For a positive integer $$n$$, let $$P(n)$$ denote the product of the digits of $$n$$ when $$n$$ is written in base 10. For example, $$P(123) = 6$$ and $$P(788) = 448$$. If $$N$$ is the smallest positive integer such that $$P(N) > 1000$$, and $$N$$ is written as $$100x + y$$ where $$x, y$$ are integers with $$0 \le x, y < 100$$, then $$x+y$$ equals
Solution
A three digit number has digit product at most $$9^3 = 729$$, so $$N$$ has four digits. A leading digit 1 forces the product to be at most 729 again, so the leading digit is 2 and the remaining three digits must have product greater than 500, the smallest such ending being 789 with $$2 \times 7 \times 8 \times 9 = 1008$$. Hence $$N = 2789$$, so $$x = 27$$, $$y = 89$$ and $$x + y = 116$$.
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