# Misc Solution: 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 $$x\in\{166, 199, 818, 919\}$$.

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