How many bits does 2 to the 100 have if written in base 2?

Not sure how to give a hint without blatantly giving the answer but...

Consider an n - digit number in base b.

That is N=an−1an−2a0=∑k=0akbk
N
=
a
n
−
1
a
n
−
2
.
.
.
.
.
a
0
=
∑
k
=
0
a
k
b
k
Note aka
k
<
b
so we can easily show NN
<
b
n
(may have to repeat and argue inductively.

And presumably to be n - digit than an−1≠0
a
n
−
1
≠
0
so N≥bn−1
N
≥
b
n
−
1
.

So we have: every n digit number is between bn−1
b
n
−
1
inclusively and bn
b
n
exclusively. This should be blindingly obvious to us if b=10
b
=
10
.

So... that's a really important and fundamental result. Remember and use it.