Reversing the digits with a subtraction How many 3-digits numbers possess the following property: 

After subtracting $297$ from such a number, we get a $3$-digit number consisting of the same digits in the reverse order.

 A: Let one of these three digit number be $\overline{abc}$ where $a,b$ and $c$ are the digits. Then the problem can be expressed as 
$$\overline{abc}-297=\overline{cba}$$
or writing it in powers of $10$ as follows
$$100a+10b+c-297=100c+10b+a\Rightarrow 99(a-c)=297\Rightarrow a-c=3$$
So we have any triple $(a,b,c)=(c+3,b,c)$ would satisfy the condition where $b\in\{0,1,2,3,4,5,6,7,8,9\}$ and $c\in\{0,1,2,3,4,5,6\}$. So in total we have $$10\times7=70$$
possible numbers if you allow for $c=0$ otherwise you would have only 
$$10\times6=60$$
three digit numbers satisfying the condition.
A: A $3$-digit number is of the form $$a_0+10a_1+100a_2\ \ \ \ \ \ \ \ \text{with }\ \ \ \ \ \ 0\leq a_0,a_1\leq9,\ \ \ \ \ \  1\leq a_2\leq9$$
Then the condition to impose is $$a_0+10a_1+100a_2-297=a_2+10a_1+100a_0\ \ \ \ \ \text{ and }\ \ \ \ a_0\neq0$$
Therefore 
$$99a_2-99a_0=297$$
or equivalently $$a_2-a_0=3$$
We get the solutions $$\begin{align}a_0&=1& a_2=4\\a_0&=2&a_2=5\\a_0&=3&a_2=6\\a_0&=4&a_2=7\\a_0&=5&a_2=8\\a_0&=6&a_2=9\end{align}$$
For each of these solutions for $a_0,a_2$, the $a_1$ can take any value $0,1,2,...,9$.
A: You have:$$\text{ }abc$$$$-297$$$$cba$$The 10's column has $b-9=b$. This implies there first must have been a borrow from the 10's column to the units column, and then a borrow from the 100's column into the 10's column. This information tells us that:$$a=c+3$$and $b$ can be any digit from $0$ to $9$.
Now you just need to calculate the number of combinations you can get with these restrictions.
