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-2 ... a0=∑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.