Find the smallest possible checksum What is the smallest possible checksum $C_{min}$ of the sum of two three-digit numbers $N_1$ and $N_2$ that are formed from the six given digits 2, 3, 4, 5, 7 and 9 (each digit shall be used only once). The checksum of a number is defined as the sum of its digits. What are the sums that have this smallest possible checksum $C_{min}$?
Here my thoughts: I believe that the smallest checksum is 3. I obtained this by simple reasoning: With $3 + 7 = 10$ there is 0 in the sum and a carry-over "1". Then, with $4 + 5 = 9$ and the carry-over "1" we obtain another "0" in the sum, creating another carry-over "1". Finally, $2 + 9 = 11$, and with the carry-over it is "12". Thus, the checksum is $1 + 2 + 0 + 0 = 3.$ But is it right? 
Examples: 
$243 + 957 = 1200,$ checksum = 3
$423 + 597 = 1020,$ checksum = 3
If the answer is at all correct, how can I prove that it is? And then, how would I be able to find all the sums of numbers? My approach: Permute the given checksum, and check if I can find N1 and N2 that have this result.
There must be a more systematic or rigorous approach, I would believe. Perhaps someone can guide me on this. Thank you.
 A: Hint1:
$$\sum_{k=0}^n 10^ka_k \equiv \sum_{k=0}^n a_k \mod{9} $$
Hint2:
Note, that the digital root for your numbers is always the same, regardless of your choice. 
A: Building on the helpful hints of @Jaroslaw Matlak, the digital root of the sum will equal the digital root of the sum of the digital roots for the summed numbers. Consider your first example:
$243 + 957 = 1200 $
The digital root of $243$ is
$2+4+3 = 9 $
and the digital root of $957$ is 
$9 + 5 + 7 = 21,$
continuing to sum until we have one digit,
$2+1 = 3 $.
So the sum of the digital roots for $243$ and $957$ is 
$9+3 = 12 $ 
giving
$1+2 = 3$. This is equal to the digital root of $1200$.
$1+2+0+0 = 3$.
You should see that the actual permutations of the numbers $2,3,4,5,7,9$ don't matter in the calculation of the digital root. The left hand side digital root will always be
$2+3+4+5+7+9=30$, $3+0=3$, the same as the digital root of the sum. 
A: The other answers have shown that the digit sum will converge to $3$ if you repeat it enough.  It appears you only want to take the digit sum once, so you can get $3,12,21$.  Once you find an example with $3$ you are done.  Finding such an example can be done with clever searching.  There is only one pair of digits, $3,7$, available that sum to $10$, so we put those in the ones place to get a $0$.  There is another pair, $4,5$, that sum to $9$ so if they are added with a carry in you will get another $0$.  The sum of two three digit numbers cannot have more than four digits, so if you get two zeros the digit sum has to be $3$ or $12$, but since the carry into the thousands is $1$ you can only get $3$.  Good work.
