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Difference between revisions of "2022 AIME I Problems/Problem 6"

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

Revision as of 17:00, 17 February 2022

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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