# MATH 433 April 24, 2015 Quiz 11: Solutions

```MATH 433
April 24, 2015
Quiz 11: Solutions
Problem 1. Find the missing digit in the following ISBN-10: 0553103 47.
Solution: 5.
Let d denote the missing digit. Then
10 &middot; 0 + 9 &middot; 5 + 8 &middot; 5 + 7 &middot; 3 + 6 &middot; 1 + 5 &middot; 0 + 4 &middot; 3 + 3d + 2 &middot; 4 + 7 ≡ 0 mod 11,
which simplifies to 3d + 139 ≡ 0 mod 11 or 3d ≡ 4 mod 11. A solution of this linear congruence is easily
found by inspection: d = 5.
Problem 2. Let f : B2 → B5 be the binary coding function that sends each two-character
word ab in the alphabet B = {0, 1} to the codeword abbax, where x is the parity bit of the word ab
(i.e., x = 0 if ab contains an even number of 1’s and x = 1 otherwise). How many errors will this
code detect? correct? Explain.
Solution: the code detects 2 errors and corrects 1 error.
We find that f (00) = 00000, f (01) = 01101, f (10) = 10011, and f (11) = 11110. The minimum distance
between these codewords is 3. It follows that the code allows to detect 2 errors and to correct 1 error.
```