2. For a positive integer of three digits, we can multiply the three digits with each other. The result of this we call the digit product of that number. So, 123 has a digit product of $1 \times 2 \times 3=6$ and 524 has a digit product of $5 \times 2 \times 4=40$. A number cannot start with the digit 0. Determine the three-digit number that is exactly five times as large as its own digit product.
B2. 175 We assume that the three-digit number consists of the digits $a, b$, and $c$, and is therefore equal to $100 a + 10 b + c$. The digit product of this number is $a \cdot b \cdot c$. Therefore, it must hold that
$100 a + 10 b + c = 5 \cdot a \cdot b \cdot c$. Since the right-hand side of this equation is a multiple of five, it must be that $c=0$ or $c=5$. If $c=0$, then $5 \cdot a \cdot b \cdot c=0$ and that cannot be, because we started with a number not equal to zero. Therefore, $c=5$. This means that $5 \cdot a \cdot b \cdot c$ is a multiple of 25 that ends in a 5, so it ends in 25 or 75, so $b=2$ or $b=7$. Moreover, it follows from the fact that $100 a + 10 b + c = 5 a b c$ ends in a 5 that $5 a b c$ is odd, so $a, b$, and $c$ are all odd, so $b=7$. Substituting the found values for $b$ and $c$ into the first equation, we get $100 a + 75 = 175 a$. The only solution to this equation is $a=1$. The only three-digit number that is exactly five times its digit product is therefore 175.