Find the smallest possible checksum

Question

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.

Answer 1

Let's start with a brief explanation of "casting out nines" as it's needed for the problem at hand. If $ABCD$ is the (at most) four-digit sum of a pair of three-digit numbers, $abc+def$, then, since $1000\equiv100\equiv10\equiv1$ mod $9$, we have

$$\begin{align} A+B+C+D &\equiv(1000A+100B+10C+D)\\ &=(100a+10b+c)+(100d+10e+f)\\ &\equiv a+b+c+d+e+f\mod 9 \end{align}$$

so, if $\{a,b,c,d,e,f\}=\{2,3,4,5,7,9\}$, we have

$$A+B+C+D\equiv2+3+4+5+7+9=(2+7)+3+(4+5)+9\equiv3\mod 9$$

which means that the digit sum for any pair $abc+def$ belongs to $\{3,12,21,\ldots\}$. The OP has already found one example with digit sum $3$, namely $243+957=1200$, so we can conclude that the smallest possible digit sum is $3$. The remaining question, how many different pairs give digit sum $A+B+C+D=3$?

Note that

$$abc+def=dbc+aef=aec+dbf=dec+abf$$

so it suffices to count the number of solutions with $a\lt d$, $b\lt e$ and $c\lt f$ and then multiply by $4$ (or by $8$, if you want to distinguish $(abc,def)$ from $(def,abc)$). Now the only realistic possibilities for $ABCD$ are $1200$, $1020$, $1002$, $1110$, $1101$, and $1011$. (That is, $500\lt abc+def\lt2000$, so we must have $A=1$ since $A=0$ would imply $B\ge5$.) Let's consider these possibilities according to what's in the ones place.

If $D=0$ (i.e., if $ABCD=1200$, $1020$, or $1110$), we can only have $c=3$ and $f=7$, which will carry a $1$ into the tens place. That makes $C=1$ impossible, but it gives $1+2+9=11$ and $1+4+5=10$ as possibilities for $C=2$ and $C=0$. Indeed, we get x two solutions (with $a\lt d$, etc.), namely $423+597=1020$ and $243+957=1200$.

If $D=1$ (i.e., if $ABCD=1101$ or $1011$), we can have $c+f=2+9=11$ or $c+f=4+7=11$. In either case the carried $1$ from $c+f$ means we need $b+d=9$ or $10$ in order to get $C=0$ or $1$. For $c+f=2+9$, both values of $C$ are attainable, each in only one way: $342+759=1101$ and $432+579=1011$. For $c+f=4+7$, neither value of $C$ is attainable, since the only digits that sum to $9$ are $2+7$ and $4+5$ and, as already remarked, the only digits that sum to $10$ is $3+7$. Thus we get just two solutions with $D=1$, namely $342+759=1101$ and $432+579=1011$.

Finally, if $D=2$ (i.e., $ABCD=1002$), we can only have $c+f=3+9$ or $c+f=5+7$. In either case we'll need $b+e=9$, which is possible only as $2+7$ or $4+5$. If $c+f=3+9$, either of these is possible, while neither is possible if $c+f=5+7$. So again we get just two solutions, namely $243+759=1002$ and $423+579=1002$.

Altogether we get six solutions with $a\lt d$, $b\lt e$ and $c\lt f$:

$$\begin{align} 423+579&=1020\\ 243+957&=1200\\ 342+759&=1101\\ 432+579&=1011\\ 243+759&=1002\\ 423+579&=1002 \end{align}$$

The total count, without the restrictions $a\lt d$ and $b\lt e$, is thus $24$; removing the restriction $c\lt f$ brings the total number of solutions to $48$.

Answer 2

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.

Answer 3

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.

Answer 4

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.