Three Digits Number

We are looking for a three-digit number that meets the following conditions:

• The third digit is the product of the first two digits
• None of the three digits is a prime
• The number sought is not divisible by any of the numbers 3 to 9

Solution

We are looking for a three-digit number $x$, that consists of the digits $a, b, c$. It follows that:

$$100\leq 100a+10b+c<1000$$

Since the third digit is the product of the first two digits, it follows:

$$c = ab < 10$$

Also, let none of the three digits be a prime, so:

$$a,b,c\notin \{2,3,5,7\}\Leftrightarrow a,b,c\in \{0,1,4,6,8,9\}$$

If the number $x$ shall not be divisible by 5 it follows the last digit $c$ is not zero or five. Since five is already excluded by the prime rule, we have $c\neq 0$. Because $100\leq x<1000$ follows that $a$ is not zero and from $c=ab$ follows that $b\neq 0$. This means $x$ does not have any zero digits.

The product of the first two digits $a$ and $b$ combined with all remaining digits $\{1,4,6,8,9\}$ shows that $c<10$ is only possible if $a=1$ or $b=1$. The final number $x=100a + 10b + ab$ can therefore only have the following format: $100 + 11b$ or $101a+10$.

Possible solutions are therefore $x\in\{111, 144, 166, 188, 199, 414, 616, 818, 919\}$.

If we test this set of possible solutions against the given divisibility restrictions, we find the solution to be $x\in\{166, 199, 818, 919\}$.