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In the sequence $$1,4,8,10,16,21,25,30,43$$, the number of blocks of consecutive terms whose sums are divisible by $$11$$ is
The consecutive blocks whose sums are divisible by $$11$$ are $$4+8+10=22$$, $$8+10+16+21=55$$, $$8+10+16+21+25+30=110$$, and $$25+30=55$$. There are no other such consecutive blocks. Hence the number of blocks is $$4$$.
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