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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?

abiessu
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justin takro
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  • 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 $00, 88$. – abiessu Aug 27 '15 at 17:17
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    Are leading zeros allowed ? – Peter Aug 27 '15 at 17:18
  • @Peter Leading zeros means A number that starts with 0? – justin takro Aug 27 '15 at 17:19
  • Yes, this is what I mean. – Peter Aug 27 '15 at 17:20
  • No,How could that be a 5 digit number then? – justin takro Aug 27 '15 at 17:21
  • @abiessu why you have considered the last 2 digits.Last 3 digits are used to check divisibility by 8 for example i was using 024,048,064,208,240,248,264,280 etc that is why i was telling it is too long. – justin takro Aug 27 '15 at 17:54
  • You only need two digits, because all the digits are even. In other words, $8\mid 200$, and therefore adding $200$ to any $2$-digit number does not change whether it is divisible by $8$. – abiessu Aug 27 '15 at 18:05

1 Answers1

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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 $8$.

Peter
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