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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, 2017 at 8:39
  • $\begingroup$ This question is good and should not vote down. $\endgroup$
    – ً ً
    Jul 21, 2017 at 8:44

3 Answers 3

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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, 2017 at 9:09
  • $\begingroup$ @MichaelHoppe I have a feeling that was left as an exercise for the reader. $\endgroup$
    – SwiftsNamesake
    Jul 21, 2017 at 9:10
  • $\begingroup$ @SwiftsNamesake Indeed $\endgroup$
    – iamwhoiam
    Jul 21, 2017 at 9:51
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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, 2017 at 9:05
  • $\begingroup$ Yes, that looks good! $\endgroup$
    – Especially Lime
    Jul 21, 2017 at 9:08
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Solution

We've to make $5$ digit numbers using given $7$ digits excluding $2$ digits every time. Now,$0+1+2+3+4+7+8=25$. As $25 \pmod 3 \equiv 1$, duplets to be excluded should have their sum $\mod 3=1$. It's not much difficult to find such duplets. Without much trouble, we get :$$(0,1),(0,4),(0,7),(1,3),(2,8),(3,4),(3,7).$$ Thus, there are such $7$ duplets & hence $7$ such groups of $5$ digits.

Now, we find that out of $7$ groups, $4$ groups have $0$ in them which isn't allowed to take 5th place ( otherwise a $4$-digit number will be formed.). Total numbers formed by these $4$ groups $=4(5!−4!)=384$.

Other $3$ groups have no zero so total numbers formed by them$=3(5!)=360$

Total possibel numbers: $384+360=744$

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