Question

The values of the numbers \(2^{2004}\) and \(5^{2004}\) are written one after another. How many digits are there in all?

• A. 4008
• B. 2003
• C. 2005
• D. None of these

Hint:

Use the logarithmic formula \(\lfloor \log_{10} N \rfloor + 1\) to calculate number of digits for large exponents.

Answer 1

The Correct Option is C

We are given: Write the values of \(2^{2004}\) and \(5^{2004}\) one after the other. Now observe: \[ 2^{2004} \times 5^{2004} = (2 \times 5)^{2004} = 10^{2004} \] So, the product of the two numbers is \(10^{2004}\), which is a 1 followed by 2004 zeroes — total \(2005\) digits. Now we need to find how many digits are there in \(2^{2004}\) and \(5^{2004}\) separately. Use the digit formula: \[ \text{Number of digits of } N = \lfloor \log_{10} N \rfloor + 1 \] \[ \log_{10} (2^{2004}) = 2004 \log_{10} 2 \approx 2004 \times 0.3010 = 603.60 \Rightarrow \text{Digits in } 2^{2004} = 604 \] \[ \log_{10} (5^{2004}) = 2004 \log_{10} 5 \approx 2004 \times 0.6990 = 1400.39 \Rightarrow \text{Digits in } 5^{2004} = 1401 \] Total digits = \(604 + 1401 = 2005\) But wait — the question says digits of both are "written one after another", which gives us: \[ 604 + 1401 = \boxed{2005} \]