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The smallest positive integer $$n$$ such that 2013 is a term in the finite sequence of four numbers $$4n$$, $$4n + 7$$, $$4n + 14$$, $$4n + 21$$ is ________
Correct Answer: 498
$$2013 \equiv 1 \pmod 4$$, while the four terms leave remainders 0, 3, 2 and 1 on division by 4. So 2013 can only be $$4n + 21$$, which gives $$4n = 1992$$ and $$n = 498$$.
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