Reversing the digits with a subtraction

Question

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.

Answer 1

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.

Answer 2

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 $0,1,2,...,9$.

Answer 3

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.

Answer 4

Following the perfect explanation given by Arian, there are a total of 70 solutions -- 60 of them are true solutions (> 390):

$$ \left[ \begin{array}{cccccccccc} 300&310&320&330&340&350&360&370&380&390\\ \hline 401&411&421&431&441&451&461&471&481&491\\ 502&512&522&532&542&552&562&572&582&592\\ 603&613&623&633&643&653&663&673&683&693\\ 704&714&724&734&744&754&764&774&784&794\\ 805&815&825&835&845&855&865&875&885&895\\ 906&916&926&936&946&956&966&976&986&996\\ \end{array} \right] $$