2. Let $n$ be a positive integer. If $n$ is divisible by 2010 and exactly one of the digits of $n$ is even, find the smallest possible value of $n$.
(1 mark)
Let $n$ be a positive integer. If $n$ is divisible by 2010 and exactly one of the digits of $n$ is even, find the smallest possible value of $n$.
2. 311550
2. The digit which is even must be the unit digit 0. Hence it remains to find the smallest positive integer \( k \) for which all digits of \( 201k \) are odd.
Clearly we must have \( k \geq 150 \), for if \( k < 100 \) then the hundreds digit of \( 201k \) is even (same as the unit digit of \( 2k \)); if \( 100 \leq k < 150 \) the ten thousands digit of \( 201k \) is 2 which is even. Furthermore \( k \) must be odd as the unit digit of \( 201k \) has to be odd. Hence we try \( 201 \times 151 = 30351 \), \( 201 \times 153 = 30753 \) and \( 201 \times 155 = 31155 \) to get the answer is 311550.