I didn't knew that but the solution made sense and I didn't have any other idea so I just assumed it. If it would have failed, it would have failed. What I'm saying is sometimes you have to go with your gut instead of proving whether the solution will work. Atleast that's what I think.

You can argue that the logic wouldnt TLE without knowing the exact number as 1344. Since the number of divisors of n = 2^a 3^b 5^c ... is given by (a+1)*(b+1)*(c+1)..., we can get a very rough estimate for the upper bound of the number of divisors as

max divisors = (log_2(10^9)+1) * (log_3(10^9)+1) * (log_5(10^9)+1) * ...
             = 33 * 20 * 13 * 9 * ...
             < 10^5

This is not something that you "prove". It is something that you intuitively know. When you unpack this statement it doesn't actually say much but the math makes it kind of convoluted so I'll explain it. To have some x*y be divisible by a*b, the set of prime divisors in a*b should be a subset of prime divisors in x*y. In other words, for every prime divisor in a*b, that prime divisor should also exist in x*y. Now if you think about it, gcd(a*b, x) simply means the set of prime divisors in x, which are also prime divisors in a*b. This means that the y that you choose only has to contain the prime divisors that is in a*b but not in x. This is given by the expression u gave ((x*y)/gcd(a*b, x)).

Both x and y should together make up a*b
=> so x should have some part of a and some part of b. 
=> so y should have some part of a and some part of b.

This some part which is I mentioned above is nothing but "factor of a given number"
Now x and y can be written as
x = afact1 * bfact1 * k
y = afact2 * bfact2 * k

Here k is some constant, afact1 and afact2 is factorials of a, bfact1 and bfact2 are factorials
of b and also note that a = afact1 * afact2 = a, b = bfact1*bfact2

so fix afact1 and bfact1 we get afact2 = a/afact1 and bfact2 = b/bfact1

Отлично)

Hi, your solution is almost correct... When choosing which indices to apply the operation on, you went from N to 1, but a bigger number doesn't necessarily mean a bigger number of twos. Take for example 5(0) and 4(2). So instead of going from N to 1, you should sort the indices by their number of twos and go from the biggest to the smallest. Hope this helps <3

I'm a bit late, but you just say that since $$$a \lt x \leq c$$$ and $$$g$$$ divides $$$x$$$, we just check if there's a multiple of $$$g$$$ that is between $$$a$$$ and $$$c$$$. This can be done similarly to the $$$y$$$ check: the largest $$$x$$$ possible is $$$\lfloor\frac{c}{g}\rfloor \cdot g = x_{max}$$$, so check if $$$a \lt x_{max}$$$.

You might think that by choosing $$$x_{max}$$$ this way the gcd could change: $$$gcd(ab, x_{max}) \gt g$$$ and that's true, but it doesn't actually matter: we assume that $$$g$$$ is the gcd, then we find some $$$y = \frac{ab}{g} \cdot k$$$, where $$$k$$$ is some constant, but since the real $$$gcd(ab, x_{max})$$$ is greater than $$$g$$$, that means that our $$$y$$$ is greater than it needed to be, which certainly doesn't hurt us (because we will check the bigger gcd at some point later on).