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2007-06-18 22:12:43 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

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 · answer #1 · answered by CJF16 3 · 1 0

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 · answer #2 · answered by Flyer 4 · 1 1

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 · answer #3 · answered by Helmut 7 · 1 1

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 · answer #4 · answered by FullMetal 2 · 0 2

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