Decimal: 3 2 5 Binary: 0011 0010 0101 so 325 = 0011 0010 0101
87
A 0, 1 system using: 5, 2, 1', 1 instead of 8, 4, 2, 1 to count binary numbers. Example: 0000 0001 0010 0101 0100 0101 1001 1100 1101 1111
It's pretty much just like multiplying in base-10 by hand, but you will not have any 'carrying over' to do, since the only possibilities are 0 & 1: 0x0 = 0, 0x1=0, 1x0=0, 1x1=1 (except when you add up the column of numbers after multiplying). An example: Five times six = thirty. So Five is 101 and Six is 110. 00101 x0110 ----- 00000 0101 101 -------- 11110 ---> in base-10: 16 + 8 + 4 + 2 = 30. that did not have any carry overs, but if you had to add two or more ones (1+1=10), then a one would carry over to the next column.
Since 16 is a power of 2, you can directly convert every hexadecimal digits to four binary digits. Look up the equivalent in a table, if you don't know it by heart. Don't forget the zeroes at the left. For example, to convert 3F5(hex), 3 = 0011, F = 1111 and 5 = 0101, so 3F5(hex) = 0011 1111 0101 (binary). In this example you may get rid of the first two zeroes, depending on the application.
What is the product of the binary numbers 0101 and 0101?
0101 hours
11001
11001
It is 13 5 5.
1110 0101 1101 1011 is E5DB
strange considering its my computer you're asking about.
0101
1:01 am
0.1012 = 0.010201
0101
CodeMale Codes 1Position #2330017A8 0001 330017AA 00012Position #3330017B0 0001 330017B2 00013Position #4330017B8 0001 330017BA 00014Position #5330017C0 0001 330017C2 00015Position #6330017C8 0001 330017CA 00016Position #7330017D0 0001 330017D2 00017Position #8330017D8 0001 330017DA 00018Position #9330017E0 0001 330017E2 0001Female Codes 9Position #2830017A8 0101 330017AA 000110Position #3830017B0 0101 330017B2 000111Position #4830017B8 0101 330017BA 000112Position #5830017C0 0101 330017C2 000113Position #6830017C8 0101 330017CA 000114Position #7830017D0 0101 330017D2 000115Position #8830017D8 0101 330017DA 000116Position #9830017E0 0101 330017E2 0001(17 codes total)