What's the first digit of 2410^2410?

Question

The first digit means the left most digit. 2410 is just an example and it can be replaced by any other numbers. Can any one help me to solve it?

Answer 1

I'll try without computer as exposed here (the idea is a combination of Prahlad and Lucian's suggestiond) but it's not really easy... (with $\,\log(x)=\log_{10}(x)=\dfrac{\ln(x)}{\ln(10)}$ and since $\,\dfrac 1{\ln(10)}\approx 0.43$) : \begin{align} 2410\cdot\log(24.1)&\approx 2410\cdot(\log(3)+3\log(2)+\log\left(1+\frac1{240}\right))\\ &\approx 2410\cdot(0.47712+3\cdot 0.30103+0.43/240)\\ &\approx 2410\cdot (0.47712+0.90309+0.0018\\ &\approx 2410\cdot 1.382\\ &\approx 241\cdot 13.82\\ &\approx 241\cdot 13+241\cdot 0.82\\ &\approx \text{####}.62\\ \end{align} The first digit is obtained by computing $10^{\text{decimal part}}\approx 4$ (since $\log(4)\approx 0.60206$ and $\log(5)\approx 0.69897\approx 1-0.30103$).

(I had some luck here since the exact computation of $2410\log(24.1)= 3330.661072\cdots$ with $10^{0.661072\cdots}= 4.58218\cdots$)

Answer 2

Using a calculatior we get that $\log_{10} (2410^{2410})=8150.66107260543188\ldots$.

Thus, $2410^{2410}$ has $8151$ digits, and the first one is $4$, since $10^{0.66107260543188}=4.582184853626742\ldots$.

In general, if we have a huge number $M$, then its first digit is the same as the first digit of $10^{\log_{10} M-\lfloor\log_{10} M\rfloor}$, where $\lfloor\cdot\rfloor$ is the integer part.

Incidentally, the second digit is 5, the third 8, the fourth 2 etc...