2021 AMC Fall 10B Problem 7

The special fractions are $\frac{a}{15-a}$ for $a = 1, \dots, 14$. In lowest terms:
$$\tfrac{1}{14},\ \tfrac{2}{13},\ \tfrac{1}{4},\ \tfrac{4}{11},\ \tfrac{1}{2},\ \tfrac{2}{3},\ \tfrac{7}{8},\ \tfrac{8}{7},\ \tfrac{3}{2},\ 2,\ \tfrac{11}{4},\ 4,\ \tfrac{13}{2},\ 14.$$

Two reduced fractions add to an integer only if they have the same denominator (a fraction with denominator $d > 1$ can only be cancelled by another with denominator $d$). So group by denominator:

• Integers $2, 4, 14$: pairwise sums (repeats allowed) are $4, 6, 8, 16, 18, 28$.
• Denominator $2$: $\frac12, \frac32, \frac{13}{2}$: sums are $1, 2, 7, 3, 8, 13$.
• Denominator $4$: $\frac14, \frac{11}{4}$: the only integer sum is $\frac14 + \frac{11}{4} = 3$.
• Denominators $3, 7, 8, 11, 13, 14$ appear once each, and doubling them gives no integer.

Collecting distinct values: $\{1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28\}$, which has $11$ elements.

The answer is $\boxed{\textbf{(C)}\ 11}$.

Sorting by reduced denominator turns a $14 \times 14$ search into a few tiny lists, because a sum of reduced fractions is an integer only when the denominators cancel each other. The remaining work is careful listing with duplicate removal.

• Forgetting that a fraction may be added to itself (such as 1/2 + 1/2 = 1), or listing a sum twice (8 arises as 4 + 4 and as 3/2 + 13/2).
• Missing the denominator-$4$ pair $\frac14 + \frac{11}{4} = 3$ (though $3$ also arises as $\frac32 + \frac32$), or not reducing $\frac{3}{12}$ and $\frac{6}{9}$ before grouping.