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Let $ A$ be the set of integers $ n$ with $ 0\leq n < 10^{2m-1}$ and satisfying property 2. Let $ B$ be the set of integers $ n$ with $ 0\leq n < 10^{2m}$ and satisfying property 2. We will form a bijection from the elements in set $ A$ to the elements of set $ B$. Take any element in set $ B$ -- a number with decimal representation $ a_{1}a_{2}...a_{2m-1}a_{2m}$. Then, there exists a permutation of the elements of set $ \{a_{1},a_{2},...,a_{2m}\}$ into a set $ \{b_{1},b_{2},...,b_{2m}\}$ so that the number $ b_{1}b_{2}...b_{2m-1}b_{2m}$ is divisible by $ 11$ so $ \sum_{i = 1}^{2m}(-1)^{i}b_{i}\equiv 0\pmod{11}$. Then $ \sum_{i = 1}^{2m}(-1)^{i}(kb_{i}+l)\equiv\sum_{i = 1}^{2m}(-1)^{i}(kb_{i})\equiv k\sum_{i = 1}^{2m}(-1)^{i}(b_{i})\equiv 0\pmod{11}$, therefore the integer $ \sum_{i = 1}^{2m}(-1)^{i}(c_{i})\equiv 0\pmod{11}$ is also divisible by $ 11$ where $ c_{i}\equiv kb_{i}+l\pmod{11}$ (*) Now, because for $ i = 1,2,...,2m$ $ a_{i}$ is one of $ 0,1,2,...,9$ then $ a_{i}+1$ is never divisible by $ 11$. Then, there exists an integer $ k$ from the set $ \{1,2,...,9,10\}$ such that $ k(a_{2m}+1)\equiv 1\pmod{11}$ so $ k(a_{2m}+1)-1\equiv 0\pmod{11}$. Also note that for $ i = 1,2,3,...,2m-1$ we have $ k(a_{i}+1)-1$ is one of $ 0,1,2,..,9$ mod $ 11$ and never $ 10$ mod $ 11$ because $ a_{i}+1$ is never divisible by $ 11$. Then, using the property (*) twice we get that $ \sum_{i = 1}^{2m}(-1)^{i}(d_{i})\equiv 0\pmod{11}$ where $ d_{i}\equiv k(b_{i}+1)-1\pmod{11}$. Then, if we take $ e_{i}= k(a_{i}+1)-1$ then $ e_{i}$ is an integer from $ 0$ to $ 9$ for $ i = 1,2,3,...,2m-1$ and $ e_{2m}= 0$. Then, because then $ d_{i}$ is an integer from $ 0$ to $ 9$ then because $ \sum_{i = 1}^{2m}(-1)^{i}(d_{i})\equiv 0\pmod{11}$, the integer $ d_{1}d_{2}...d_{2m}$ is divisible by $ 11$ and $ \{d_{1},d_{2},...,d_{2m}\}$ is a permutation of $ \{e_{1},e_{2},...,e_{2m}\}$ (because $ \{b_{1},b_{2},...,b_{2m}\}$ is a permutation of $ \{a_{1},a_{2},...,a_{2m}\}$). Therefore, the digits of $ e_{1}e_{2}...e_{2m}$ can be permuted to get an integer divisible by $ 11$, and we can interchange odd and even positions in order to get $ e_{2m}= 0$ being the right-most digit. Therefore, the digits of the number $ e_{1}e_{2}...e_{2m-1}$ can be permuted to get an integer divisible by $ 11$ (because the $ 0$ on the right end can be just removed). So we get a one-to-one bijection from $ a_{1}a_{2}...a_{2m}$ to $ e_{1}e_{2}...e_{2m-1}$. Now we can reverse the procedure to go from every integer $ e_{1}e_{2}...e_{2m-1}$ in set $ B$ and add choose any $ a_{2m}$ from the set $ \{0,1,2,...,9\}$ and then find the required $ k$ such that $ k(a_{2m}+1)-1\equiv 0\pmod{11}$, and then reverse the process described above (solve for $ a_{i}$ in $ k(a_{i}+1)-1\equiv e_{i}\pmod{11}$ and then we get a unique solution $ \pmod{11}$ with $ a_{i}$ belonging to the set $ \{0,1,2,...,9\}$) and reconstruct the number $ a_{1}a_{2}...a_{2m}$ belonging to $ A$. There are ten ways to choose $ a_{2m}$ for every element in $ B$, and each yields a unique element from $ A$, and we will get every element in $ A$ because every element in $ A$ can be mapped on an element in $ B$ by the reverse procedure. So, for every element in $ B$ there are $ 10$ elements from $ A$ so $ f\left(2m\right) = 10 f\left(2m-1\right)$ QED.

