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Summary Free E071 Discrete Math Preview | Modular Arithmetic Explained

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See modular arithmetic as a cyclic counter, work through an example, and check two original questions with explained answers. A two-page sample of Accelerated Reviews' independent E071 study guide.

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ACCELERATED REVIEWS E071 / FREE PREVIEW




Applied Discrete Mathematics: Study Guide Preview
This independent preview shows the teaching approach in the full E071 topic review. It is original learning
material, not an official university publication or an assessment item. The full guide covers logic,
sets/functions/relations, Boolean algebra, counting, probability, modular arithmetic, and graphs/trees. It is a topic
review; it does not claim to match every current assessment item or guarantee a passing result.


Sample lesson: modular arithmetic as a cyclic counter
Suppose a service labels successive work slots 0,1,2,3,4 and then returns to 0. The label after k advances from
label s is (s+k) mod 5. The expression a mod 5 means the unique remainder in {0,1,2,3,4} when a is divided by
5. For example, 12=2*5+2, so 12 mod 5=2. This definition also works with negative inputs: -3=(-1)*5+2, so -3 mod
5=2.

Two integers are congruent modulo 5 when they have the same remainder. We write a congruent to b (mod 5)
precisely when 5 divides a-b. Since 12-2=10 is divisible by 5, 12 congruent to 2 (mod 5). You can reduce a large
number before adding or multiplying. For example, (12+19) mod 5 equals (2+4) mod 5, which is 1. The direct
check is 31=6*5+1.

Complete worked example. The current work-slot label is 4. What label appears after 17 advances? Write the
model first: (4+17) mod 5 = 21 mod 5. Divide 21 by 5: 21=4*5+1. Therefore the label is 1. To check it another way,
17 advances are three complete five-slot cycles plus two additional steps. From 4, one step reaches 0 and the
second reaches 1. Both methods agree. The cyclic model assumes exactly five labels, one advance per step, and
no skipped or reset states; change those assumptions and the answer may change.

When solving a congruence such as 3x congruent to 1 (mod 5), be careful with "division." Modular cancellation
requires a multiplicative inverse. Here 3*2=6 congruent to 1 (mod 5), so multiply both sides by 2 to get x
congruent to 2 (mod 5). Verify: 3*2=6, whose remainder is 1. The inverse exists because 3 and 5 have greatest
common divisor 1.

Common error. Reporting a negative remainder for a negative input. With positive modulus 5, the least
nonnegative remainder must be one of 0,1,2,3,4. Show a quotient-remainder equation to make the answer
unambiguous.

Recall check. Explain why -8 mod 5=2 without using a calculator. Then explain what must be true before you
cancel a factor in a modular equation.




Edition 1.0 - October 2026 1

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29 de octubre de 2026
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