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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?

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    $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ Commented Mar 13, 2019 at 13:03
  • $\begingroup$ All the numbers formed from the five digits $1, 2, 3, 4, 5$ are divisible by $3$ since their digit sum is $15$, so you just have to check if the numbers are also divisible by $4$. $\endgroup$ Commented Mar 13, 2019 at 13:06
  • $\begingroup$ @BJKShah If a number is divisible by $3$ and $4$ , and you divide it by $3$, it's still divisible by $4$ ... $\endgroup$
    – Matti P.
    Commented Mar 13, 2019 at 13:06
  • $\begingroup$ I understand it now cause the numbers will have both the factors 3 and 4 and so 12 would also divide it. $\endgroup$
    – Bijay
    Commented Mar 13, 2019 at 13:07

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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.

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  • $\begingroup$ Like if the factors were 6 and 3 then it is not necessary that 18 would divide the numbers, but 12 would? $\endgroup$
    – Bijay
    Commented Mar 13, 2019 at 13:10
  • $\begingroup$ @BJKShah Indeed. $\endgroup$ Commented Mar 13, 2019 at 13:11

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