130 = 128+2 = 1000 0010 17 = 16+1 = 0001 0001 70 = 64+4+2 = 0100 0110 83 = 64+16+2+1 = 0101 0011So the binary equivalent of 130.17.70.83 is the 32-bit number:
1000 0010 0001 0001 0100 0110 0101 0011
1000 0000 0000 1010 0000 0010 0000 0011 (destination address) 1111 1111 1111 1111 0000 0000 0000 0000 (address mask)The boundary between the 1's and 0's in the mask can be used to signify that the prefix/suffix boundary in the destination address is that same location.
N==(D&M)