Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"
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~lopkiloinm | ~lopkiloinm | ||
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/p9_RH4s-kBA?t=810 | ||
| + | |||
| + | ~Interstigation | ||
==See Also== | ==See Also== | ||
Revision as of 01:21, 9 June 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.
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 ![\[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\]](http://latex.artofproblemsolving.com/e/c/d/ecde03213fbc8d54fa2100cbeb7d1fb0b247568b.png)
We rewrite them in the simplest form: ![\[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\]](http://latex.artofproblemsolving.com/c/0/a/c0a0344dcd8b190127f1714b22d3a570e06b5498.png)
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: ![\[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\]](http://latex.artofproblemsolving.com/d/a/2/da2a5842412da3a6f3cd4cd4425f0fb73bdbc957.png)
For the set 
two elements (not necessarily different) can sum to 
For the set 
two elements (not necessarily different) can sum to 
For the set 
two elements (not necessarily different) can sum to 
Together, there are 
distinct integers that can be written as the sum of two, not necessarily different, special fractions: ![\[1,2,3,4,6,7,8,13,16,18,28.\]](http://latex.artofproblemsolving.com/b/d/7/bd7a67e7e903be4e6dd126b8fd6d2a05e440df7b.png)
~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
Solution 2
Let 
so the special fraction is ![\[\frac ab = \frac{15-b}{b} = \frac{15}{b}-1.\]](http://latex.artofproblemsolving.com/a/b/1/ab175ffc6abe6a6dd7d9eedd4c2818449d8b3300.png)
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
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=810
~Interstigation
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. 
