Product of integers with N and M digits 
    N = 2, M = 3:  10 * 100   =    1000,   D = 4 = [N + M - 1] 
	N = 2, M = 3:  99 * 100   =    9900,   D = 4 = [N + M - 1]
	N = 2, M = 3:  10 * 999   =    9990,   D = 4 = [N + M - 1]
	N = 2, M = 3:  99 * 999   = 98901,   D = 5 = [N + M]
 
	N = 2, M = 4:  10 * 1000 =    10000,  D = 5 = [N + M - 1] 
	N = 2, M = 4:  99 * 1000 =    99000,  D = 5 = [N + M - 1]
	N = 2, M = 4:  10 * 9999 =    99990,  D = 5 = [N + M - 1]
	N = 2, M = 4:  99 * 9999 = 989901,  D = 6 = [N + M]
 
	Thus: the answer is N+M-1 and N+M.
	
	  
	    Note: 
	  3^4 = 81  
	  3^5 = 243  
	  3^6 = 729  
	  3^7 = 2187  
	  3^8 = 6561  
Thus: 
         3^n = 1 if remainder(n/4) = 0 
               = 3 if remainder(n/4) = 1 
               = 9 if remainder(n/4) = 2 
               = 7 if remainder(n/4) = 3 
 
	  Similarly, last digit of 4^n is 6 is n is an even number and 4 if n is an odd number.
	  
 
Similarly, last digit of 7^n 
	  
 
Thus: 
         7^n = 1 if remainder(n/4) = 0 
               = 7 if remainder(n/4) = 1 
               = 9 if remainder(n/4) = 2 
               = 3 if remainder(n/4) = 3