This one come from newspapers in France, where this week is the "mathematics week", and some problems are proposed to kids. This single one is easy and funny:

Find the lowest 8-digit number, all distinct figures, such that when 2 figures are consecutive, one is the multiple of the other.

All figures 0..9 are allowed, but we can propose 2 solutions: 0 allowed, 0 disallowed.

  • $\begingroup$ The phrasing is unclear to me. I assume by "figures" you mean "digits" but what does "one is the multiple of the other" mean? Do you mean that every two consecutive digits have to be divisible without remainder? $\endgroup$
    – Hinton
    Mar 10, 2023 at 12:01
  • 1
    $\begingroup$ @Hinton yes by figure I mean digit. For the second part of your question, I mean that in (1st digit, 2nd digit), one is divisible by the other without remainder. Same for (2nd digit, 3rd digit), same for (3rd digit, 4 digit) and so all..... We don't consider only 1st/2nd then 3rd/4th.... $\endgroup$ Mar 10, 2023 at 13:34

1 Answer 1


Solution (0 allowed)



If we build the number from the front and assume non-leading zero then the smallest possible first two digits are "10".

If we continue with a 2 then the fourth digit must be 4,6 or 8. With 4 or 8 we would have to put the other one as the fifth digit and there is no possible way of completing the number. With 6 the sequence is forced to continue as 102639 and there is then no option for the seventh digit.

If the third digit is 3 then the fourth digit must be 6 or 9. There is no way to continue after 9 but if we pick 6 then we must continue as follows - 1036248 or 1036284 - but then there is no option for the eighth digit.

Hence the smallest possible value for the third digit is 4 and the next digit is either 2 or 8. If the fourth digit is a 2 then the fifth digit is either an 8, which leaves no possibility for the sixth digit, or 6 which forces the continuation 1042639 but then there is no possibility for the eighth digit.

Thus the fourth digit must be 8 and choosing the smallest available value for each digit in the rest of the number gives 10482639

Solution (0 disallowed)



For 0 disallowed we must include either 5 or 7. These two digits can only be adjacent to 1 so must be at the beginning or end and we can only choose one of them (and 5 is smaller). If we can start the number with a digit less than 5 then 5 must be at the end. Hence the number will end with 15.

The other digits that will be used are 2,4,6,8,3,9. Now, 9 can only be adjacent to 3 or 1 so must be at the beginning or next to the 1. To make the number as small as possible, we must put it, therefore, next to the 1. Then 3 must be to the left of this and 6 must be to the left of that. This means the third digit must be 2 and the smallest possibility for the first two digits is "48".

Hence the smallest possible value in this case is 48263915


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