2021 Fall AMC 12B Problems/Problem 5

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: $\frac{3}{12} + \frac{11}{4}$ , $\frac{5}{10} + \frac{5}{10}$ , $\frac{5}{10} + \frac{9}{6}$ , $\frac{5}{10} + \frac{13}{2}$ , $\frac{9}{6} + \frac{9}{6}$ , $\frac{9}{6} + \frac{13}{2}$ , $\frac{10}{5} + \frac{10}{5}$ , $\frac{10}{5} + \frac{12}{3}$ , $\frac{10}{5} + \frac{14}{1}$ , $\frac{12}{3} + \frac{12}{3}$ , $\frac{12}{3} + \frac{14}{1}$ , $\frac{13}{2} + \frac{13}{2}$ , $\frac{14}{1} + \frac{14}{1}$ .