[Math] How many $3$ digit different number that will be divisible by $5$ can be formed from the digit $0,2,3,4,5,6$ lying between $100$ and $1000$.

Question

How many $3$ digit different number that will be divisible by $5$ can be formed from the digit $0,2,3,4,5,6$ lying between $100$ and $1000$.

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My attempt:

Divisible by $5$ is possible only when unit digit will be either $0$ or $5$, means possibility for unit digit is $2$.

And possible number of choices for third place digit is $5$ (except $0$).

And possible number of choices for tenth place digit is $6$ (all).

So, total possible such numbers are $=5\times6\times2=60$ (answer).

Can you explain in formal/alternative way, please?

Answer 1

Here is a formal way (though not much different than yours)...

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Add up the following:

• The amount of numbers of the form $\underbrace{2,3,4,5,6}_{5\text{ options}}|\underbrace{0,2,3,4,5,6}_{6\text{ options}}|\underbrace{0}_{1\text{ option}}$
• The amount of numbers of the form $\underbrace{2,3,4,5,6}_{5\text{ options}}|\underbrace{0,2,3,4,5,6}_{6\text{ options}}|\underbrace{5}_{1\text{ option}}$

The total amount is $5\cdot6\cdot1+5\cdot6\cdot1=60$.