Well I’ll just give you 2 cookies, beacuse you didn’t explain the formula. That makes three cookies for a single problem, you should be more than happy I’ll demonstrate it in your place.
We have a number of prime divisors, (namely 2, 3, 5 and 7), each wth his own exponent, and to form a divisor, we can combine any available power of those divisors freely. We can take 2^3, 3 and 5 to form 835=120, take only 2^7 and form 128, and so on.
To form all possible divisors, we just have to multiply all possibilities for each prime divisor, and if that divisor has n as its maximum exponent, that means we have n+1 choices for it. (IE for 5^2, we can ignore it and take 5^0, take a single 5, or take both and get 5^2=25, three total choices). Multiplying all choices (that are completely independent) we get (8+1)(4+1)(2+1)(1+1) = 9532 = 270 total divisors. (Notice how it is completely irrelevant from the kind of prime divisors we have)
It is indeed mathematically impossible to do it the standard way, because there are 4 points from which an odd number of lines is leaving. (And the maximum number to make it possible is 2). So, we could draw extra lines here and there to help us, perhaps over another paper so we don’t ruin the drawing. Is it so?
But if you didn’t say not to fold the paper, I would have immediately thought of it first