New answers tagged arithmetic-progression
The problem is a little harder than it seems at first glance. Pick m large, say m > 8. There are 3k=3^m numbers with first ternary digit 1 and m other ternary digits. Suppose 0<= a < 3^m is smallest such that 3k +a= 3*2^l. Then 2^(l+2)=4k +4a/3, so there will be either at least k/2 numbers before 3k+a or after 4k+ 4a/3 which are part of the 3k ...
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