3. Find a positive integer $k$ whose product of digits is equal to $\frac{11 k}{4}-199$.求一個正整數 $k$, 使其數字之積等於 $\frac{11 k}{4}-199$ 。
3. 84
3. We must have $\frac{11 k}{4}-199 \geq 0$ and so $k \geq 73$. On the other hand, the product of digits of $k$ must not exceed $k$ (to see this, suppose $k$ is an $n$-digit number with leftmost digit $b$; then $k \geq b \times 10^{n-1}$ but the product of digits of $k$ is at most $k \times 9^{b-1}$ ), so we have $\frac{11 k}{4}-199 \leq k$ and thus $k \leq 113$. Since $\frac{11 k}{4}-199$, being the product of digits of $k$, must be an integer, $k$ is divisible by 4 . Hence the product of digits of $k$ is also even, i.e. $\frac{11 k}{4}$ is odd. Therefore $k \equiv 4(\bmod 8)$, so the possible values of $k$ include $76,84,92,100$ and 108 . It is easy to check that only 84 works.