2. For a four-digit positive integer $A B C D$, where $A, B, C$, and $D$ are digits with $A \neq 0$ and $D \neq 0$, we define the Reverse Digit sum (RD sum) of $A B C D$ to be the sum of $A B C D$ and $D C B A$. For example, the RD sum of 1205 is $1205+5021=6226$, while the integer 2300 does not have an RD sum.
Note that the four-digit positive integer $A B C D$ is equal to $1000 A+100 B+10 C+D$.
(a) Determine the RD sum of 4281 .
(b) There are positive integers $m$ and $n$ with the property that the RD sum of the integer $A B C D$ is always equal to $m \times(A+D)+n \times(B+C)$. State the value of $m$ and the value of $n$.
(c) Determine the number of four-digit integers whose RD sum is 3883 .
(d) Determine the number of four-digit integers that are equal to the RD sum of a four-digit integer.
2. (a) The RD sum of 4281 is $4281+1824=6105$.
(b) The integer with digits $A B C D$ is equal to $1000 A+100 B+10 C+D$. The integer with digits $D C B A$ is equal to $1000 D+100 C+10 B+A$. The RD sum of $A B C D$ is
$$
\begin{array}{rlr}
A B C D+D C B A & =(1000 A+100 B+10 C+D)+(1000 D+100 C+10 B+A) \\
& =1001 A+110 B+110 C+1001 D & \text { (group like terms) } \\
& =1001(A+D)+110(B+C) & \text { (factor) }
\end{array}
$$
(group like terms)
(factor)
and so $m=1001$ and $n=110$.
(c) Suppose the RD sum of $A B C D$ is 3883 .
From part (b), we have that $1001(A+D)+110(B+C)=3883$.
If $A+D$ were 4 or greater, then $1001(A+D)$ would be at least 4004 , which is larger than 3883 .
Therefore, $A+D$ is no larger than 3 .
Since $A$ and $D$ are both nonzero digits, they are each at least 1 , so $A \geq 1$ and $D \geq 1$, which implies $A+D \geq 1+1=2$.
This means $A+D=2$ or $A+D=3$.
If $A+D=2$, then $1001(A+D)=2002$.
Substituting this into $1001(A+D)+110(B+C)=3883$, we get $2002+110(B+C)=3883$ or $110(B+C)=1881$.

We also know that $B+C$ is an integer, so $110(B+C)$ is a multiple of 10 , so it cannot be equal to 1881 which is not a multiple of 10 .
Therefore, it is not possible that $A+D=2$, so we must have $A+D=3$.
Using that $A+D=3$, we get $1001(A+D)=3003$, so
$$
110(B+C)=3883-1001(A+D)=3883-3003=880
$$

