Example:
1011x0111=1001101
(11x7=77)
break 0111 down into 0001+0010+0100:
1011*0001=001011
1011*0010=010110 (just add a zero to the end)
1011*0100=101100 (just add two zeros to the end)
it's just like multiplying by 1, 10, and 100 in decimal. Then add them together.
101100
+10110
+01001
=1001011
When adding, two 1's carry over to the next digit... for instance...1+1=10, 1+11=100, 100+110=1010, etc.
2007-06-19 00:29:41
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answer #1
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answered by CJF16 3
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example:
1011*0111=1001101
(11*7=77)
breaking 0111 down into 0001+0010+0100:
1011*0001=001011
1011*0010=010110 (just add a zero to the end)
1011*0100=101100 (just add two zeros to the end)
it's just like multiplying by 1, 10, and 100 in decimal. Then add them together.
101100
+10110
+01001
=1001011
When adding, two 1's carry over to the next digit... for instance...1+1=10, 1+11=100, 100+110=1010, etc.
2007-06-19 05:40:42
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answer #2
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answered by Flyer 4
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Example:
21*29
10101*11101
Since you have two 5-bit numbers, the answer will be 10 or 11 bits, so set up for 12 bits. Flip the two numbers as you would in decimal to make an easier problem:
0000 0001 1101
0000 0001 0101
------- ------- -------
0001 1101 0000
0000 0111 0100 +
------- ------- -------
0010 0100 0100
0000 0001 1101 +
------- ------- --------
0010 0110 0001 = 512 + 64 + 32 + 1 = 609
Listing the products before adding just makes it unnecessarily difficult.
2007-06-19 06:45:51
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answer #3
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answered by Helmut 7
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I would just go ahead and put them into base-10 and multiply them and then change them back into binary.
2007-06-19 05:20:17
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answer #4
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answered by FullMetal 2
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