SECTION 5: Linear Congruence Equations
Comments to the Instructor
(included in session notes below)
A. INSTRUCTOR MATERIALS
 transparency of Mod 11 and Mod 12 multiplication tables
B. PARTICIPANT MATERIALS
 course notebook (Mod 11 & 12 multiplication tables)
 calculators
 Handout: Section 5 outline
C. SESSION NOTES
Section 5: Linear Congruence Equations
Consider: General linear congruence equation ax  b mod m , where solutions are
congruence classes identified by the element in least residue (i.e. least residue
solutions, the names of the “boxes”). What can we expect when we solve them?
Work through some simple, specific examples (such as 2x ≡ 6 mod 7 and 2x ≡ 1
mod 5) by trying each least residue element (a brute force method). Explain that
while larger numbers may make the congruence equation true, only least residue
elements are needed to identify solutions.
Example: Simultaneously display the following congruence equations: 2 x  1 mod 6 ,
2 x  1 mod 3 , 2 x  2 mod 6 , 3x  7 mod 12 and 3x  6 mod 12 and make
conjectures about when congruence equations have no solutions, unique least residue
solutions, or multiple least residue solutions. (If they have multiple solutions, how
many do they have?) Arrive at Theorems 5.1 – 5.3.
Theorem 5.1: The congruence equation ax  b mod m has solutions if and only if
(a , m)|b .
*Example: Compare congruence equation ax  b mod m to linear Diophantine equation
ax  my  b .
Theorem 5.2: If (a, m)  1 then ax  b mod m has exactly one least residue solution.
Theorem 5.3: If (a , m)  d and d |b then ax  b mod m has exactly d least residue
solutions. (Test the conjecture with another example, but note that this
course will focus on those equations with ONE least residue solution.)
Explain the multiplication tables for Mod 11 & 12. Connect to the mod 12 congruence
equations in the example above.
Note: While there are other approaches that can be utilized to solve linear congruence equations
(examples 1 – 4 use trial an error of items in least residue or locating the solution on an appropriate
modular multiplication table), for this course we focus only on the method of multiplying by the
multiplicative inverse of the coefficient a of x in mod m, and then reducing the result to least
residue. To make a comparison with concepts they already understand, demonstrate on an algebraic
Copyright 2007. Number Theory and Cryptology for Middle Level Teachers. Developed by the Math in the Middle Institute
Partnership, University of Nebraska, Lincoln.
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equation such as 8x=36, that instead of thinking of “dividing by 8” one can multiply both sides of
the equation by 1 , the multiplicative inverse of 8, and then reducing the result.
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Examples: 1) 8x  1 mod 11 (refer to Mod 11 multiplication table in Misc. section)
2) 8x  3 mod 11
3) 2x  5 mod 7 (use trial and error to exhaust all least residue possibilities)
4) 4x  3 mod 6 (exhaust all possibilities as a reminder that not all equations
have solutions)
5) 5x  1 mod 12 (solve using multiplicative inverses from Z12 table)
6) 5x  11 mod 12 (solve using multiplicative inverses from Z12 table)
7) 7 x  8 mod 12 (solve using multiplicative inverses from Z12 table)
8) Can 8x  4 mod 12 be solved using multiplicative inverses? (Note when
coefficients do not have multiplicative inverses in mod m.)
Examples: For larger numbers some of the above approaches to solving congruence
equations are inefficient. (A multiplication table in mod 83 is an unlikely resource!)
a) Consider 29 x  1 mod 83 . An efficient way to solve this is to find the
multiplicative inverse of 29 in mod 83: write as 29 x  83y  1 and compare
to Diophantine equations. Use Euclidean algorithm to find that x   20
and y  7 satisfy 29 x  83y  1 , which implies 29( 20)  1 mod 83
(a multiplicative inverse), so that x  63 mod 83 is the solution.
b) Now consider 29 x  17 mod 83 . We again use the fact that -20 is the
multiplicative inverse of 29 in mod 83 and multiply both sides of the
congruence equation 29 x  17 by (-20) and determine least residue
solution: x  75 mod 83.
c) 91x  25 mod 136
d) 132 x  25 mod 253 (no solution)
Work Problem Session 5 problems in class.
D. ASSIGNMENT
 Work Evening Homework problems for Section 5
Copyright 2007. Number Theory and Cryptology for Middle Level Teachers. Developed by the Math in the Middle Institute
Partnership, University of Nebraska, Lincoln.
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