What is the minimum value of $n$ , such that we are able to place one each of $1\times 1, 1\times 2, 1\times 3, \dots, 1\times n$ tiles inside a $10\times 10$ board in such a manner that we do not have sufficient space to place an $1\times (n+1)$ tile inside the same board?
Details and Assumptions:
The tiles are not allowed to stick out of the board or overlap.
We get to choose the placement of the tiles.
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The figure above shows that $n\leq 6$ .
Now I will prove that 6 satisfies the conditions.
In the figure above 16 $1\times 6$ rectangles are highlighted. So we need to cover at least one square from each $1\times 6$ rectangle. These together only contain $1+2+3+4+5<16$ squares, so at least one $1\times 6$ rectangle will be free.
Therefore the answer is $\boxed{6}$ .