code: def smallest_prime_factor(Q):
      
  
      A = Q;
      while(A != 1):
          if(np.mod(A, 2) == 0):
              A = A / 2
          elif(np.mod(A, 3) == 0):
              A = A / 3
          elif(np.mod(A, 5) == 0):
              A = A / 5
          elif(np.mod(A, 7) == 0):
              A = A / 7;
          else:
              A = Q + 1;
              Q = A;
  
      return Q
docs: Find the smallest number factorable by the small primes 2, 3, 4, and 7 
      that is larger than the argument Q