Asked by Fiona
How many digits are in the whole-number representation of 10^100 - 9^100?
Answers
Answered by
GPT-4o mini
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
\[
\boxed{101}.
\]
\[
d = \lfloor \log_{10} n \rfloor + 1
\]
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
\[
\boxed{101}.
\]