Problem on base-two numeral Can anyone help me on this? I don't know where to start.

A base-two numeral consists of 15 digits, all of which are ones. When tripled and written in base two, how many digits does this number contain?

 A: HINT: You could simply work it out longhand, doing the arithmetic in base two: when you triple the number, you’re multiplying by $11_{\text{two}}$. 
Alternatively, you could notice that
$$111,111,111,111,111_{\text{two}}=1,000,000,000,000,000_{\text{two}}-1=2^{15}-1\;,$$
so three times this number is 
$$3\left(2^{15}-1\right)=3\cdot 2^{15}-3=(2+1)2^{15}-3=2^{16}+2^{15}-3\;.$$
When you write $2^{16}+2^{15}$ in base two, you get
$$11,000,000,000,000,000\;.$$
When you subtract $3$ (i.e., $11_{\text{two}}$), what do you get? How many digits does it have? (You should actually be able to answer the last question even without actually doing the subtraction.)
A: A base two number x multiplied by three is the same as 2*x + x.
In base two, multiplying by 2 simply adds a 0 to the end of the number, just like how multiplying by 10 acts in base 10.
When we add these numbers, we are only concerned about the number of digits of the result making it only necessary to compute that 11 + 1 = 100 (base two) to see how many digits get carried. This is followed by 14 digits that are irrelevant so we can conclude that the final sum must have 17 digits.
