Let's determine

$\left(\frac{A \times B}{C}\right)^{n}$

the number of digits in the expression, if the number of digits in $A, B, C$ are $a, b$ and $c$ respectively.
We need to determine the lower and upper limits of the number of digits.

In the product of two numbers, the lower limit of the number of digits is obtained by adding the number of digits of the factors and subtracting 1 from this sum, while the upper limit is obtained by simply adding the number of digits of the factors.

In the product $A \times B$, the lower limit of the number of digits is thus $a+b-1$, and the upper limit is $a+b$.

For the number of digits in the quotient, the lower limit is the number of digits of the dividend minus the number of digits of the divisor; the upper limit is the difference in the number of digits plus 1; so in this case, the lower limit is $a+b-1-c$, and the upper limit is $a+b-c+1$.

In $\left(\frac{A \times B}{C}\right)^{2}$, the lower limit is $2(a+b-c-1)-1$,

and the upper limit is $2(a+b-c-1)+1$,

In $\left(\frac{A \times B}{C}\right)^{3}$, the lower limit is $3(a+b-c-1)-2$,

and the upper limit is $3(a+b-c+1)$,

In $\left(\frac{A \times B}{C}\right)^{n}$, the lower limit is $n(a+b-c-1)-(n-1)$, and the upper limit is $n(a+b-c+1)$.

Thus, the sought lower limit is $n(a+b-c-2)+1$, and the upper limit is $n(a+b-c+1)$.

The problem was solved by: Freibauer E., Kornis Ö., Krisztián Gy., Lukhaub Gy., Prohászka J.