{13}{2} + \frac{13}{2}</math>, <math>\frac{14}{1} + \frac{14}{1}</math>.
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This leads to the following distinct integers: 3, 1, 2, 7, 8, 4, 6, 16, 18, 13, 28.
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Therefore, the answer is <math>\boxed{\textbf{(C) }11}</math>.
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~Steven Chen (www.professorchenedu.com)

Revision as of 21:21, 25 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 (Enumeration)

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

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

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

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

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


~Wilhelm Z

Solution 3

All special fractions are: $\frac{1}{14}$, $\frac{2}{13}$, $\frac{3}{12}$, $\frac{4}{11}$, $\frac{5}{10}$, $\frac{6}{9}$, $\frac{7}{8}$, $\frac{8}{7}$, $\frac{9}{6}$, $\frac{10}{5}$, $\frac{11}{4}$, $\frac{12}{3}$, $\frac{13}{2}$, $\frac{14}{1}$.

Hence, the following numbers are integers: $\frac{3}{12} + \frac{11}{4}$, $\frac{5}{10} + \frac{5}{10}$, $\frac{5}{10} + \frac{9}{6}$, $\frac{5}{10} + \frac{13}{2}$, $\frac{9}{6} + \frac{9}{6}$, $\frac{9}{6} + \frac{13}{2}$, $\frac{10}{5} + \frac{10}{5}$, $\frac{10}{5} + \frac{12}{3}$, $\frac{10}{5} + \frac{14}{1}$, $\frac{12}{3} + \frac{12}{3}$, $\frac{12}{3} + \frac{14}{1}$, $\frac{13}{2} + \frac{13}{2}$, $\frac{14}{1} + \frac{14}{1}$.

This leads to the following distinct integers: 3, 1, 2, 7, 8, 4, 6, 16, 18, 13, 28.

Therefore, the answer is $\boxed{\textbf{(C) }11}$.

~Steven Chen (www.professorchenedu.com)

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