Find the number of digit of $\sum_{n=0}^{99} 3^n$.
You may use $\log_{10} 3=0.4771$.

2012 Tokyo Institute of Technology entrance exam, problem 2-A
1. First, we recognize that the given sum is a geometric series:
   \[
   \sum_{n=0}^{99} 3^n = 3^0 + 3^1 + 3^2 + \cdots + 3^{99}
   \]
   The sum of a geometric series \( \sum_{n=0}^{k} ar^n \) is given by:
   \[
   S = \frac{a(r^{k+1} - 1)}{r - 1}
   \]
   Here, \( a = 1 \), \( r = 3 \), and \( k = 99 \). Therefore, the sum is:
   \[
   \sum_{n=0}^{99} 3^n = \frac{3^{100} - 1}{3 - 1} = \frac{3^{100} - 1}{2}
   \]

2. To find the number of digits of a number \( N \), we use the formula:
   \[
   \text{Number of digits} = \left\lfloor \log_{10} N \right\rfloor + 1
   \]
   Here, \( N = \frac{3^{100} - 1}{2} \).

3. We need to compute \( \log_{10} \left( \frac{3^{100} - 1}{2} \right) \). Using the properties of logarithms:
   \[
   \log_{10} \left( \frac{3^{100} - 1}{2} \right) = \log_{10} (3^{100} - 1) - \log_{10} 2
   \]

4. We approximate \( \log_{10} (3^{100} - 1) \) by noting that for large \( n \), \( 3^{100} - 1 \approx 3^{100} \). Thus:
   \[
   \log_{10} (3^{100} - 1) \approx \log_{10} (3^{100}) = 100 \log_{10} 3
   \]
   Given \( \log_{10} 3 = 0.4771 \):
   \[
   100 \log_{10} 3 = 100 \times 0.4771 = 47.71
   \]

5. Now, we need to subtract \( \log_{10} 2 \). We know that \( 0 < \log_{10} 2 < 0.3010 \). For a more precise calculation, we use \( \log_{10} 2 \approx 0.3010 \):
   \[
   \log_{10} \left( \frac{3^{100} - 1}{2} \right) \approx 47.71 - 0.3010 = 47.409
   \]

6. Finally, the number of digits is:
   \[
   \left\lfloor 47.409 \right\rfloor + 1 = 47 + 1 = 48
   \]

The final answer is \(\boxed{48}\).