Multiplication of (10101) by (10000) in GF(25) with mod 100101Group of answer choices101111111110101110111
Question
Multiplication of (10101) by (10000) in GF(25) with mod 100101Group of answer choices101111111110101110111
Solution
The multiplication of two binary numbers in GF(2^5) involves a few steps. Here's how you can do it:
-
Convert the binary numbers to their decimal equivalents. In this case, (10101) is 21 and (10000) is 16.
-
Multiply the decimal numbers together. 21 * 16 = 336.
-
Convert the result back to binary. 336 in binary is (101010000).
-
Now, we need to perform the modulo operation with the given binary number (100101). This is equivalent to the decimal number 37.
-
Convert the result from step 3 back to decimal and perform the modulo operation. 336 mod 37 = 14.
-
Convert the result back to binary. 14 in binary is (1110).
So, the result of the multiplication of (10101) by (10000) in GF(2^5) with mod 100101 is (1110).
Please note that this is a simplified explanation and the actual process of multiplication in GF(2^5) involves polynomial multiplication and reduction, which is a bit more complex.
Similar Questions
Which expression is the same as 111 ten, 333 ones, and 999 thousandths??question markChoose 1 answer:Choose 1 answer:(Choice A) (1×10)+(3×1)+(9×1100)(1×10)+(3×1)+(9× 1001 )left parenthesis, 1, times, 10, right parenthesis, plus, left parenthesis, 3, times, 1, right parenthesis, plus, left parenthesis, 9, times, start fraction, 1, divided by, 100, end fraction, right parenthesisA(1×10)+(3×1)+(9×1100)(1×10)+(3×1)+(9× 1001 )left parenthesis, 1, times, 10, right parenthesis, plus, left parenthesis, 3, times, 1, right parenthesis, plus, left parenthesis, 9, times, start fraction, 1, divided by, 100, end fraction, right parenthesis(Choice B) (1×10)+(3×1)+(9×11000)(1×10)+(3×1)+(9× 10001 )left parenthesis, 1, times, 10, right parenthesis, plus, left parenthesis, 3, times, 1, right parenthesis, plus, left parenthesis, 9, times, start fraction, 1, divided by, 1000, end fraction, right parenthesisB(1×10)+(3×1)+(9×11000)(1×10)+(3×1)+(9× 10001 )left parenthesis, 1, times, 10, right parenthesis, plus, left parenthesis, 3, times, 1, right parenthesis, plus, left parenthesis, 9, times, start fraction, 1, divided by, 1000, end fraction, right parenthesis(Choice C) (1×10)+(3×1)+(9×110)(1×10)+(3×1)+(9× 101 )left parenthesis, 1, times, 10, right parenthesis, plus, left parenthesis, 3, times, 1, right parenthesis, plus, left parenthesis, 9, times, start fraction, 1, divided by, 10, end fraction, right parenthesisC(1×10)+(3×1)+(9×110)(1×10)+(3×1)+(9× 101 )
The binary AND operation of the numbers “100101” and “110011” results (in decimal) 100001 2 33 63
The greatest integer which divides the number 101100 −1 is
The remainder obtained when 12345678987654321 is divided by 1001 is
What is the binary representation of the decimal number 25?Group of answer choices11001111001010011010
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.