## (Math Quiz 10 – CT) From the digits 0, 1, 2, 3, 4, 5, how many different 3-digit natural numbers can be formed that are divisible by 9?

• Question:

From the digits 0, 1, 2, 3, 4, 5, how many different 3-digit natural numbers can be formed that are divisible by 9?

Reference explanation:

Call the number to look up as $$\overline {abc}$$ with $$a,b,c \in \left\{ {0;1;2;3;4;5} \right\}$$.

Since $$\overline {abc} \vdots 9$$ the sum of the digits $$a + b + c \vdots 9$$ is inferred.

Then $$a,b,c \in \left\{ {\left( {0;4;5} \right),\left( {2;3;4} \right),\left( {1; 3;5} \right)} \right\}$$.

TH1. With $$a,b,c \in \{ 0;4;5\}$$ it follows that there are 2.2 = 4 numbers that satisfy the requirements.

TH2. With $$a,b,c \in \left\{ {2;3;4} \right\}$$ there are 3! = 6 numbers that satisfy the requirement.

TH3. With $$a,b,c \in \left\{ {1;3;5} \right\}$$ there are 3! = 6 numbers that satisfy the requirement.

So it is possible to generate 16 natural numbers satisfying the problem.