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How to find number of 4 digit numbers divisible by 4 that can be obtained using 1,2,3,4,5,6 with no numbers repeating?

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2 Answers 2

up vote 7 down vote accepted

For non-repeating: number is divisible by 4 iff its last two digits are divisible by 4. IN your case, it will be 12 16 24 32 36 52 56 64

So you have 8 possible endings.

First two digits can be chosen in 4 * 3 ways, so overall answer will be 12 * 8 = 96

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The first two digits can only be chosen in $4\cdot3=12$ ways, so the overall answer is $12\cdot8=96$. –  joriki Aug 18 '11 at 17:28
    
Thanks, edited my post. –  Oleksandr Kuvshynov Aug 18 '11 at 17:30

[This is an answer to the original question, in which repetition was not excluded.]

For the number to be divisible by $4$, the number formed by the last two digits must be divisible by $4$. With the digits $1$ through $6$, you can use the $3$ last digits $2$, $4$ and $6$, and each of these can be combined with half of the digits to form a $2$-digit number divisible by $4$, so there are $9$ combinations for the last two digits. The other two digits can be chosen arbitrarily, so there are $36$ combinations for those, for a total of $9\cdot 36=324$ different numbers.

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I forgot to say that none of the nos should be repeated... –  S.M.09 Aug 18 '11 at 17:24
    
Then having your $9$ possibilities for the last two digits, you can choose $4$ numbers for the first digit and $3$ for the second, giving $9\cdot 4 \cdot 3=108$ –  Ross Millikan Aug 18 '11 at 17:27
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@Ross: There's only $8$ for the last two because one of the $9$ is $44$ with repetition. –  joriki Aug 18 '11 at 17:29
    
@joriki: right you are. So $96$ total. –  Ross Millikan Aug 18 '11 at 18:16

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