   Chapter 2.5, Problem 41E

Chapter
Section
Textbook Problem

In the congruences a x ≡ b ( m o d   n ) in Exercises 40 − 53 , a and n may not be relatively prime. Use the results in Exercises 38 and 39 to determine whether there are solutions. If there are, find d incongruent solutions modulo n . 6 x ≡ 33   ( m o d  27 )

To determine

Whether there are solutions of congruence 6x33(mod27). If there are, find d incongruent solutions modulo 27.

Explanation

Formula used:

i) For d=(a,n), where n>1 if there is a solution to axb(modn), then d divides b.

ii) Suppose that n>1 and that d=(a,n) is divisor of b. Let a=a0d,b=b0d and n=n0d,

where a0,b0,n0 are integers the following statements are hold:

a) axb(modn) if and only if a0xb0(modn0).

b) If x1 and x2 are any two solutions of a0xb0(modn0), then it follows that x1x2(modn0).

c) Let x1 be a fixed solution to a0xb0(modn0), and each of the d integers in the list

x1,x1+n0,x1+2n0,x1+(d1)n0 is a solution to axb(modn).

d) No two of the solutions listed in part c) are congruent modulo n.

e) Any solution to axb(modn) congruent to one of the numbers listed in part c).

iii) The Euclidean Algorithm:

a=bq0+r1,0r1<bb=r1q1+r2,0r2<r1r1=r2q2+r3,0r3<r2rk=rk+1qk+1+rk+2,0rk+2<rk+1

Since the integers r1,r2,,rk+2 are decreasing and are all non-negative, there is a smallest integer n, such that rn+1=0: rn1=rnqn+rn+1,0=rn+1.

Explanation:

Given 6x33(mod27)

Comparing 6x33(mod27) with axb(modn)

Then, a=6,n=27,b=33

Now, d=(a,n)=(6,27)=3

By using result, for d=(a,n), where n>1 if there is a solution to axb(modn), then d divides b

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