Here is another solution which uses a different bijection.

Suppose that we allow the digit $A$, which has value $10$, inside our $k$-digit numbers. Also, define $f(k,d)$ to be the number of good strings of length $k$ which exclude the digit $d$. Hence, $f(k)=f(k,A)$. Now, note that the number of good strings in $f(2m,A)$ which end in a digit $d$ is equivalent to the number of good strings with $2m-1$ digits which can be split into two groups whose sums differ by $d$. However, subtracting $d\pmod{11}$ from these $2m-1$ digits, we see that there is a bijection between good strings in $f(2m-1,A-d)$ and good strings in $f(2m,A)$ which end in $d$. So, $f(2m,A)=f(2m-1,1)+\ldots +f(2m-1,A)$. Now, we claim $f(2m-1,d)$ is constant for nonzero $d$. To see this, we can just map every digit $k$ in our string to $k\zeta\pmod{11}$, where $\zeta$ is a fixed primitive root. Then, $f(2m-1,d)=f(2m-1,d\zeta)=\ldots$, as desired. In particular, this means that $f(2m)=10*f(2m-1)$, as desired.

How about a solution that doesn't require that $11$ is prime? A comment after the first official solution says that $10$ can be replaced by any even base $b$ (as long as $11$ and $9$ are replaced by $b+1$ and $b-1$), yielding $f\left(2m\right) = b f\left(2m-1\right)$.

We fix a positive integer $m$. Call a $2m$-digits number (with leading zeroes allowed) tasty if and only if its digits can be permuted in a way that yield an integer divisible by $11$. Let $A=f(2m-1)$ and $B_i$ be the number of tasty numbers starting with digit $i$, where $i=0,1,\hdots,9$. Claim: $A=B_0$ Proof: See that $\overline{x_1x_2\hdots x_{2m-1}}\mapsto \overline{0x_1x_2\hdots x_{2m-1}}$ is indeed a working bijection. $\blacksquare$ Thus, we restrict our attention to $B_0,B_1,\hdots,B_9$. It turns out that they are all equal. The critical observation to prove this is the following. Claim: If we allow $10$ as a digit, then for any $a,b\in\mathbb{F}_{11}$, if $\overline{x_1x_2\hdots x_{2m}}$ is tasty, then its component-wise linear transformation $$\overline{(ax_1+b)(ax_2+b)\hdots (ax_{2m}+b)}$$is also tasty, where each number is treated as a residue modulo $11$. Proof: As $\overline{x_1x_2\hdots x_{2m}}$ is tasty, one can partition $x_1,x_2\hdots,x_{2m}$ to two sets $T_1, T_2$ such that $|T_1|=|T_2|=m$ and $$\sum_{t\in T_1} t \equiv \sum_{t\in T_2} t \pmod{11}.$$Thus, $$\sum_{t\in T_1} at+b \equiv \sum_{t\in T_2} at+b\pmod{11}$$which yields the desired partition of the transformed number. $\blacksquare$ All that remains is to find a linear transformation in $\mathbb{F}_{11}$ which fix $10$ and maps $0$ to $k$. It's easy to see that the desired transformation is $x\mapsto (k+1)(x+1)-1$. This completes the problem.