2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 7 Easy

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?(2021 AMC Fall 10B, Question #7)

  • A.

    9

  • B.

    10

  • C.

    11

  • D.

    12

  • E.

    13

Answer:C

Solution 1:

The 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} We rewrite them in the simplest form: \frac{1}{14}, \frac{2}{13}, \frac{1}{4}, \frac{4}{11}, \frac{1}{2}, \frac{2}{3}, \frac{7}{8}, 1 \frac{1}{7}, 1 \frac{1}{2}, 2,2 \frac{3}{4}, 4,6 \frac{1}{2}, 14 . Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once: \frac{1}{4}, \frac{1}{2}, 1 \frac{1}{2}, 2,2 \frac{3}{4}, 4,6 \frac{1}{2}, 14 For the set \{2,4,14\}, two elements (not necessarily different) can sum to 4,6,8,16,18,28. For the set \left\{\frac{1}{2}, 1 \frac{1}{2}, 6 \frac{1}{2}\right\}, two elements (not necessarily different) can sum to 1,2,3,7,8,13.

For the set \left\{\frac{1}{4}, 2 \frac{3}{4}\right\}, two elements (not necessarily different) can sum to 3 . Together, there are (C) 11 distinct integers that can be written as the sum of two, not necessarily different, special fractions: 1,2,3,4,6,7,8,13,16,18,28 \text {. }

Solution 2:

Let a=15-b, so the special fraction is \frac{a}{b}=\frac{15-b}{b}=\frac{15}{b}-1 . We can ignore the -1 part and only focus on \frac{15}{b}. The integers are \frac{15}{1}, \frac{15}{3}, \frac{15}{5}, which are 15,5,3, respectively. We get 30,20,18,10,8,6 from this group of numbers.

The halves are \frac{15}{2}, \frac{15}{6}, \frac{15}{10}, which are 7 \frac{1}{2}, 2 \frac{1}{2}, 1 \frac{1}{2}, respectively. We get 15,10,9,5,4,3 from this group of numbers. The quarters are \frac{15}{4}, \frac{15}{12}, which are 3 \frac{3}{4}, 1 \frac{1}{4}, respectively. We get 5 from this group of numbers. Note that 10 and 5 each appear twice. Therefore, the answer is (C) 11 .