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by ahmedittihad » Mon Feb 20, 2017 3:59 pm
Solution to problem 2
We will show that $k=3$ is the greatest we can achieve.
Construction for $k=3$,
We let,
the first three terms of $A_1 = \{7,8,9\}$. The next terms are in the form of $3n+1$ where $n=\{3,4,5....\}$.
the first three terms of $A_2 = \{4,5,6\}$. The next terms are in the form of $3n+2$ where $n=\{3,4,5...\}$.
the first three terms of $A_3 = \{1,2,3\}$. The next terms are in the form of $3n$ where $n=\{4,5,6....\}$.
This construction indeed covers all integers more than $14$.
Now, we show that $k=3$ is the best we can do,
Assume that there is a partition for $k=4$.
We will add elements to $A_1$ to make sums of $15$, $16$, $17$....
To get $15$ in a partition, we need $A_1=\{a_1, 15-a_1\}$.
To get $16$ in a partition, we need \begin{align*}
A_1=&\{a_1, a_1+1, 15-a_1\}\\
\text{ or }&\{a_1, 15-a_1, 16-a_1\}
\end{align*}
To get $17$ in a partition, we need \begin{align*}
A_1=&\{a_1, a_1+1, 15-a_1, 16-a_1\}\\
\text{ or }&\{a_1, a_1+1, 15-a_1, 17-a_1\}\\
\text{ or }&\{a_1, a_1+1, a_1+2, 15-a_1\}\\
\text{ or }&\{a_1, 15-a_1, 16-a_1, 17-a_1\}\\
\text{ or }&\{a_1, a_1+2, 15-a_1, 16-a_1\}
\end{align*}
As there are $4$ partitions, all the numbers from $1$ to $16$ are used to get sum of $15, 16, 17$.
From casework, we get that the only possibility of $$\{A_1, A_2, A_3, A_4\}=\{\{1, 14, 15, 16\}, \{2, 3, 4, 13\}, \{5, 10, 11, 12\}, \{6, 7, 8, 9\}\}$$.
Here, we cannot add any numbers to the partitions to make them have sum $18$. We get a contradiction.
Frankly, my dear, I don't give a damn.