# First decimal digit of a very large number. [duplicate]

Find the first digit (the left one) of the number $2016^{2016}$, not by actually compute it. I know the solution is 7, thanks to Wolfram Alpha's power, but I did not succeeded in finding it.

Question number two: how may i calculate log values used in solving this?

• Pardon for the question. I didn't find it. Thanks – Blumer Jan 19 '17 at 18:59
• Its ok. You can edit your question with some new doubts. Like how to calculate log values used in solving this. – Kanwaljit Singh Jan 19 '17 at 19:00
• And give reference to that question too. – Kanwaljit Singh Jan 19 '17 at 19:01
• math.stackexchange.com/questions/1136486/… – Kanwaljit Singh Jan 19 '17 at 19:01

The leftmost digit of a positive integer $x$ is $d$ if the fractional part of $\log_{10}(x)$ is in the interval $[\log_{10}(d), \log_{10}(d+1))$.
Lin this case $\log_{10}(2016^{2016}) = 2016 \log_{10}(2016) \approx 6661.8529$. The fractional part $0.8529\ldots$ is between $\log_{10}(7) \approx .8451$ and $\log_{10}(8) \approx 0.9031$, so the first digit is indeed $7$.