Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
+ | The special fractions are <cmath>\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}.</cmath> | ||
+ | We rewrite them in the simplest form: <cmath>\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.</cmath> | ||
+ | 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: <cmath>\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.</cmath> | ||
~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com) | ~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com) |
Revision as of 06:27, 5 January 2022
- The following problem is from both the 2021 Fall AMC 10B #7 and 2021 Fall AMC 12B #5, so both problems redirect to this page.
Contents
[hide]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 1
The special fractions are We rewrite them in the simplest form: 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:
~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
Solution 2
Let so the special fraction is We can ignore the part and only focus on
The integers are which are respectively. We get from this group of numbers.
The halves are which are respectively. We get from this group of numbers.
The quarters are which are respectively. We get from this group of numbers.
Note that and each appear twice. Therefore, the answer is
~lopkiloinm
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 |
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.