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GATE | Gate IT 2005 | Question 78
  • Difficulty Level : Easy
  • Last Updated : 28 Dec, 2015
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Consider the following message M = 1010001101. The cyclic redundancy check (CRC) for this message using the divisor polynomial x5 + x4 + x2 + 1 is :

(A) 01110
(B) 01011
(C) 10101
(D) 10110

Answer: (A)


M = 1010001101
Divisor polynomial: 1.x5 +1.x4+0.x3+1.x2+0.x2+1.x0   
Divisor polynomial bit= 110101
Bits to be appended to message= (divisor polynomial bits – 1) = 5
Append 5 zeros to message bits, modified message: 101000110100000

Now, divide and XOR the message with divisor polynomial bits. Make resultant reminder to 5 bit again and that is the CRC send along with the message.


This explanation has been contributed by Sandeep Pandey.

Please visit the following links to learn more on CRC and its calulation:

Wikipedia article: Cyclic Redundancy Check
GeeksforGeeks article: Error Detection | Computer Networks

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