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Given all the digits $0,\dots,9$ and using each digit exactly once,

What is the smallest possible number you can make whose digits are all $1$'s?

The only operations that are allowed are:

  • addition ($+$)
  • concatenation

and you can concatenate after a sum, or practically anywhere.

So, for example, $87+24=111$ - let us call this an answer of length 3 - and then $\operatorname{concat}(111,1)$ to give $1111$, an answer of length 4.

In this example, I have only used the digits $2, 4, 7, 8$ and $1$, but in your answer you must use ALL the digits from $0,\dots,9$.

You can also concatenate during a sum, for example:

  • $\operatorname{concat}(3+4,\operatorname{concat}(1,2))$

gives $712$

whereas:

  • $\operatorname{concat}(\operatorname{concat}(1,2),3+4)$

gives $127$

and:

  • $\operatorname{concat}(\operatorname{concat}(3,4),1+2)$

gives $343$.

Good luck!!!

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  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – EKons Sep 20 '18 at 16:45
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    $\begingroup$ This question has been closed as "too broad", because it doesn't provide clear, objective criteria for choosing one best answer. If you edit the question to provide a way to objectively select one best answer, you can nominate the question for reopening. $\endgroup$ – GentlePurpleRain♦ Sep 20 '18 at 18:30
  • $\begingroup$ I made some "aimless" corrections on the post. Can OP check and see if it still maintains the same meaning? $\endgroup$ – Hugh Sep 20 '18 at 21:31
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    $\begingroup$ Also, in your example, you have used 1 twice. Once in 13 and once in $+1$. You might want to change it. $\endgroup$ – Hugh Sep 20 '18 at 21:33
  • $\begingroup$ Can you give more details on the concatenation operation, please? Is "after" positional or just chronological? You gave an example with concat(sum, digit), but are concat(digit, sum) and concat(sum, sum) allowed? $\endgroup$ – xhienne Sep 20 '18 at 23:56
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How about this?

$(9+2)(8+3)(7+4)(6+5)(1+0) = (11)(11)(11)(11)(1) = 111111111$

where consecutive brackets represent concatenation.

or write:

$\operatorname{concat}(9+2, 8+3, 7+4, 6+5, 1+0)=111111111$

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  • $\begingroup$ I'm afraid that the rules specify that you cannot concatenate two sums together, just one or more digits to a prior sum. $\endgroup$ – xhienne Sep 20 '18 at 22:32
  • $\begingroup$ @xhienne No, this is allowed. The rules state: ...and you can concatenate after a sum... it does not specifically state that you cannot concatenate two sums. $\endgroup$ – Hugh Sep 20 '18 at 23:19
  • $\begingroup$ @Hugh What is not stated is not necessarily allowed. I just asked JonMark Perry. $\endgroup$ – xhienne Sep 21 '18 at 0:01
  • $\begingroup$ I've tweaked the q... $\endgroup$ – JonMark Perry Sep 21 '18 at 3:02
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    $\begingroup$ This also proves to be minimal, as the final answer has to be divisible by 9, and needs to be positive. $\endgroup$ – elias Sep 21 '18 at 6:55
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Since the question has changed since hexomino's answer, here is an answer that addresses the specific requirement of the result being the smallest possible number whose digits are all 1's.

From any sum of arbitrary numbers $S = n1 + n2 + n3 + n4 + ...$
let us study the concatenation of any two of its members (say n1 and n2):

If d is the number of digits of n2, appending n2 to n1 gives a new number $n' = n1 * 10^d + n2$

The resulting sum is then:
$S' = S - (n1 + n2) + n1 * 10^d + n2$
$=> S' = S + (10^d - 1) * n1$

$(10^d - 1)$ is a multiple of 9, so is $(10^d - 1) * n1$, and so is $S'$ if $S$ is a multiple of 9 initially.

We can then conclude that, whatever the number of additions and concatenations among the members of $S$, if $S$ is a multiple of 9, then $S'$ remains a multiple of 9.

In this puzzle, $S = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$ which is a multiple of 9. Therefore, after any number of the allowed operations, the result will still remain a multiple of 9.

Since the sum of the digits of a multiple of 9 is itself a multiple of 9, the smallest multiple of 9 whose digits are all 1's is 111111111. This means that hexomino has found the optimal answer.

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