Question

The numbers \(2^{2024}\) and \(5^{2024}\) are expanded and their digits are written out consecutively on one page. The total number of digits written on the page is:

• A. \(1987\)
• B. \(2025\)
• C. \(2065\)
• D. \(2000\)

Hint:

To calculate the number of digits of large powers, use logarithms. The number of digits is \( \lfloor \log_{10} (n) \rfloor + 1 \).

Answer 1

The Correct Option is B

Step 1: Use logarithms to estimate the number of digits. The number of digits \(d\) in a number \(n\) is given by the formula: \[ d = \lfloor \log_{10} n \rfloor + 1. \] For the number \(2^{2024}\), we have: \[ \log_{10} (2^{2024}) = 2024 \log_{10} 2 \approx 2024 \times 0.3010 = 609.224. \] Thus, the number of digits in \(2^{2024}\) is: \[ d_2 = \lfloor 609.224 \rfloor + 1 = 610. \] Similarly, for \(5^{2024}\): \[ \log_{10} (5^{2024}) = 2024 \log_{10} 5 \approx 2024 \times 0.6990 = 1414.776. \] Thus, the number of digits in \(5^{2024}\) is: \[ d_5 = \lfloor 1414.776 \rfloor + 1 = 1415. \] Step 2: Add the digits of both numbers. The total number of digits is: \[ d_{\text{total}} = d_2 + d_5 = 610 + 1415 = 2025. \] Thus, the total number of digits written on the page is 2025.