Let $t$ be TNYWR.

Suppose that

$$
\frac{1}{2^{12}}+\frac{1}{2^{11}}+\frac{1}{2^{10}}+\cdots+\frac{1}{2^{t+1}}+\frac{1}{2^{t}}=\frac{n}{2^{12}}
$$

(The sum on the left side consists of $13-t$ terms.)

What is the value of $n$ ?
Multiplying both sides of the given equation by $2^{12}$, we obtain

$$
1+2^{1}+2^{2}+\cdots+2^{12-(t+1)}+2^{12-t}=n
$$

Since the answer to (b) is 6 , then $t=6$ and so we have

$$
1+2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}=n
$$

Therefore, $n=1+2+4+8+16+32+64=127$.

ANSwer: $3,6,127$