Make all the ones

Question

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!!!

Answer 1

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:

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

Answer 2

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.