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How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if each digit is to be used at most once?

The total number of $5$ digit numbers using the digits $0,1,2,3,4,7$ and $8$ is $6\times6\times5\times4\times3=2160.$

Now I found the numbers not divisible by $3$, i.e. even numbers ending in $2,4,8.$

Even numbers from the digits $0,1,2,3,4,7$ and $8$ are $5\times5\times4\times3\times3=930.$

So the numbers divisible by $3$ are $2160-930=1230$ but the answer is $744.$ Where I am wrong?

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    $\begingroup$ $78432,$ $74328$ and $78234$ are divisible by $3.$ $\endgroup$ – mfl Jul 21 '17 at 8:39
  • $\begingroup$ This question is good and should not vote down. $\endgroup$ – ً ً Jul 21 '17 at 8:44
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The numbers which are divisible by $3$ are those where the sum of the digits is a multiple of $3$; it doesn't matter what the last digit is. So which sets of $5$ digits are possible? Once you know all of these you can use your method above to say how many numbers contain each set of $5$.

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  • $\begingroup$ I got it,the possible numbers are 21084 and its 96 arrangements,21078 and its 96 arrangements,31047 and its 96 arrangements,24078 and its 96 arrangements,12387 and its 120 arrangements,12348 and its 120 arrangements,23478 and its 120 arrangements=744 possibilities $\endgroup$ – learner_avid Jul 21 '17 at 9:05
  • $\begingroup$ Yes, that looks good! $\endgroup$ – Especially Lime Jul 21 '17 at 9:08
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You are wrong in assuming that even numbers can't be divisible by 3. As a counter example, 6 is divisible by 3.

The divisibility rule for 3 states that the if the sum of the digits of the number is divisible by 3, the number itself is divisible 3.

Out of the seven digits given, you need to find groups of 5 for which the sum is divisible by 3. For example, $(1, 2, 3, 4, 8)$ is one such group. For every such group you need to find the number of 5 digit numbers that can be formed and get the total.

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  • $\begingroup$ Which is $5!$ for all tuples not containing zero. $\endgroup$ – Michael Hoppe Jul 21 '17 at 9:09
  • $\begingroup$ @MichaelHoppe I have a feeling that was left as an exercise for the reader. $\endgroup$ – SwiftsNamesake Jul 21 '17 at 9:10
  • $\begingroup$ @SwiftsNamesake Indeed $\endgroup$ – iamwhoiam Jul 21 '17 at 9:51

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