Solve the equation modulo 5. All solutions that exist should be a whole number strictly less than 5. 2x+3=1

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter7: Applying Fractions
Section: Chapter Questions
Problem 26CLR
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Solve the equation modulo 5. All solutions that exist should be a whole number strictly less than 5.

2x+3=1

Expert Solution
Step 1

Formula: If ab (mod n), then for any integer c, a±cb±c (mod n)

 

The given modulo equation is:

2x+31 (mod 5)

Subtract -3 from the both sides of the equation:

2x+3-31-3 (mod 5)2x-2 (mod 5)

Step 2

Result: If ab (mod n), then ab+n (mod n)

2x-2 (mod 5)

From the result,

2x-2+5 (mod 5)2x3 (mod 5)

Step 3

Multiplication inverse:

Let a be an element in modulo n. Then a-1 is said to be the multiplication inverse of a, if

aa-11 (mod n)

a-1 exists only if gcd (a, n)=1.

 

Find the multiplicative inverse of 2 in modulo 5.

Since 2 and 5 has no common divisor other than 1,

gcd (2, 5)=1

Hence, multiplicative inverse of 2 exists.

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