13.377 The difference of the logarithms of the hundreds and tens digits of a three-digit number is equal to the logarithm of the difference of the same digits, and the sum of the logarithms of the hundreds and tens is equal to the logarithm of the sum of the same digits, increased by 4/3 times. If this three-digit number is subtracted from the number with the reverse order of digits, then their difference will be equal to a positive number, the hundreds digit of which coincides with the tens digit of the given number. Find this number.
Solution. Let $\overline{x y z}$ be the desired number. According to the condition, $\log _{a} x-\log _{a} y=\log _{a}(x-y) \Rightarrow x=y(x-y) \Rightarrow x: y \Rightarrow x=n y, n-$ integer. From here $n y=y(n y-y) \Rightarrow n=y(n-1) \Rightarrow n:(n-1)$

But $n$ and $n-1$ are coprime, hence $n-1=1, y=2, x=4$.

According to the problem condition $\overline{x y z}-\overline{z y x}=\overline{y^{* *}} \Rightarrow$
$\Rightarrow(400+20+z)-(100 z+20+4)=\overline{2 * *} \Rightarrow 99(4-z)=\overline{2 * *} \Rightarrow 4-z=3 \Rightarrow z=1$.

Answer: 421.

Note. In the condition of problem 13.377, there is redundant information that $\log x+\log y=\log \frac{4}{3}(x+y)$ In the solution provided above, we did without this condition.