How many five digit numbers formed from digits $1,2,3,4,5$ (used exactly once) are divisible by $12$?
My answer is $24$ but I doubt if it's right or not.
Sum of all the digits is $15$, so all the numbers are divisible by $3$. Also there are $24$ numbers divisible by $4$. I have found this by
- Fixing $4$ at units place , so I must place $2$ at tens place and number divisible by $4$ is $3!=6$
- Fixing $2$ at units place, so I have $1,3$ or $5$ at tens place and number divisible by $4$ is $3!×3=18$
Since $12=3×4$ and all numbers are divisible by $3$ so numbers divisible by $12$ is $24$.
Is the reason valid?
Your decomposition of the problem is valid, and only works because those two divisors are co-prime (there is no number bigger than $1$ dividing both divisors). This means that if a number is divisible by $3$ and $4$ it is automatically divisible by $12$, and you can check each condition independently – which you did.