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Find the smallest number that is made up of each of the digits $1$ through $8$ exactly once and is divisible by $88$.

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$12437568$ should work, but what did you try? –  Kirthi Raman Mar 20 '12 at 14:55
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What have you tried? –  user5137 Mar 20 '12 at 14:55
    
I know that in order to be divisible by 11, the alternating digit sums has to be divisible by 11.But didn't know where to start –  Sam Mar 20 '12 at 14:59
    
I get it, I start with two sets $\{1,3,5,7\}, \{2,4,6,8\}$ and since the sums are $16$ and $20$, I have to find a way to exchange a digit from each of these sets –  Sam Mar 20 '12 at 15:05
    
Yes. good keep looking you will find the answer –  Kirthi Raman Mar 20 '12 at 15:07

2 Answers 2

Looks like this is a modified version of the problem at MathsChallenge (on page 29) You should go read that and try solving yourself.

If someone else comes with a different approach, that probably will be helpful to you.

You should also mention what was your approach (You would have started somewhere, didn't you?)

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It's been a long day for me so far but if the number is divisible by $88$ shouldn't it therefore be divisible by $11$? If that's the case shouldn't the last two digits be $\ldots 21$?

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I think he is looking for smallest number that uses all the digits from $1$ to $8$ that is divisible by 88 –  Kirthi Raman Mar 20 '12 at 15:03
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An odd number divisible by $88$? –  Arthur Fischer Mar 20 '12 at 15:04
    
I know, I know I'm easily confused! Thanks to both - I see now that it needs to be the \emph{smallest} number! –  Autolatry Mar 20 '12 at 15:07

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