RMO part 1!

Since $x$ has $10$ digits and is of the form $(10^5-1)y$ , we see that $10^4 < y \le 10^5+1$ . We can rule out $y = 10^5$ and $10^5+1$ since it leads to an $x$ with repeated digits, so this means that $y$ has $5$ digits.

Note that $x = \overline{(y-1)(10^5-y)}$ , so the two five-digit halves of $x$ are numbers with no repeated digits that add up to $99999$ . For example $y = 23986$ , $x = 2398576014$ . Note that the digits that are five spaces apart are paired up with each other; each pair sums to $9$ .

The number of choices for $x$ is the number of choices for the left five digits of $x$ . We can choose one of two digits from each pair of digits that add to $9$ , and then we can permute those digits however we want; so the number of such $x$ is almost $2^5 \cdot 5!$ ; but we cannot have a leading 0, so we must subtract the $2^4 \cdot 4!$ choices that involve picking a first digit of $0$ . So the answer is $2^5 \cdot 5! - 2^4 \cdot 4! = \fbox{3456}$ .

Almost the same solution as @Patrick Corn but I explain it differently.

The number can be written as $10^{5}.\overline{a_{1}a_{2}a_{3}a_{4}a_{5}} + \overline{a_{6}a_{7}a_{8}a_{9}a_{10}}$ .

Therefore , $99999 \mid \overline{a_{1}a_{2}a_{3}a_{4}a_{5}} + \overline{a_{6}a_{7}a_{8}a_{9}a_{10}}$ .

Now clearly , $\overline{a_{1}a_{2}a_{3}a_{4}a_{5}} + \overline{a_{6}a_{7}a_{8}a_{9}a_{10}} < 199998$

$\Rightarrow \overline{a_{1}a_{2}a_{3}a_{4}a_{5}} + \overline{a_{6}a_{7}a_{8}a_{9}a_{10}} = 99999$

$\Rightarrow a_{i} + a_{5 + i} = 9$ , where ${i}\in{1 , 2 , 3 , 4 , 5}$ .

Therefore it is equivalent to finding the number of possible 5 pairs such that the sum of the elements in a pair = 9 and all the ten digits are used.

It is equal to $2^{5}.5! - 2^{4}.4!$ ....... (Subtracting cases when $a_{1} = 0$ )

Therefore , the number of magic numbers = $\fbox{3456}$