Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"
Lopkiloinm (talk | contribs) (→Solution) |
(→Solution) |
||
| Line 4: | Line 4: | ||
<math>\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13</math> | <math>\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13</math> | ||
| − | ==Solution== | + | ==Solution 1== |
<math>a=15-b</math> so the fraction is <math>\frac{15-b}{b}</math> which is <math>\frac{15}{b}-1</math>. We can just ignore the <math>-1</math> part and only care about <math>\frac{15}{b}</math>. Now we just group <math>\frac{15}{1}, \frac{15}{3}, \frac{15}{5}</math> as the integers and <math>\frac{15}{2}, \frac{15}{6}, \frac{15}{10}</math> as the halves. We get <math>30, 20, 18, 10, 8, 6</math> from the integers group and <math>15, 10, 9, 5, 4, 3</math> from the halves group. These are both <math>6</math> integers and we see that <math>10</math> overlaps, so the answer is <math>\boxed{\textbf{(C)}\ 11}</math>. | <math>a=15-b</math> so the fraction is <math>\frac{15-b}{b}</math> which is <math>\frac{15}{b}-1</math>. We can just ignore the <math>-1</math> part and only care about <math>\frac{15}{b}</math>. Now we just group <math>\frac{15}{1}, \frac{15}{3}, \frac{15}{5}</math> as the integers and <math>\frac{15}{2}, \frac{15}{6}, \frac{15}{10}</math> as the halves. We get <math>30, 20, 18, 10, 8, 6</math> from the integers group and <math>15, 10, 9, 5, 4, 3</math> from the halves group. These are both <math>6</math> integers and we see that <math>10</math> overlaps, so the answer is <math>\boxed{\textbf{(C)}\ 11}</math>. | ||
~lopkiloinm | ~lopkiloinm | ||
| + | |||
| + | ==Solution 2 (Bash)== | ||
| + | |||
| + | WORKING IN PROGRESS | ||
| + | |||
| + | ~Wilhelm Z | ||
Revision as of 07:51, 24 November 2021
Problem 5
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
so the fraction is 
which is 
. We can just ignore the 
part and only care about 
. Now we just group 
as the integers and 
as the halves. We get 
from the integers group and 
from the halves group. These are both 
integers and we see that 
overlaps, so the answer is 
.
~lopkiloinm
Solution 2 (Bash)
WORKING IN PROGRESS
~Wilhelm Z
