All Red and Blue
Define the set consists of a consecutive integers from 1 to .
Let and be two subsets of such that
- ,
- ,
- If and are distinct elements in , then the value of is not an element is , and
- If and are distinct elements in , then the value of is not an element is .
Find the largest possible value of .
The answer is 8.
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Great Question!
We start by assorting the first three numbers 1 , 2 , 3 in Sets S 1 , S 2 . We see that there are three ways to do it:
\[\begin{array} {c|c} S_{1} &1 \\ \hline S_{2} & 2 & 3 \end{array}
\quad \quad \quad
\begin{array} {c|c} S_{1} &2 \\ \hline S_{2} & 1 & 3 \end{array}
\quad \quad \quad
\begin{array} {c|c} S_{1} &3 \\ \hline S_{2} & 1 & 2 \end{array}\]
Arrangement 1:
The general strategy to follow is to rule out the sum of elements and place it in the other Set. To be specific, in S 2 , the number 5 cannot be placed. Following this, we make the arrangements. At one particular instance, we will be able to observe that a number cannot be placed in any of the set and that will be a dead end.
\[\begin{array} {c|c} S_{1} &1 \\ \hline S_{2} & 2 & 3 \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} &1 &\blue{5} \\ \hline S_{2} & \blue{2} & \blue{3} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} &\blue{1} &\blue{5} \\ \hline S_{2} & 2 & 3 &\blue{6} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} &\red{1} &\red{5} & \cancel{\red{4}} \\ \hline S_{2} & \red{2} & 3 &\red{6} & \cancel{\red{4}} \end{array} \]
Greatest number that we could place was 3
Arrangement 2:
We follow the same thing done in Arrangement 1.
\[\begin{array} {c|c} S_{1} &2 \\ \hline S_{2} & 1 & 3 \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} &2 & \blue{4} \\ \hline S_{2} & \blue{1} & \blue{3} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} & \blue{2} & \blue{4} \\ \hline S_{2} & 1 & 3& \blue{6} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} & 2 & 4 & \blue{5} \\ \hline S_{2} & \blue{1} & 3& \blue{6} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} & \red{2} & 4 & \red{5} & \cancel{\red{7}} \\ \hline S_{2} & \red{1} & 3& \red{6} & \cancel{\red{7}} \end{array}
\]
Greatest number that we could place was 6
Arrangement 3:
In arrangement 3, 1 + 2 = 3 has already been placed. So we see two subcases for placing the number 4
\[\begin{array} {c|c} S_{1} &{3} & {4} \\ \hline S_{2} & {1} & {2} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} & \blue{3} & \blue{4} \\ \hline S_{2} & {1} & {2} & \blue{7} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} & {3} & {4}& \blue{6} \\ \hline S_{2} & \blue{1} & {2} & \blue{7} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} & {3} & {4}& 6 & \blue{5} \\ \hline S_{2} & {1} & \blue{2} & \blue{7} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} & \red{3} & {4}& 6 & \red{5} & \cancel{\red{8}} \\ \hline S_{2} & \red{1} & {2} & \red{7} & \cancel{\red{8}} \end{array} \]
\[\begin{array} {c|c} S_{1} &{3} \\ \hline S_{2} & {1} & {2} & {4} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} &{3} & \blue{6} \\ \hline S_{2} & {1} & \blue{2} & \blue{4} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} &{3} & 6&\blue{5} \\ \hline S_{2} & \blue{1} & {2} & \blue{4} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} &\blue{3} & 6& \blue{5} & \\ \hline S_{2} & {1} & {2} & {4}& \blue{8} \end{array}
\quad \implies \quad
\begin{array} {c|c} S_{1} & \red{3} & \red{6}& {5} & \cancel{\red{9}} \\ \hline S_{2} & \red{1} & {2} & {4}& \red{8} & \cancel{\red{9}} \end{array}
\]
Greatest number that we could place was 8
Conclusion
The sets are:
\[\begin{align} S_{1} &= \{3, 5, 6, 7\} \\ S_{1} &= \{1, 2, 4, 8\} \end{align}
\quad \text{ OR } \quad
\begin{align} S_{1} &= \{3, 5, 6\} \\ S_{1} &= \{1, 2, 4, 7, 8\} \end{align}\]
⟹ N = 8