Difference between revisions of "2021 Fall AMC 12B Problems/Problem 5"
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==Solution 2 (Bash)== | ==Solution 2 (Bash)== | ||
− | + | Consider all the cases where <math>a+b=15</math>, and construct the following table: | |
+ | |||
+ | <cmath>\begin{array}{|c|c|c|} | ||
+ | & & \ [-2ex] | ||
+ | \textbf{a} & \textbf{b} & \textbf{a/b} \ [0.5ex] | ||
+ | \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}</cmath> | ||
+ | |||
+ | Let <math>\frac{a}{b}=n</math>. | ||
+ | Now, we list all the possible integers obtained from an addition of two values of <math>n</math>: | ||
+ | |||
+ | <cmath>\begin{array}{|c|c|c|c|} | ||
+ | & & &\ [-2ex] | ||
+ | \textbf{n1} & \textbf{n2} & \textbf{sum} & \textbf{condition} \ [0.5ex] | ||
+ | \hline | ||
+ | & & &\ [-2ex] | ||
+ | \ \ \ \ 2 \ \ \ \ & \ \ \ \ 2 \ \ \ \ & \ \ \ \ 4 \ \ \ \ & \ \ [1ex] | ||
+ | & 4 & 6 & \ [1ex] | ||
+ | & 14 & 16 & \ [1ex] | ||
+ | 4 & 4 & 8 & \ [1ex] | ||
+ | & 14 & 18 & \ [1ex] | ||
+ | 14 & 14 & 28 & \ [1ex] | ||
+ | 1/2 & 1/2 & 1 & \ [1ex] | ||
+ | & 3/2 & 2 & \ [1ex] | ||
+ | & 13/2 & 7 & \ [1ex] | ||
+ | 3/2 & 3/2 & 3 & \ [1ex] | ||
+ | & 13/2 & 8 & \ \ Rep.\ \ \ [1ex] | ||
+ | 13/2 & 13/2 & 13 & \ [1ex] | ||
+ | 1/4 & 11/4 & 3 & Rep. \ [1ex] | ||
+ | \end{array}</cmath> | ||
+ | |||
+ | Although <math>13</math> terms are found in total, <math>8</math> and <math>3</math> have appeared twice respectively. Taken repetition into account, we have a total of <math>\boxed{\textbf{(C)}\ 11}</math> terms. | ||
~Wilhelm Z | ~Wilhelm Z |
Revision as of 08:55, 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:
Let . Now, we list all the possible integers obtained from an addition of two values of :
Although terms are found in total, and have appeared twice respectively. Taken repetition into account, we have a total of terms.
~Wilhelm Z