MathLabs

Problem 5

Is it possible to choose 19831983 distinct positive integers, all less than or equal to 10510^5, no three of which are consecutive terms of an arithmetic progression? Justify your answer.
Step 3 of 4: Bound the largest constructed value
Previous stepNext step
1983=1111011111121983=11110111111_2
Detailed analysis

We have 1983=1111011111121983=11110111111_2, so a1983=111101111113a_{1983}=11110111111_3. Every ana_n for n≤1983n\le1983 has at most 1111 ternary digits, hence an≤111111111113=(311−1)/2<105a_n\le11111111111_3=(3^{11}-1)/2<10^5.