2021 AMC 12B Fall Problem 5

Attempt Problem 5 of the 2021 AMC 12B Fall below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Fall solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

Call a fraction ab,\dfrac{a}{b}, not necessarily in the simplest form, special if aa and bb are positive integers whose sum is 15.15. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
99
1010
1111
1212
1313
Answer: C
Concepts:fractioncasework
Small hint:
List the special fractions and note which are integers, half-integers, or quarter-integers
Big hint:
Two specials sum to an integer only when their fractional parts cancel
Solution:
Listing a15−a\frac{a}{15-a} for 1≤a≤14,1\le a\le14, the integers are 2,4,14;2,4,14; the fractions with fractional part 12\frac{1}{2} are 12,32,132;\tfrac12,\tfrac32,\tfrac{13}{2}; and 14,114\tfrac14,\tfrac{11}{4} have complementary fractional parts 14,34.\frac{1}{4},\frac{3}{4}.
The remaining fractional parts are 114,213,411,23,78,17,\frac{1}{14},\frac{2}{13},\frac{4}{11},\frac{2}{3},\frac{7}{8},\frac{1}{7}, and none of their complements occurs. Thus only the integer, half-integer, and quarter cases can give integer sums.
Integer pairs give 4,6,8,16,18,28.4, 6, 8, 16, 18, 28. Half-integer pairs (12,32,132)\left(\tfrac12, \tfrac32, \tfrac{13}{2}\right) give 1,2,3,7,8,13.1, 2, 3, 7, 8, 13. The quarter pair 14+114\tfrac14 + \tfrac{11}{4} gives 3.3.
The distinct integers are 1,2,3,4,6,7,8,13,16,18,28,1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28, a total of 11.11.
Thus, the correct answer is C.
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