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In how many ways can five-digit numbers be formed by using digits $0,2,4,6,8$ such that the numbers are divisible by $8$?

Assume the case in which repetition is not allowed

Our Approach:

Case1: When repetition is not allowed.

We start to make pairs of combinations such that they are divisible by $8$ and left over number can be selected in the following ways also.

Is there any other approach (without making cases as it can be too long) through which I can solve this problem?

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  • $\begingroup$ The approach by cases isn't really that long, for instance the only possible last digits are $0,4,8$ and the only possible last two digits combinations are $08, 24, 40, 48, 64, 80$, of which all have exactly the same possible statement that the remaining digits can be in any combination. With repeated digits, the only additions to the possible-last digits list are $00, 88$. $\endgroup$ – abiessu Aug 27 '15 at 17:17
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    $\begingroup$ Are leading zeros allowed ? $\endgroup$ – Peter Aug 27 '15 at 17:18
  • $\begingroup$ @Peter Leading zeros means A number that starts with 0? $\endgroup$ – justin takro Aug 27 '15 at 17:19
  • $\begingroup$ Yes, this is what I mean. $\endgroup$ – Peter Aug 27 '15 at 17:20
  • $\begingroup$ No,How could that be a 5 digit number then? $\endgroup$ – justin takro Aug 27 '15 at 17:21
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Hint : Since all the digits have to be even, any possible number is divisible by $8$, if and only if the last two digits form a number divisible by $8$.

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  • $\begingroup$ for divisibility by 8 it should be last 3 digit number $\endgroup$ – justin takro Aug 27 '15 at 17:20
  • $\begingroup$ Consider, that $200$ is divisibel by $8$, so here two digits are sufficient. $\endgroup$ – Peter Aug 27 '15 at 17:23
  • $\begingroup$ But Repeatition of digits is not allowed $\endgroup$ – justin takro Aug 27 '15 at 17:42
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    $\begingroup$ I mean, two digits are sufficient to check the divisibility because $200$ is divisible by $8$. For example $20648$ is divisible by $8$ because $48$ is divisible by $8$. $\endgroup$ – Peter Aug 27 '15 at 17:53
  • $\begingroup$ Okay you mean using 0,2,4,6,8 by 2 is sufficient? $\endgroup$ – justin takro Aug 27 '15 at 17:55

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