Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"
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<cmath>\begin{array}{|c|c|c|} | <cmath>\begin{array}{|c|c|c|} | ||
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\textbf{a} & \textbf{b} & \textbf{a/b} \\ [0.5ex] | \textbf{a} & \textbf{b} & \textbf{a/b} \\ [0.5ex] | ||
| − | \hline | + | \hline \hline |
& & \\ [-2ex] | & & \\ [-2ex] | ||
\ \ \ \ 1 \ \ \ \ & \ \ \ \ 14 \ \ \ \ & \ \ \ \ 1/14 \ \ \ \ \\ [1ex] | \ \ \ \ 1 \ \ \ \ & \ \ \ \ 14 \ \ \ \ & \ \ \ \ 1/14 \ \ \ \ \\ [1ex] | ||
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| − | & & &\\ [-2ex] | + | \hline |
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\textbf{n1} & \textbf{n2} & \textbf{sum} & \textbf{condition} \\ [0.5ex] | \textbf{n1} & \textbf{n2} & \textbf{sum} & \textbf{condition} \\ [0.5ex] | ||
| − | \hline | + | \hline \hline |
& & &\\ [-2ex] | & & &\\ [-2ex] | ||
\ \ \ \ 2 \ \ \ \ & \ \ \ \ 2 \ \ \ \ & \ \ \ \ 4 \ \ \ \ & \ \\ [1ex] | \ \ \ \ 2 \ \ \ \ & \ \ \ \ 2 \ \ \ \ & \ \ \ \ 4 \ \ \ \ & \ \\ [1ex] | ||
& 4 & 6 & \\ [1ex] | & 4 & 6 & \\ [1ex] | ||
& 14 & 16 & \\ [1ex] | & 14 & 16 & \\ [1ex] | ||
| + | \hline | ||
| + | & & &\\ [-2ex] | ||
4 & 4 & 8 & \\ [1ex] | 4 & 4 & 8 & \\ [1ex] | ||
& 14 & 18 & \\ [1ex] | & 14 & 18 & \\ [1ex] | ||
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14 & 14 & 28 & \\ [1ex] | 14 & 14 & 28 & \\ [1ex] | ||
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1/2 & 1/2 & 1 & \\ [1ex] | 1/2 & 1/2 & 1 & \\ [1ex] | ||
& 3/2 & 2 & \\ [1ex] | & 3/2 & 2 & \\ [1ex] | ||
& 13/2 & 7 & \\ [1ex] | & 13/2 & 7 & \\ [1ex] | ||
| + | \hline | ||
| + | & & &\\ [-2ex] | ||
3/2 & 3/2 & 3 & \\ [1ex] | 3/2 & 3/2 & 3 & \\ [1ex] | ||
& 13/2 & 8 & \ \ Rep.\ \ \\ [1ex] | & 13/2 & 8 & \ \ Rep.\ \ \\ [1ex] | ||
| + | \hline | ||
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13/2 & 13/2 & 13 & \\ [1ex] | 13/2 & 13/2 & 13 & \\ [1ex] | ||
| + | \hline | ||
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1/4 & 11/4 & 3 & Rep. \\ [1ex] | 1/4 & 11/4 & 3 & Rep. \\ [1ex] | ||
| + | \hline | ||
\end{array}</cmath> | \end{array}</cmath> | ||
Revision as of 09:00, 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)
Consider all the cases where 
, and construct the following table:
![\[\begin{array}{|c|c|c|} \hline & & \\ [-2ex] \textbf{a} & \textbf{b} & \textbf{a/b} \\ [0.5ex] \hline \hline & & \\ [-2ex] \ \ \ \ 1 \ \ \ \ & \ \ \ \ 14 \ \ \ \ & \ \ \ \ 1/14 \ \ \ \ \\ [1ex] 2 & 13 & 2/13 \\ [1ex] 3 & 12 & 1/4 \\ [1ex] 4 & 11 & 4/11 \\ [1ex] 5 & 10 & 1/2 \\ [1ex] 6 & 9 & 2/3 \\ [1ex] 7 & 8 & 7/8 \\ [1ex] 8 & 7 & 8/7 \\ [1ex] 9 & 6 & 3/2 \\ [1ex] 10 & 5 & 2 \\ [1ex] 11 & 4 & 11/4 \\ [1ex] 12 & 3 & 4 \\ [1ex] 13 & 2 & 13/2 \\ [1ex] 14 & 1 & 14 \\ [1ex] \end{array}\]](http://latex.artofproblemsolving.com/6/2/8/628fd7735ef5dcd0f83e569807b048d5cb3814a7.png)
Let 
.
Now, we list all the possible integers obtained from an addition of two values of 
:
![\[\begin{array}{|c|c|c|c|} \hline & & &\\ [-2ex] \textbf{n1} & \textbf{n2} & \textbf{sum} & \textbf{condition} \\ [0.5ex] \hline \hline & & &\\ [-2ex] \ \ \ \ 2 \ \ \ \ & \ \ \ \ 2 \ \ \ \ & \ \ \ \ 4 \ \ \ \ & \ \\ [1ex] & 4 & 6 & \\ [1ex] & 14 & 16 & \\ [1ex] \hline & & &\\ [-2ex] 4 & 4 & 8 & \\ [1ex] & 14 & 18 & \\ [1ex] \hline & & & \\ [-2ex] 14 & 14 & 28 & \\ [1ex] \hline & & &\\ [-2ex] 1/2 & 1/2 & 1 & \\ [1ex] & 3/2 & 2 & \\ [1ex] & 13/2 & 7 & \\ [1ex] \hline & & &\\ [-2ex] 3/2 & 3/2 & 3 & \\ [1ex] & 13/2 & 8 & \ \ Rep.\ \ \\ [1ex] \hline & & &\\ [-2ex] 13/2 & 13/2 & 13 & \\ [1ex] \hline & & &\\ [-2ex] 1/4 & 11/4 & 3 & Rep. \\ [1ex] \hline \end{array}\]](http://latex.artofproblemsolving.com/8/b/d/8bd2807648b99ff981b5f748369274c81c05e0a6.png)
Although 
terms are found in total, 
and 
have appeared twice respectively. Taken repetition into account, we have a total of terms.
~Wilhelm Z
