Difference between revisions of "2021 Fall AMC 10B Problems/Problem 7"

(Solution)
(Solution)
Line 13: Line 13:
 
Simplifying and grouping based on their denominators gives
 
Simplifying and grouping based on their denominators gives
  
<math>\{14, 4, 2}, \{\frac{1}{2}, \frac{3}{2}, \frac{13}{2}</math>}<math>, \{\frac{1}{4}, \frac{11}{4}</math>}
+
<math>\{14, 4, 2</math>}<math>, \{\frac{1}{2}, \frac{3}{2}, \frac{13}{2}</math>}<math>, \{\frac{1}{4}, \frac{11}{4}</math>}
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=8|num-b=6}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=8|num-b=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:43, 23 November 2021

Problem

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

Listing out all special fractions, we get: $\{\frac{1}{14}, \frac{2}{13}, \frac{3}{12}, \frac{4}{11}, \frac{5}{10}, \frac{6}{9}, \frac{7}{8}, \frac{8}{7}, \frac{9}{6}, \frac{10}{5}, \frac{11}{4}, \frac{12}{3}, \frac{13}{2}, \frac{14}{1}$}

Simplifying and grouping based on their denominators gives

$\{14, 4, 2$}$, \{\frac{1}{2}, \frac{3}{2}, \frac{13}{2}$}$, \{\frac{1}{4}, \frac{11}{4}$}

See Also

2021 Fall AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png