2021 AMC Fall 10B Problem 7

P1-10~5 minPrint

Fractions & Decimals Basic Counting Organized listing / direct enumeration Split into exhaustive cases and handle each

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?

Show solution without answering

Key insight

Reduce all 14 special fractions. Two of them can only add to a whole number if both are whole numbers, or both have the same bottom (2 or 4).

Solution

List the special fractions. The top and bottom add to $15$ , so they are $\frac{1}{14}, \frac{2}{13}, \ldots, \frac{14}{1}$ . Reduce each one:

$a$	1	2	3	4	5	6	7	8	9	10	11	12	13	14
$\frac{a}{15-a}$	$\frac{1}{14}$	$\frac{2}{13}$	$\frac{1}{4}$	$\frac{4}{11}$	$\frac{1}{2}$	$\frac{2}{3}$	$\frac{7}{8}$	$\frac{8}{7}$	$\frac{3}{2}$	$2$	$\frac{11}{4}$	$4$	$\frac{13}{2}$	$14$

2. When can two of them add to a whole number? For example, $\frac23 + \frac78 = \frac{37}{24}$ , which is not whole. In lowest terms, a fraction with bottom $4$ can only be "completed" to a whole number by another fraction with bottom $4$ . So group the fractions by their bottom number.
3. Whole numbers $2, 4, 14$ . Add any two, repeats allowed:
$2+2 = 4$ , $2+4 = 6$ , $2+14 = 16$ , $4+4 = 8$ , $4+14 = 18$ , $14+14 = 28$ .
4. Halves $\frac12, \frac32, \frac{13}{2}$ . Two halves always make a whole number:
$\frac12+\frac12 = 1$ , $\frac12+\frac32 = 2$ , $\frac12+\frac{13}{2} = 7$ , $\frac32+\frac32 = 3$ , $\frac32+\frac{13}{2} = 8$ , $\frac{13}{2}+\frac{13}{2} = 13$ .
5. Quarters $\frac14, \frac{11}{4}$ . Only $\frac14 + \frac{11}{4} = 3$ works. ( $\frac14+\frac14$ and $\frac{11}{4}+\frac{11}{4}$ give halves.)
6. The rest (bottoms $3, 7, 8, 11, 13, 14$ ) each appear once. Doubling one of them, like $\frac23+\frac23 = \frac43$ , is not a whole number.
7. Collect and remove repeats:
$1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28.$
( $2$ , $3$ , $4$ and $8$ each came up twice.) That is $11$ different whole numbers.

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

Why this works

Sorting the fractions by their bottom number turns a search through all $14 \times 14$ pairs into a few short lists, because the fractional parts can only cancel between fractions with the same bottom. After that, the only danger is careless listing, so write every sum down and cross off repeats at the end.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(C)}\ 11}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 11: Decimals and Fractions
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)