Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
$12437568$ should work, but what did you try? – Kirthi Raman Mar 20 '12 at 14:55
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

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?)

share|cite|improve this answer

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 $\ldots 21$?

share|cite|improve this answer
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
An odd number divisible by $88$? – arjafi 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.