To determine the number of digits in the whole-number representation of \( 10^{100} - 9^{100} \), we start by finding an expression for the number of digits in a number \( n \). The number of digits \( d \) in \( n \) can be calculated using the formula:
\[ d = \lfloor \log_{10} n \rfloor + 1 \]
Now, we need to compute \( n = 10^{100} - 9^{100} \).
Let's first evaluate \( \log_{10} (10^{100} - 9^{100}) \):
Using properties of logarithms and inequalities, we know that:
\[ \log_{10} (a - b) \approx \log_{10} a \quad \text{when } a \gg b \]
In our case, \( a = 10^{100} \) and \( b = 9^{100} \). Since \( 10^{100} \) is significantly larger than \( 9^{100} \), we can approximate:
\[ \log_{10} (10^{100} - 9^{100}) \approx \log_{10} (10^{100}) = 100 \]
Now to find \( 9^{100} \):
\[ 9^{100} = (10 - 1)^{100} \]
Using the Binomial Theorem, we can expand \( (10 - 1)^{100} \):
\[ (10 - 1)^{100} = \sum_{k=0}^{100} \binom{100}{k} 10^{100-k} (-1)^k \]
The first term in this expansion is \( 10^{100} \) and the second term is \( -100 \cdot 10^{99} \); as we keep adding terms, they will contribute smaller and smaller amounts, but we mainly need to know that the total value remains slightly less than \( 10^{100} \).
Let’s evaluate \( 9^{100} \) approximately:
\[ 9^{100} = 10^{100} \left( \frac{9}{10} \right)^{100} = 10^{100} \cdot 0.9^{100} \]
Calculating \( 0.9^{100} \):
Using logarithms:
\[ \log_{10}(0.9^{100}) = 100 \cdot \log_{10}(0.9) \]
From known log values, we have \( \log_{10}(0.9) \approx -0.045757 \). Therefore,
\[ 100 \cdot \log_{10}(0.9) \approx 100 \cdot -0.045757 \approx -4.5757 \]
Thus,
\[ 9^{100} \approx 10^{100} \cdot 10^{-4.5757} \approx 10^{95.4243} \]
This shows that \( 9^{100} \) is much smaller than \( 10^{100} \) but still significantly contributes to the subtraction. We actually find that:
\[ 10^{100} - 9^{100} \approx 10^{100}(1 - 0.9^{100}) \approx 10^{100}(1 - 10^{-4.5757}) \approx 10^{100} \text{ since } 9^{100} \text{ is small.} \]
Now we have:
\[ \log_{10} (10^{100} - 9^{100}) \approx 100 \]
Finally, we compute the number of digits:
\[ \lfloor \log_{10} (10^{100} - 9^{100}) \rfloor + 1 \approx \lfloor 100 \rfloor + 1 = 101 \]
Therefore, the number of digits in the whole-number representation of \( 10^{100} - 9^{100} \) is
\[ \boxed{101}. \]