Dividing $110(B+C)=880$ on both sides by 110 , we conclude that $B+C=8$.
Since $A$ and $D$ are positive digits that have a sum of 3 , we must have $A=1$ and $D=2$ or $A=2$ and $D=1$.
Since $B$ and $C$ are digits that have a sum of 8 but there is no restriction that they must be positive, we have that $(B, C)$ can be any of the pairs $(0,8),(1,7),(2,6),(3,5),(4,4)$, $(5,3),(6,2),(7,1)$, or $(8,0)$.
There are 2 possibilities for the values of $A$ and $D$, and 9 possibilities for the values of $B$ and $C$, Therefore, there are $2 \times 9=18$ integers with an RD sum of 3883 . They are listed below.
$1082,1172,1262,1352,1442,1532,1622,1712,1802$
2081,2171,2261,2351,2441,2531,2621,2711,2801
(d) Suppose that $n$ is a four-digit integer and that $A B C D$ is an integer with RD sum equal
to $n$.
From part (b), this means $1001(A+D)+110(B+C)=n$. Note that $A$ and $D$ are both
non-zero.
Suppose there is some other integer, $E F G H$, with an RD sum equal to $n$.
Again by part (b), this means $1001(E+H)+110(F+G)=n$.
Therefore, we have $1001(A+D)+110(B+C)=1001(E+H)+110(F+G)$.
For convenience, set $w=A+D, x=B+C, y=E+H$, and $z=F+G$.
The equation above simplifies to $1001 w+110 x=1001 y+110 z$.
Rearranging this equation, we get $1001 w-1001 y=110 z-110 x$.
After factoring, we have $1001(w-y)=110(z-x)$, and after dividing both sides of this equation, we get $91(w-y)=10(z-x)$.
Observe that $91=7 \times 13$.
The quantities $w-y$ and $z-x$ are both integers, which means $91(w-y)$ and $10(z-x)$ are both integers.
Moreover, since $91(w-y)$ is a multiple of 13 , we must also have that $10(z-x)$ is a multiple of 13 .
The integer 13 is prime and 10 does not have a factor of 13 , so $z-x$ must have a factor of 13 .
By similar reasoning, $z-x$ must have a factor of 7 , which means $z-x$ is a multiple of $7 \times 13=91$.
Now recall that each of $x$ and $z$ is the sum of two digits.
This means that each of $x$ and $z$ is an integer from 0 to 18 inclusive.
The difference between two integers that are at least 0 and at most 18 must lie between
$0-18=-18$ and $18-0=18$.
We have shown that $z-x$ is a multiple of 91 that lies between -18 and 18 .
The only such integer is 0 , which shows that $z-x=0$ or $z=x$.
Since $z-x=0,10(z-x)=0$, which means $91(w-y)=0$.
Therefore, $w-y=0$ as well, so $w=y$.

We have shown that if $A B C D$ and $E F G H$ have the same RD sum, then $A+D=E+H$ and $B+C=F+G$.
In other words, if $n$ is the RD sum of a four digit integer, then the values of $A+D$ and $B+C$ are uniquely determined.
This means, to count the number of four-digit RD sums, we can count the number of pairs $(w, x)$ with the property that $1001 w+110 x \leq 9999$, subject to the condition that $w$ and $x$ are the sum of digits, with $w$ being the sum of positive digits.
Thus, the answer to the question is the number of pairs $(w, x)$ with $2 \leq w \leq 18$ and $0 \leq x \leq 18$ that satisfy $1001 w+110 x \leq 9999$.
If $w=2$, then $1001 w+110 x=2002+110 x$, so we want $2002+110 x \leq 9999$ or $110 x \leq$ 9999 - 2002 or $110 x \leq 7997$.
The largest that $x$ can be is 18 , so the largest that $110 x$ can be is $18 \times 110=1980$, which is less than 7997 .
Therefore, if $w=2$, then $x$ can be any integer from 0 through 18 .
Thus, we get 19 pairs $(w, x)$ when $w=2$.
If $w=3$, then $1001 w+110 x=3003+110 x$, which means we need $110 x \leq 6996$.
As with the case when $w=2$, every possible value of $x$ from 0 through 18 satisfies this inequality, so we get 19 pairs in this case as well.
If $w=4$, similar reasoning implies that $110 x \leq 5995$, so we again get 19 pairs.
Continuing this reasoning, we find that we get 19 pairs for each of $w=5, w=6, w=7$, and $w=8$.
If $w=9$, then we must have that $9009+110 x \leq 9999$, so $110 x \leq 9999-9009$ or $110 x \leq 990$. It is not difficult to see that $x$ can be any integer from 0 through 9 inclusive, but if $x \geq 10$, then $110 x \leq 990$ is not true. Therefore, we get 10 pairs in this case.
If $w \geq 10$, then $1001 w$ is at least 10010 , so there is no way to choose $x$ so that $1001 w+110 x$ has four digits.
Therefore, there are no more cases to consider, and the number of possible four-digit RD sums is
$$
19+19+19+19+19+19+19+10=143
$$