# Task
 Let's consider a table consisting of `n` rows and `n` columns. The cell located at the intersection of the i-th row and the j-th column contains number i × j. The rows and columns are numbered starting from 1.

 You are given a positive integer `x`. Your task is to count the number of cells in a table that contain number `x`.

# Example

 For `n = 5 and x = 5`, the result should be `2`.
 
 The table looks like:
 ```
   1  2  3  4  (5)
   2  4  6  8  10
   3  6  9  12 15
   4  8  12 16 20
 (5) 10 15 20 25```
 There are two number `5` in it.
  
 For `n = 10 and x = 5`, the result should be 2.
 
 For `n = 6 and x = 12`, the result should be 4.
 ```
   1  2   3   4   5   6
   2  4   6   8   10 (12)
   3  6   9  (12) 15  18
   4  8  (12) 16  20  24
   5  10  15  20  25  30
   6 (12) 18  24  30  36
 ```
 
# Input/Output


 - `[input]` integer `n`

  `1 ≤ n ≤ 10^5.`


 - `[input]` integer `x`

  `1 ≤ x ≤ 10^9.`


 - `[output]` an integer

  The number of times `x` occurs in the table.

Difficulty: introductory

Starter code:

def count_number(n, x):
	

Solutions:
def count_number(n, x):
    return len([j for j in range(1,n+1) if x%j==0 and x/j <= n]) 

from math import ceil

def count_number(n, x):
    return sum(x % i == 0 for i in range(max(int(ceil(x / n)), 1), min(n,x) + 1))

def count_number(n, x):
    #your code here
    return sum(1 for i in range(1,n+1) if x//i in range(1,n+1) and x%i==0) 


count_number=lambda n,x:sum(x%e==0for e in range(1,n+1)if e*n>=x)

def count_number(n, x):
    return sum(range(i, i * n + 1, i).count(x) for i in range(1, n+1))

from math import ceil
def count_number(n, x):
    return sum(x % a == 0 for a in range(ceil(x / n), n + 1))

def count_number(n, x):
    k = [1, x // n][x > n]
    return sum(1 for i in range(k, n + 1) if x/i == x//i and x/i in range(1, n + 1))

count_number=lambda n,x:sum(x-n*r<=x%r<1for r in range(1,n+1))

from math import ceil

def count_number(n, x):
    return sum((x%i == 0) and (x/i != i) + 1 for i in range(ceil(x/n), int(x**0.5)+1))

count_number=lambda n,x: (lambda r: len([i for i in range(1,int(r)+1) if x%i==0 and x/i<=n])*2-(1 if r%1==0 and r<=n else 0))(x**0.5)