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2022 AIME I Problems/Problem 6

Revision as of 16:00, 17 February 2022 by Bluesoul (talk | contribs)

solution 1

divide cases into $7\leq a<20; 21\leq a\leq28$. There are three cases that arithmetic sequence forms: $3,12,21,30;4,16,28,40;3,5,7,9$. So the answer is $22+...+10+1+2+...+8-13-3=228$

~bluesoul

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