The number `1035` is the smallest integer that exhibits a non frequent property: one its multiples, `3105 = 1035 * 3`, has its same digits but in different order, in other words, `3105`, is one of the permutations of `1035`.
The number `125874` is the first integer that has this property when the multiplier is `2`, thus: `125874 * 2 = 251748`
Make the function `search_permMult()`, that receives an upper bound, nMax and a factor k and will output the amount of pairs bellow nMax that are permuted when an integer of this range is multiplied by `k`. The pair will be counted if the multiple is less than `nMax`, too
Let'see some cases:
```python
search_permMult(10000, 7) === 1 # because we have the pair 1359, 9513
search_permMult(5000, 7) === 0 # no pairs found, as 9513 > 5000
search_permMult(10000, 4) === 2 # we have two pairs (1782, 7128) and (2178, 8712)
search_permMult(8000, 4) === 1 # only the pair (1782, 7128) 
search_permMult(5000, 3) === 1 # only the pair (1035, 3105)
search_permMult(10000, 3) === 2 # found pairs (1035, 3105) and (2475, 7425)
```
Features of the random Tests:
```
10000 <= nMax <= 100000
3 <= k <= 7
```
Enjoy it and happy coding!!
def search_permMult(n_max, k):
    return sum(1 for n in range(1, n_max // k) if sorted(str(n)) == sorted(str(n * k)))
    
    
    
    
    #pairs = 0
    #for n in range(1, n_max // k):
    #    if sorted(str(n)) == sorted(str(n * k)):
    #        pairs = pairs + 1
    #return pairs

def search_permMult(n_max, k):
    return sum(1 for n in range(1, n_max // k) if sorted(str(n)) == sorted(str(n * k)))
    
    
    
    
    #pairs = 0
    #for n in range(1, n_max // k):
    #    if sorted(str(n)) == sorted(str(n * k)):
    #        pairs = pairs + 1
    #return pairs
