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Weekly Problem 45 - 2010

Stage: 3 Short Challenge Level: Challenge Level:1

One way to proceed is to regard the pattern as four arms, each two squares wide, with four corner pieces of three squares each. So for the nth pattern, we have $4\times 2\times n + 4\times 3 = 8n+12$. For $n = 10$, we need $ 8 \times 10 +12$ i.e. $92$ squares.

Alternatively, it is possible to see the patter as a complete square with corners and a central square removed. So for the nth pattern, we have a complete $(n+4)(n+4)$ square with the four corners and a central $n\times n$ square removed. Hence the number of squares is $(n+4)^2 - n^2 -4 = 8n +12$.

This problem is taken from the UKMT Mathematical Challenges.

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