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A question I encountered recently :

A six digit number divisible by $3$ is to be formed using the digits $0,1,2,3,4$ and $5$ without repetition. How many number of ways can this be done ?

If it asked for numbers divisible by $2$, I know how to proceed -- the last digit could be $0, 2$ or $4$. But I have no idea how to do this problem.

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Hint: A number is divisible by $3$ if the sum of the digits is divisible by $3$.

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  • $\begingroup$ So, I've to choose the numbers only ? $\endgroup$ – H G Sur Nov 26 '15 at 11:50
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    $\begingroup$ yep! All the permutation of the 6 digits give you a number divisible by $6$. But remember $0$ cannot be the first digit. $\endgroup$ – Jack Frost Nov 26 '15 at 11:51
  • $\begingroup$ Thank you! Now I've got it :) $\endgroup$ – H G Sur Nov 26 '15 at 11:54
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What would the answer be?

Is it $6! -5! = 600$?

We should use all the given numbers because it's a 6 digit numbers and given that it must be divisible by 3 we could have a clue if it's solvable or not, which it is.

$1+2+3+4+5+0 = 15$ (Divisibility rule of 3)

There are $6!$ permutations for the $6$ digit places given that no repetitions are allowed. And then since $0$ could not be on the hundred thousands place, you must exclude it from the rest.

The permutations to be excluded can be determined by dividing $6!$ by $6$ or simply $5!$.

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