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
The correct option is (c).

The number $N=2^{48}-1$ is very large, but we can still find many of its divisors. For this, we use the equality

$$
(x-1)\left(x^{a-1}+x^{a-2}+\cdots+x+1\right)=x^{a}-1
$$

We notice that the divisors of 48 are $1,2,3,4,6,8,12,24$ and 48. Taking $x=2^{2}$ and $a=24$ in the above equality, we establish that $x-1=2^{2}-1=3$ is a divisor of $\left(2^{2}\right)^{24}-1=2^{48}-1=N$. Similarly, taking $x=2^{3}$ and $a=18$, we establish that $x-1=2^{3}-1=7$ is another divisor of $\left(2^{3}\right)^{18}-1=2^{48}-1=N$. Proceeding in this manner, we verify that for any divisor $d$ of 48, the number $2^{d}-1$ is a divisor of $N$. Thus, we conclude that $2^{2}-1=3, 2^{3}-1=7, 2^{4}-1=17, 2^{6}-1=63, 2^{8}-1=251$, and so on, are divisors of $N$.

In addition to these, we can obtain other divisors of $N$ by considering the even divisors $d=2n$ of 48 and using the notable product

$$
2^{d}-1=\left(2^{n}\right)^{2}-1=\left(2^{n}+1\right)\left(2^{n}-1\right)
$$

Since $2^{d}-1$ is a divisor of $N$, $2^{n}+1$ is also a divisor of $N$. For example, $d=4=2 \times 2$ provides the divisor $2^{2}+1=5$ of $N$, $d=6=2 \times 3$ provides the divisor $2^{3}+1=9$ of $N$, $d=8=2 \times 4$ provides the divisor $2^{4}+1=17$ of $N$, $d=12=2 \times 6$ provides the divisor $2^{6}+1=65$ of $N$, and so on.

Note that we have already obtained two divisors of $N=2^{48}-1$ between 60 and 70, namely, 63 and 65.

Observation: With the help of a computer, we can see that $N$ is indeed a very large number, as $N=2^{48}-1=281474976710655$, and obtain its factorization, given by

$$
N=2^{48}-1=3^{2} \times 5 \times 7 \times 13 \times 17 \times 97 \times 241 \times 257 \times 673
$$

Using the argument presented in the solution to this exercise, it is possible to find many divisors of quite large numbers.