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 01:43, 23 November 2021
Problem
Call a fraction 
, not necessarily in the simplest form special if 
and 
are positive integers whose sum is 
. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
Solution
Listing out all special fractions, we get:
}
Simplifying and grouping based on their denominators gives
}
}
}
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
