Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"
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<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 08:51, 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