Ecole Polytechnique Federale de LausanneDepartement de mathematiques

Prof. E. Bayer Fluckiger

Algebra for digital communicationsSection de syste`mes de communication winter semester 2004-2005

Solutions to exercise sheet 2 10.28.04

Exercise 1(i) First of all, we need to compute the gcd of 59 and 36.

59 = 36 1 + 23 ( 23 = 1 59 1 36 )36 = 23 1 + 13 ( 13 = 1 59 + 2 36 )23 = 13 1 + 10 ( 10 = 2 59 3 36 )

During the computation, we find that 10 = 2 593 36, so in order to get 10 cm3 of water,we put in the big tank twice a full 59 cm3s tank and then remove three times 36 cm3 ofwater.

(ii) Computation of the gcd of 36 and 21.

36 = 21 1 + 15 ( 15 = 1 36 1 21 )21 = 15 1 + 6 ( 6 = 1 36 + 2 21 )15 = 6 2 + 3 ( 3 = 3 36 5 21 )6 = 3 2 + 0

Multiply the last equation by 3 to obtain :

9 = 9 36 15 21.

(iii) For each x, y Z, the number 36x + 21y is divisible by 3, so the equation 36x + 21y = 7has no solution in Z.

(iv) Once again, we have to use Euclides algorithm.

89 = 24 3 + 17 ( 17 = 1 89 3 24 )24 = 17 1 + 7 ( 7 = 1 89 + 4 24 )17 = 7 2 + 3 ( 3 = 3 89 11 24 )7 = 3 2 + 1 ( 1 = 7 89 + 26 24 )3 = 1 3 + 0

Thus the inverse of 24 modulo 89 is 26.

Exercise 21. See the online course, chapter 2.

2. (125) = (53) = 4 52 = 100.(63) = (7 32) = (7) (32) = 6 2 3 = 12.

Exercise 31. dK(eK(x)) = dK(x+K) = (x+K)K = x.

2. VKRIMHZKTIAR.

3. VERY GOOD JOB.

Challenge exercise 11. Since a is prime to 10, a is prime to 10n for each n > 1. For each n > 1, take a Bezout

relation9 9 n

= un a+ vn 10n,

with |vn| < a. Therefore, for some n and m, n > m, we will have that vn = vm. If wemultiply the equality 9 9

m

= um a+ vm 10m by 10nm, and substract it to the equality9 9 n

= un a+ vn 10n, we get that

9 9 nm

= (un 10nmum)a,

which is a multiple of a.

2. The proof is the same as above, except that you must replace each 9 by a digit 7.