Limited Factors

Factoring by DoS gives $(2^{10}-1)(2^{10}+1)$ . Further expanding, $(2^5-1)(2^5+1)(2^{10}+1)$ . We can compute this quickly to get $2^{20}-1=(31)(33)(1025)$ . Luckily for us, $31 \times 33 > 1000$ , and $1025 > 1000$ , which knocks out a lot of difficult casework. Factoring the entire thing gives

$3 \times 5^2 \times 11 \times 31 \times 41.$

The tricky part is counting. We can isolate factors and construct a list. Factors that are divisible by $3$ : $3, 15, 33, 75, 93, 123, 165, 465, 615, 825$ . Factors divisible by $5$ that aren't already accounted for: $5, 25, 55, 155, 205, 275, 775$ . Factors divisible by $11$ that aren't already accounted for: $11, 341, 451$ . Last but not least, we have just $1$ , $31$ , and $41$ . Thus, there are $\boxed{23}$ integers less than $1000$ that divide $2^{20}-1$ .

Seeing no clever way of doing this I used the Python sympy library:

Using Java Netbeans Programmer..

This Is the Programming screen

Output Screen

And Output Is 23..

Here is my solution:

${ 2 }^{ 20 }-1=1048575=(3)({ 5 }^{ 2 })(11)(31)(41)$

Now we can count the number of factors from the prime factorization, but we must remember: our factors must be less than 1000. Generating our factors less than 1000, we have:

$\boxed{ 1, 3, 5, 11, 15, 25, 31, 33, 41, 55, 75, 93, 123, 155, 165, 205, 275, 341,}$ $\boxed{451, 465, 615, 775, 825}$

Java solution -

int n = 0;
    for(int i = 1; i < 1000; i++)
        if((Math.pow(2,20) - 1) % i == 0) n++;
System.out.println(n);

I did it using java code

1, 3, 5, 11, 15, 25, 31, 33, 41, 55, 75, 93, 123, 155, 165, 205, 275, 341, 451, 465, 615, 775, 825 -- in all 23 factors.