Forgive me if my question is slightly incorrect.
I once saw a note, where it was stated that to compute $\tan\left(10^{100}\right)$ (by hand, without calculator) we need to know the $100$-th digit of $\pi$.
Is it true, and if it indeed is, how is this $100$-th number related to the task?
On
Suppose you want to find $\tan(1000)$ by hand. Since you're Feynman and you're really smart you can do this fairly accurately for numbers in the interval $(-\frac \pi 2, \frac \pi 2)$ but since the tangent function is periodic you can reduce it to that case by using that $1000-318\pi=0.973 \dotsb$ but if your approximation of $\pi$ were of by $10^{-3}$, say $3.141$ you would get $1000-318\times 3.141 = 1.161$ and since the error in the ratio of $\tan(1.161)$ and $\tan(0.973)$ is more than 10% you lost your bet.
On
The digits needed are actually the 101st digit and 102nd digit after the decimal point of $1\over\pi$. These two digits happen to be "12" (courtesy of Wolframalpha) so $\tan\left(10^{100}\right)\approx\tan(0.12\pi)$.
To see why this works, let's follow the entire calculation:
Fetching the first 102 digits of $1\over\pi$ from Wolframalpha, we get:
0.318309886183790671537767526745028724068919291480912897495334688117793595268453070180227605532506171912... (note the "12" in the end)
So $10^{100}\over\pi$ is:
3183098861837906715377675267450287240689192914809128974953346881177935952684530701802276055325061719.12...
Which we will write as $N+0.12...$ (where $N$ is an integer).
Now we can write:
$\tan\left(10^{100}\right)=\tan\left[\left(10^{100}\over\pi\right)\cdot\pi\right]\approx\tan\left[\left(N+0.12\right)\cdot\pi\right]=\tan\left(N\pi+0.12\pi\right)$
And since $\tan(x)$ has a period of $\pi$, we get:
$\tan(N\pi+0.12\pi)=\tan(0.12\pi)$
Which can further be approximated as:
$\tan(0.12\pi)\approx\tan(0.38)\approx0.38$ (the last step uses the approximation of $\tan(x)\approx{x}$ for small $x$)
A user in the comments answers this in the following question here.
To quote
edit: link to the answer is here