subtract two 4-digit numbers and obtain the sum of the digits always 18

Question

Let (abcd) and (dcba) be 4-digit numbers and

(abcd)-(dcba)= (xyzt)

show that the sum of the number (xyzt) is always 18.

I think we will use divisibility rules but i could not

succeed...

Answer 1

Without additional condition, the claim is not true. For example, $9551-1559=7992$ and $1+9+9+2=27\ne 18$; or: $7317-7137=180$ and $0+1+8+0=9\ne 18$. However, if we impose the additional condition that $a\ne d$ and $b\ne c$ in the four-digit number $\overline{abcd}$ (and of course that $\overline{abcd}>\overline{dcba}$), then the claim becomes true:

Recall that the digit sum of a number has the same remainder modulo $9$ as the original number. Thus $\overline{abcd}$ and $\overline{dcba}$ have the same remainder and their difference $\overline{xyzt}$ as well as its digit sum $s:=x+y+z+t$ must be multiples of $9$.

From $a\ne d$ and $\overline{abcd}>\overline{dcba}$ we find $a-d\ge 1$. Since at most one borrow can occur in each place, we conclude $a-d\ge x\ge a-d-1$. On the other hand, $t=10+d-a$ (and here a borrow must occur), so that $$9\le x+t\le 10.$$

Similarly, $y\in\{b-c-1,b-c,b-c+9,b-c+10\}$ and (as a borrow from the one's place must occur) $z\in\{c-b-1,c-b+9\}$ so that together with $0\le x+y\le 18$ we find $$ y+z\in\{8,9,18\}$$ Thus $s\le 9$ is not possible and $s=27$ is possible only if $x+t=9$ and $y=z=9$. But $z=9$ is only possible with $b=c$, hence is excluded. We conclude that $$s=9.$$