Just for fun!-XV

It's the same method as the one posted by Raj. But for those who are new to these kind of problems I'd like to post a more detailed solution.

At first, let's take a look at some explicit examples :

Smallest $1, 2, 3 \text{ \& }4$ digit positive integer are 1, 10, 100 \text { &} \, 1000 respectively. Now, $\log1=0$ $\log10=1$ $\log 100=2$ $\log1000=3$ Notice that, for $n$ digits it's always $(n-1)$ . As, $\log x$ always increases as $x$ increases towards positive infinity, it's always true that, for $x$ to be a $n$ digit positive integer, $n-1\leq \log x<n$ $\Rightarrow \lfloor \log x \rfloor =n-1$ $\Rightarrow n=\lfloor \log x \rfloor+1$

Here, $\lfloor. \rfloor$ represents the floor function. $\lfloor x\rfloor$ is the greatest integer not greater than $x$ . For example, $\lfloor 3.98\rfloor =3$ & $\lfloor 4\rfloor=4$

Thus, for this question, $n=\lfloor \log2^{100} \rfloor+1$ $\Rightarrow n=\lfloor 100 \log 2\rfloor+1$ $\Rightarrow n=\lfloor 100\times 0.3010\rfloor+1$ $\Rightarrow n=\lfloor 30.1 \rfloor +1=30+1=\color{#0C6AC7} {\boxed {31}}$

log 2¹⁰⁰ = 100 log 2 = 30.1 ... ... ... => 2¹⁰⁰ = 10^30.1 ... ... ... => 31 digit

Did it mentally