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  2. Mathematics
  3. Number Theory
  4. Equations
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Equations

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  • nayeemjoy59
     + 3 comments

    Let's take a look here ,

    (1/x) + (1/y) =(1/n!) number of solutions

    let's n! as p okk ? Now ..

    (1/x) + (1/y) =(1/p)

    (x+y)/(xy) = (1/p)

     px + py =xy

- px -py +xy=0

 p^2 - px -py +xy =p^2   [ adding p^2 both sides ]

 p(p-x) -y(p-x)=p^2

 (p-y) (p-x) =p^2

 Now , think (p-y) as A and (p-x) as B

 So ,   A*B = p^2 

 Now , if anybody asks u the number of solutions of this solution .
 Then It will make U common sense that the number of divisors will be unique solutins Because Look At this . Let's think p^2 =36

 So solutions are (value of A and B) 

 a * b

 1*36=36
 2*18=36
 3*12=36
 4*9=36
 6*6=36 


 a * b

 36*1=36
 18*2=36
 12*3=36
 9*4=36

 Total 9 unique solutions . 

 Now , put n! in p So , U have to find out the number of divisors of (n!)^2 . That's all !!! 

 Hope it helps 

 My github link

    https://github.com/joy-mollick/Problem-Solving-Solutions-Math-Greedy-

  • TChalla
     + 0 comments
    #!/bin/python3

import os
import sys

# n = N!
# x + y = xy / n
# xn + yn = xy
# xy - xn - yn + n^2 = n^2
# (x-n)(y-n) = n^2
# XY = n^2
# answer is number of factors of N!^2

# Complete the solve function below.
def count(n, p):
    count = 0
    n_copy = n
    while n_copy > 1:
        count += n_copy // p
        n_copy //= p
    return count

def solve(n):
    pr = 1
    isprime = [True] * (n + 1)
    for num in range(2, n + 1):
        if not isprime[num]: 
            continue
        pr = pr * (2 * count(n, num) + 1) % 1000007
        for multiple in range(2 * num, n + 1, num):
            isprime[multiple] = False
    return pr

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')
    n = int(input())
    result = solve(n)
    fptr.write(str(result) + '\n')
    fptr.close()

  • xdavidliu
     + 1 comment

    hint: in order to see the "divisors of n!^2" thing, do not complete the partial fractions, e.g. 1/x + 1/y = xy/(x+y). that doesn't help all. Instead, solve for x.

    also, from the editorial, the proof that y-n! cannot be negative is kind of glossed over and slightly nontrivial. It can be done by starting from x>0 and showing that y-n!<0 would lead to a contradiction.

  • hungthai
     + 1 comment

    • The key point is that the number of positive integral solutions of the given equation is equal to the number of factors of (n!)^2. (Hint: rearrage the equation. You may need to check out Diophantine Equation)

    • Another important point is to compute the number of factors of (n!)^2 with a good time complexity. It turns out that you could build each factor from the more fundamental element: the prime factors of (n!)^2. (Hint: determine all the prime factors of (n!)^2, which is all the prime numbers not greater than n, and their number of appearances)

    • Once you have all the prime factors of (n!)^2 and their number of appearances, the last step is just a simple Combinatorial Counting Problem.

  • leewangzhong
     + 3 comments

    Because I personally think that small answers help clarify "what is the question", which is important. These answers shouldn't give a hint about the solution. f(0) = 1; f(1) = 1; f(2) = 3; f(3) = 9

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