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Find the number of 2-digit positive integers $$n$$ such that $$n = 26 + (a \times b)$$, where $$a$$ and $$b$$ are the two digits of $$n$$.
Correct Answer: 3
With $$a$$ the tens digit and $$b$$ the units digit, the condition $$10a + b = 26 + ab$$ rearranges to $$(a - 1)(10 - b) = 16$$. Since $$1 \le a - 1 \le 8$$ and $$1 \le 10 - b \le 10$$, the only factor pairs are $$(2, 8)$$, $$(4, 4)$$ and $$(8, 2)$$, giving $$(a, b) = (3, 2), (5, 6), (9, 8)$$. The three numbers are 32, 56 and 98, so the count is 3.
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