The number $2^{48}$ - 1 is divisible by two numbers between 60 and 70. What are these numbers?
(a) 61 and 63
(b) 61 and 65
(c) 63 and 65
(d) 63 and 67
(e) 67 and 69
Remember that

$$
a^{4}-1=(a-1)\left(a^{3}+a^{2}+a+1\right)
$$

Therefore, if $a=2^{12}$, we have:

$$
2^{48}-1=\left(2^{12}\right)^{4}-1=\left(2^{12}-1\right)\left(2^{36}+2^{24}+2^{12}+1\right)
$$

and $2^{12}-1=\left(2^{6}+1\right)\left(2^{6}-1\right)=65 \times 63$. The correct option is (c).