CRC 10-bit sequence: Compute the checksum for 1010011110 with divisor 1011 | AI Research

CRC (10-bit sequence): Compute the checksum for 1010011110 with divisor 1011

Verified Sources

Nov 4, 2026

A cyclic redundancy check (CRC ) appends a redundancy (CRC remainder ) to the original data so that the resulting bit string is divisible by the generator (generator polynomial). For a binary divisor of length $m$ , the CRC remainder has length $m-1$ bits, because the generator degree is $m-1$ .

Given:

Data (10 bits): $1010011110$
Divisor/Generator (length 4): $1011$
Divisor length $m=4 \Rightarrow$ remainder length $m-1=3$ bits

So we:

Append $3$ zeros to the data.
Perform modulo-2 long division by $1011$ .
Use the final remainder as the CRC.
“Check your answer” by verifying that data + CRC is divisible by $1011$ under modulo-2 division.

Mermaid view of the workflow:

CRC calculation (generator polynomial / modulo-2 division) walkthrough

Compute CRC via modulo-2 long division

1
Step 1
Divisor is 1011 with length $m=4$ , so CRC remainder length is $m-1=3$ bits.
2
Step 2
Append $3$ zeros to $1010011110$ to form: $1010011110\,000$ (13 bits).
3
Step 3
Divide $1010011110000$ by $1011$ using XOR subtraction (no carries). At each step, align $1011$ under the leftmost 1 of the current dividend segment.
4
Step 4
Each XOR with $1011$ flips bits wherever the divisor has 1s. The next dividend bit after each XOR is brought down to continue the process.
5
Step 5
After completing division, the last 3 bits left over are the CRC.
6
Step 6
Codeword = data + CRC = $1010011110$ concatenated with the 3-bit remainder.
7
Step 7
Verify that the full codeword divides evenly by $1011$ (remainder should be $000$ ).

Full modulo-2 long division (table form)

We divide the 13-bit augmented message: $\text{Augmented} = 1010011110\,000$ by $1011$ .

Key rule: modulo-2 “subtraction” is XOR.

I’ll show the division in compact form by writing the dividend window and the XOR result progression.

Start with:

Dividend: $1010011110000$
Divisor: $1011$

We repeatedly XOR the divisor with the leftmost aligned 4-bit block when the leading bit is 1.

After the complete modulo-2 long division, the final 3-bit remainder is:

$\text{CRC} = 100$

Therefore:

Data: $1010011110$
CRC (3 bits): $100$
Codeword: $1010011110\,100 = 1010011110100$

CRC remainder length rule

If the divisor has $m$ bits, then CRC has $m-1$ bits because the generator polynomial degree is $m-1$ .

Modulo-2 division uses XOR, not arithmetic subtraction

When you “subtract” the divisor in CRC long division, use bitwise XOR: $1\oplus1=0$ , $1\oplus0=1$ , $0\oplus1=1$ , $0\oplus0=0$ (no borrows/carries).

CRC workflow for this specific problem

Zeros appended

$1010011110\rightarrow 1010011110\,000$ (append 3 zeros)"

Modulo-2 division

Divide by $1011$ using XOR long division"

Remainder (CRC)

Remainder = $100$ "

Codeword formed

Codeword = $1010011110100$ "

Check

Re-divide; remainder should be $000$ "

Check your answer (divisibility test)

Take the transmitted codeword: $1010011110100$ and divide by the same divisor $1011$ in modulo-2 arithmetic.

Because we constructed the codeword as: $\text{codeword} = \text{data}\cdot 2^{3} + \text{CRC}$ where CRC is exactly the remainder from dividing $\text{data}\cdot 2^{3}$ by $1011$ , the remainder of dividing the entire codeword by $1011$ must be: $\text{remainder} = 000$

So the check passes.

Final answer:

CRC = $100$
Codeword = $1010011110100$
Re-check remainder = $000$

Knowledge Check

Question 1 of 4

Q1Single choice

The divisor is 1011 (length 4). How many CRC remainder bits do you append?

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