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2024 SSMO Tiebreaker Round Problems/Problem 1

Problem

Compute the exact value of $2^2+0^2+2^2+4^2+20^2+22^2+24^2+40^2+42^2+202^2.$

Solution

We have \[2^2+0^2+2^2+4^2+20^2+22^2+24^2+40^2+42^2+202^2 =\]\[4+0+4+16+400+484+576+1600+1764+40804 = \boxed{45652}.\]

~SMO_Team

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