# Finding the first digit of $2015^{2015}$

Can anyone help me find the first digit of $2015^{2015}$?

It is easy to find the last digit but I have no idea for the first digit.

• Use logarithms to approximately calculate the result. – Mario Carneiro Feb 6 '15 at 14:19
• Is this a question from an on-going contest? – Joel Reyes Noche Mar 17 '15 at 4:20
• Similar question now on Puzzling: First digit of 3^2020 – Glorfindel Apr 16 '20 at 13:55

## 3 Answers

Now, you didn't say "by hand" as these calculations often do, but I'll assume that at least you're from before 1950 so that solutions like just asking Wolfram Alpha about a mere 8000-digit number are beyond you.

We can calculate $\log_{10}(2015^{2015})=2015(\log_{10}(2.015)+3)$, and consulting my trusty tables I find that $$\log_{10}(2.015)\approx\frac{\log_{10}(2.01)+\log_{10}(2.02)}2\approx\frac{0.30320+0.30535}2\approx0.30428,$$ so $\log_{10}(2015^{2015})\approx6658.1$. (I need to do this calculation well enough to show it is between $6658$ and $6658+\log_{10} 2\approx6658.3$ to prove the claim.)

Then, $6658\le\log_{10}(2015^{2015})<6658+\log_{10} 2$ gives $10^{6658}\le2015^{2015}<2\cdot 10^{6658}$, so the first digit is $1$.

• $1$ is the most likely leading digit anyway :) – Hagen von Eitzen Sep 23 '15 at 18:29

The units digit is of course a 5.

For the leading digit, see that $n^n = 10^{n\cdot log_{10}(n)} \approx 10^{2015\cdot3.304275} \approx 10^{6658.114} = 10^{6658}10^{0.114} \approx 1.3008\cdot 10^{6658}$. Therefore, the first digits of your number are 13008...

• I wasn't sure which digit the OP was referring to, so I wrote about both. Usually to avoid confusion I call them "leading" and "units" digit. – Luigi D. Feb 6 '15 at 14:45

Take the base-ten logarithm.
By the log laws, that will equal $2015\log2015$. Suppose that is $x+y$, where $x$ is a whole number and $0<y<1$. The number has $x+1$ digits.
$y$ tells you the first digit. Or rather, $10^y$ starts with the same digits $2015^{2015}$ does.