Problem 1336

To find the number of digits in $2015!$, we use the formula for the number of digits of a factorial, which is given by:

$$\text{Number of digits of } n! = \left\lfloor \log_{10}(n!) \right\rfloor + 1$$

Using Stirling's approximation for $n!$:

$$n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$$

Taking the logarithm base 10 of both sides:

$$\log_{10}(n!) \approx \log_{10}(\sqrt{2 \pi n}) + \log_{10}\left(\left(\frac{n}{e}\right)^n\right)$$

$$\log_{10}(n!) \approx \log_{10}(\sqrt{2 \pi n}) + n \log_{10}\left(\frac{n}{e}\right)$$

$$\log_{10}(n!) \approx \frac{1}{2} \log_{10}(2 \pi n) + n \log_{10}(n) - n \log_{10}(e)$$

$$\log_{10}(n!) \approx \frac{1}{2} \log_{10}(2 \pi) + \frac{1}{2} \log_{10}(n) + n \log_{10}(n) - n \log_{10}(e)$$

For $n = 2015$:

$$\log_{10}(2015!) \approx \frac{1}{2} \log_{10}(2 \pi) + \frac{1}{2} \log_{10}(2015) + 2015 \log_{10}(2015) - 2015 \log_{10}(e)$$

Calculating each term separately:

$$\frac{1}{2} \log_{10}(2 \pi) \approx \frac{1}{2} \log_{10}(6.2832) \approx \frac{1}{2} \times 0.798 \approx 0.399$$

$$\frac{1}{2} \log_{10}(2015) \approx \frac{1}{2} \times 3.304 \approx 1.652$$

$$2015 \log_{10}(2015) \approx 2015 \times 3.304 \approx 6659.36$$

$$2015 \log_{10}(e) \approx 2015 \times 0.434 \approx 874.51$$

Adding these together:

$$\log_{10}(2015!) \approx 0.399 + 1.652 + 6659.36 - 874.51 \approx 5786.901$$

The number of digits is:

$$\left\lfloor 5786.901 \right\rfloor + 1 = 5786 + 1 = 5787$$

The final answer is $\boxed{5787}$