**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:**

**Correct answer: A**

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.

ADSENSE

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