Invariance problem dealing with the sums of units digits

53 Views Asked by At

We may write all the digits from 1 to 9 in a row in any order we like, and then we write plus signs between some digits (as many plus signs as we like). Finally, we evaluate the obtained expression. Prove that there is no way to get the value 100, or 101, or 102, or 103 and so on. What is the smallest three-digit number that can be obtained in this game?

1

There are 1 best solutions below

0
On

Answer is 108.

$1 + 2 + 3 + 4 + 5 + 6 + 78 + 9 = 108$.

If we use any digit $X$ in number with more than one digit it gives $10^k * X$ to the final sum. It increases $1 + 2 + ... + 9 = 45$ by $9....9 * X$. Cause of this final sum has the same remainder of the division by 9. So, final sum is always divided by 9. Numbers from $100$ to $107$ are not divided by 9, so there is no way to get them.