Mathematical-Reasoning · Unit 10 · Video 2 · Interactive Practice

One Quotient, One Remainder: The Division Algorithm

IKey Formulas

Formula	Name	Meaning
$a = bq + r$ , $\ 0 \le r < b$	Division Algorithm (taken as an axiom)	For all integers $a$ and $b$ with $b > 0$ , exactly one pair of integers $q$ , $r$ satisfies both conditions
$r = a - bq$	Finding the remainder	$bq$ is the largest multiple of $b$ that does not exceed $a$
$a = b(q - 1) + (r + b)$	Shifting the quotient	Lowering $q$ by $1$ raises $r$ by $b$
$r \in \{0, 1, \dots, b - 1\}$	Possible remainders	Exactly $b$ of them; for $b = 2$ , $a = 2q$ (even) or $a = 2q + 1$ (odd)

Key Insight: Many pairs satisfy $a = bq + r$ , and the condition $0 \le r < b$ keeps exactly one. For $a = -17$ and $b = 5$ that pair is $-17 = 5 \cdot (-4) + 3$ , not the calculator's $-17 = 5 \cdot (-3) + (-2)$ .

IIVisualization 1 — Only One Pair Survives

Every integer $q$ gives an $r$ with $a = bq + r$ , but only one of those $r$ satisfies $0 \le r < b$ .

Division Quotient q 5

IIIVisualization 2 — Remainders Cycle Through 0 to b − 1

The quotient comes from the largest multiple of $b$ not exceeding $a$ , so $r$ is always one of $0, 1, \dots, b - 1$ .

Dividend a −17 Divisor b 5

IVQuiz Questions

Problem 1 · Which Pair Is the Division Algorithm's?

Given: $a = 94$ and $b = 8$ . Each equation below is true. Which one gives the quotient and the remainder guaranteed by the Division Algorithm?

✅ Correct!

q = 11

and

r = 6

are integers with

0 \le 6 < 8

, so the quotient is

11

and the remainder is

6

❌ Not quite. The arithmetic is right, but

r = -2

is negative, so it fails

0 \le r < 8

. Lower

q

1

and

r

rises by

8

❌ Not quite. The arithmetic is right, but

r = 14

is not less than

8

. Raise

q

1

and

r

drops by

8

❌ Not quite. The quotient must be an integer, and

11.75

is not one. The Division Algorithm works inside the integers.

Show solution

The largest multiple of $8$ that does not exceed $94$ is $88 = 8 \cdot 11$ , so

r = 94 - 88 = 6, \qquad 94 = 8 \cdot 11 + 6.

Both $q = 11$ and $r = 6$ are integers and $0 \le 6 < 8$ . The other equations are true but fail a condition: $r = -2 < 0$ , $r = 14 \ge 8$ , and $q = 11.75$ is not an integer.

Problem 2 · A Negative Dividend

Given: $a = -38$ and $b = 6$ . Find the quotient $q$ and the remainder $r$ guaranteed by the Division Algorithm.

✅ Correct!

-38 = 6 \cdot (-7) + 4

and

0 \le 4 < 6

❌ Close, but… This is the calculator's answer:

-38/6 \approx -6.33

truncates to

-6

, giving

-38 = 6 \cdot (-6) + (-2)

. The remainder

-2

is negative, so lower

q

1

❌ Not quite. This copies

38 = 6 \cdot 6 + 2

and flips only the quotient, but

6 \cdot (-6) + 2 = -34

, not

-38

❌ Not quite.

6 \cdot (-8) + 10 = -38

is true, but

10 \ge 6

. The quotient went one step too low.

Show solution

The multiples of $6$ near $-38$ are $-42$ and $-36$ . Since $-36 > -38$ , the largest multiple of $6$ that does not exceed $-38$ is $-42 = 6 \cdot (-7)$ . Then

r = -38 - (-42) = 4, \qquad -38 = 6 \cdot (-7) + 4,

and $0 \le 4 < 6$ . So $q = -7$ and $r = 4$ . Choosing $q = -6$ gives $r = -2$ , which is negative and not allowed.

Problem 3 · Remainder of an Expression

Given: When the integer $a$ is divided by $7$ , the remainder is $5$ . Find the remainder when $3a + 4$ is divided by $7$ .

✅ Correct!

3a + 4 = 7(3q + 2) + 5

with

0 \le 5 < 7

, so the remainder is

5

❌ Not quite.

3a + 4 = 21q + 19

, but

19

is not less than

7

, so it cannot be the remainder. Split

19 = 14 + 5

❌ Not quite.

1

is the remainder of

3a

alone, since

3 \cdot 5 = 15 = 14 + 1

. The

+4

still has to be added.

❌ Not quite. The

2

7(3q + 2) + 5

belongs to the quotient

3q + 2

, not to the remainder.

Show solution

By the Division Algorithm there is an integer $q$ with $a = 7q + 5$ . Then

3a + 4 = 3(7q + 5) + 4 = 21q + 19 = 21q + 14 + 5 = 7(3q + 2) + 5.

Here $3q + 2$ is an integer and $0 \le 5 < 7$ . Because the pair is unique, the quotient is $3q + 2$ and the remainder is $5$ . Check with $a = 12$ : $3 \cdot 12 + 4 = 40 = 7 \cdot 5 + 5$ .

Problem 4 · Cases by Remainder

Given: With $b = 3$ , the Division Algorithm says every integer $a$ is exactly one of $3q$ , $3q + 1$ , $3q + 2$ for some integer $q$ . Which statement is true?

✅ Correct! The three cases give remainders

0

1

and

1

, so

2

never occurs.

❌ Not quite. The case

a = 3q

breaks it:

a = 3

gives

a^2 = 9 = 3 \cdot 3 + 0

❌ Not quite. The case

a = 3q + 2

breaks it:

a = 2

leaves remainder

2

, but

a^2 = 4 = 3 \cdot 1 + 1

leaves remainder

1

❌ Not quite. Square each of the three forms and rewrite as

3k + r

with

0 \le r < 3

: the remainder

2

never appears.

Show solution

Use proof by cases on the remainder of $a$ .

Case $a = 3q$ : $a^2 = 9q^2 = 3(3q^2) + 0$ , remainder $0$ .

Case $a = 3q + 1$ : $a^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1$ , remainder $1$ .

Case $a = 3q + 2$ : $a^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1$ , remainder $1$ .

In each case the quotient is an integer and the remainder lies in $\{0, 1, 2\}$ , so by uniqueness these are the remainders. Every integer falls in one case, so $a^2$ always leaves remainder $0$ or $1$ , never $2$ .

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