2007 AMC 10B Problem 24

Since $n$ is divisible by $9,$ its digit sum is a multiple of $9.$ With $k$ fours and $m$ nines, the digit sum is $4k+9m,$ so $4k$ is divisible by $9,$ forcing $k$ to be divisible by $9.$ Thus $k\ge 9,$ and with at least one $9,$ the number has at least ten digits. For divisibility by $4,$ the last two digits must form a multiple of $4,$ and among $44,$ $49,$ $94,$ and $99$ only $44$ works, so $n$ ends in $44.$ The smallest such ten-digit number places the single $9$ in the lowest available position, giving $4{,}444{,}444{,}944.$ Its last four digits are $4944.$ Thus, the correct answer is C.