Difference between revisions of "2021 Fall AMC 10B Problems/Problem 7"
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12 \qquad\textbf{(E)}\ 13</math> | 12 \qquad\textbf{(E)}\ 13</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Listing out all special fractions, we get: | Listing out all special fractions, we get: | ||
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~KingRavi | ~KingRavi | ||
| + | |||
| + | == Solution 2 == | ||
| + | All special fractions are: <math>\frac{1}{14}</math>, <math>\frac{2}{13}</math>, <math>\frac{3}{12}</math>, <math>\frac{4}{11}</math>, <math>\frac{5}{10}</math>, <math>\frac{6}{9}</math>, <math>\frac{7}{8}</math>, <math>\frac{8}{7}</math>, <math>\frac{9}{6}</math>, <math>\frac{10}{5}</math>, <math>\frac{11}{4}</math>, <math>\frac{12}{3}</math>, <math>\frac{13}{2}</math>, <math>\frac{14}{1}</math>. | ||
| + | |||
| + | Hence, the following numbers are integers: <math>\frac{3}{12} + \frac{11}{4}</math>, <math>\frac{5}{10} + \frac{5}{10}</math>, <math>\frac{5}{10} + \frac{9}{6}</math>, <math>\frac{5}{10} + \frac{13}{2}</math>, <math>\frac{9}{6} + \frac{9}{6}</math>, <math>\frac{9}{6} + \frac{13}{2}</math>, <math>\frac{10}{5} + \frac{10}{5}</math>, <math>\frac{10}{5} + \frac{12}{3}</math>, <math>\frac{10}{5} + \frac{14}{1}</math>, <math>\frac{12}{3} + \frac{12}{3}</math>, <math>\frac{12}{3} + \frac{14}{1}</math>, <math>\frac{13}{2} + \frac{13}{2}</math>, <math>\frac{14}{1} + \frac{14}{1}</math>. | ||
| + | |||
| + | This leads to the following distinct integers: 3, 1, 2, 7, 8, 4, 6, 16, 18, 13, 28. | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\textbf{(C) }11}</math>. | ||
| + | |||
| + | ~Steven Chen (www.professorchenedu.com) | ||
| + | |||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/p9_RH4s-kBA?t=810 | https://youtu.be/p9_RH4s-kBA?t=810 | ||
Revision as of 22:20, 25 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 1
Listing out all special fractions, we get:
{
}
Simplifying and grouping based on their denominators gives
{
}, {
}, {
}
Note all other special fractions have denominators that no other special fraction has, and therefore cannot be added with another special fraction to produce an integer. Furthermore, integers can only be produced by adding in these groupings because the denominators are equal so they could simplify to a denominator of 1 after being added. By simple inspection, the integers that can be expressed as the sum of two special fractions are:
{28, 18, 16, 8, 6, 4, 1, 2, 7, 3, 13}
And there are 11 of them, or 
~KingRavi
Solution 2
All special fractions are: 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
.
Hence, the following numbers are integers: 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
.
This leads to the following distinct integers: 3, 1, 2, 7, 8, 4, 6, 16, 18, 13, 28.
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=810
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. 
