Summing 10 Digits

Let us consider the feasible cuts. We know that when adding two numbers, the last number is always either the same number of digits as the largest number in the sum (e.g. $75+1968=2043$ ) or 1 digit larger if there is a carry-over (e.g. $349+857=1206$ ). This leaves us with the following options:

• If the first number is 1 digit, the second would have to be 4 digits and the third 5 digits, resulting in something like this: $5+9997=10002$ . We can see that in any situation like this, there will be too many 9s and 0s, and so this is not feasible.
• If the first number is 2 digits, the second must be 4 digits and the third must be 4 digits (e.g. $75+1968=2043$ ). We will call this $(2,4,4)$ .
• If the first number is 3 digits, the second must be 3 digits and the third must be 4 digits (e.g. $349+857=1206$ ). We will call this $(3,3,4)$ .
• If the first number is 4 digits, the second must be 2 digits and the third must be 4 digits (e.g. $4926+87=5013$ ). We will call this $(4,2,4)$ .

If we let the numbers in the arrangement be the letters $a$ through $j$ , these three options give us the following equations:

$\begin{aligned}&(2,4,4):&(10a+b)+(1000c+100d+10e+f)&=(1000g+100h+10i+j)\\ &(3,3,4):&(100a+10b+c)+(100d+10e+f)&=(1000g+100h+10i+j)\\ &(4,2,4):&(1000a+100b+10c+d)+(10e+f)&=(1000g+100h+10i+j)\end{aligned}$

We can determine whether or not these are possible by solving simultaneously, and since all three have $(1000g+100h+10i+j)$ on the right hand side, we can simply let the left hand sides of each equation equal each other to solve them.

First, let us solve $(2,4,4)$ and $(3,3,4)$ :

$(10a+b)+(1000c+100d+10e+f)=(100a+10b+c)+(100d+10e+f)\\ \therefore 999c=90a+9b\\ \therefore 10a+b=111c$

This is clearly impossible, because $10a+b$ cannot be large enough to match $111c$ . Even if $10a+b$ is maximised and $111c$ is minimised, we get $10a+b=10(9)+8=98$ and $111c=111(1)=111$ . Note that $c$ cannot be zero because it is the first digit of the number $(1000c+100d+10e+f)$ .

Next, consider $(4,2,4)$ and $(3,3,4)$ : $(1000a+100b+10c+d)+(10e+f)=(100a+10b+c)+(100d+10e+f)\\ \therefore 900a+90b+9c=99d\\ \therefore 100a+10b+c=11d$

This is impossible for the same reason: the maximum value of $11d$ is $99$ , which is still not big enough to match $100a$ where $a$ is not zero.

Finally, consider $(2,4,4)$ and $(4,2,4)$ : $(10a+b)+(1000c+100d+10e+f)=(1000a+100b+10c+d)+(10e+f)\\ \therefore 990c+99d=990a+99b\\ \therefore 10a+b=10c+d$

Here, $10a+b$ and $10c+d$ are equal two digit numbers, where the tens digit is given by $a$ or $c$ and the units digit is given by $b$ or $d$ . From this we can conclude, since all the variables are single digit numbers, that $a=c$ and $b=d$ , which is not possible as all the variables must be distinct.

Thus, no two equations can be satisfied at the same time, and therefore the number of arrangements of all ten digits that can be cut in more than one way such that the equation works is zero .

First we need to know the possible $(a,b,c)$ , where $a$ , $b$ and $c$ are the number of digits of the first, second and the third number, respectively. One can discover (after a bit of observation) that the only possibilities are $(3,3,4)$ , $(2,4,4)$ and $(4,2,4)$ . The three possibilities would be called structures of the cuts, from now on.

Then, the key is to pay attention to "more than one way". Obviously, an arrangement can not be cut in two ways such that both cuts have the same structure (i.e. both are of type $(3,3,4)$ ). Therefore, if such arrangement exists, then one cut is of structure $(3,3,4)$ and the other one is of structure $(2,4,4)$ or $(4,2,4)$ . First, we treat $(3,3,4)$ and $(2,4,4)$ .

Take the arrangement to be $\overline{abcdefghij}$ . With $(3,3,4)$ structure, we have the equation

$10^2a+10b+c+10^2d+10e+f=10^3g+10^2h+10i+j$

With $(2,4,4)$ structure, we have the equation

$10a+b+10^3c+10^2d+10e+f=10^3g+10^2h+10i+j$

The right hand sides of both equations are the same. So, we put the left hand sides to be equal.

$10a+b+10^3c+10^2d+10e+f= 10^2a+10b+c+10^2d+10e+f \implies 10^2a+10b+c=10a+b+10^3c$

$10^2a+10b+c=10a+b+10^3c \implies 10a(a-1)+9b=999c$

We need to have $9 \mid a(a-1)$ . Since $a$ and $a-1$ are consecutive integers, not both can b multiples of $3$ . Therefore, either one of them should be a multiple of $9$ . $a-1$ cannot be a multiple of $9$ , since $a$ would not be a single digit number anymore. Therefore, $a=9$ and the last equation simplifies to

$111c-b=80$

It is a linear diophantine equation, for which there is no solution $(x,y)$ such that both $x$ and $y$ are single digit numbers at the same time.

One can do the same for structures $(3,3,4)$ and $(4,2,4)$ (also to the case $(2,4,4)$ and $(4,2,4)$ ), which would be a bit more complicated. However the same ideas would be applied throughout to realise there is not such arrangement that gives more than one cut that satisfy the mentioned property.