Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"
MRENTHUSIASM (talk | contribs) (About to merge the solutions in the two pages while keeping the page clean. I will credit all contributors for the other solution. Also, reformatted Sol 2.) |
MRENTHUSIASM (talk | contribs) (→Solution 2) |
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The halves are <math>\frac{15}{2},\frac{15}{6},\frac{15}{10},</math> which are <math>7\frac12,2\frac12,1\frac12,</math> respectively. We get <math>15,10,9,5,4,3</math> from this group of numbers. | The halves are <math>\frac{15}{2},\frac{15}{6},\frac{15}{10},</math> which are <math>7\frac12,2\frac12,1\frac12,</math> respectively. We get <math>15,10,9,5,4,3</math> from this group of numbers. | ||
− | Note that <math>10</math> | + | The quarters are <math>\frac{15}{4},\frac{15}{12},</math> which are <math>3\frac34,1\frac14,</math> respectively. We get <math>5</math> from this group of numbers. |
+ | |||
+ | Note that <math>10</math> and <math>5</math> each appear twice. Therefore, the answer is <math>6+6-1=\boxed{\textbf{(C)}\ 11}.</math> | ||
~lopkiloinm | ~lopkiloinm |
Revision as of 06:46, 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
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
~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.