Mathematical-Reasoning · Unit 29 · Video 1 · Interactive Practice

Remainders of Sums, Products and Digit Sums Mod 6 and Mod 9

IKey Formulas

Formula	Meaning
$x \in [3] \iff x \equiv 3 \pmod 6 \iff 6 \mid (x - 3)$	membership in a congruence class modulo $6$
$x \equiv a,\ y \equiv b \pmod n \ \Rightarrow\ x + y \equiv a + b,\ \ xy \equiv ab,\ \ x^2 \equiv a^2 \pmod n$	congruence arithmetic (proved earlier): the remainder depends only on the classes
$r = n - 9 \cdot \operatorname{int}(n/9)$ , $\ 0 \le r < 9$	the remainder of $n$ on division by $9$ , from $n = 9q + r$
$n \equiv s(n) \pmod 9$	a number and its digit sum leave the same remainder — observed here, proved in the last video

Key Insight: A remainder lies between $0$ and $n - 1$ : $-5 = 6 \cdot (-1) + 1$ , so $-5$ leaves remainder $1$ on division by $6$ , not $-5$ .

IIAny Members, Same Remainders

The remainders of $x + y$ , $xy$ and $x^2$ modulo $6$ depend only on the classes of $x$ and $y$ .

Class of x Class of y

IIIA Number and Its Digit Sum Modulo 9

For every test number, $n$ and its digit sum $s(n)$ leave the same remainder on division by $9$ .

💡 Six matching rows are evidence, not a proof; the last video of this unit proves $n \equiv s(n) \pmod 9$ for every natural number $n$ .

IVQuiz Questions

Problem 1 · Remainders From the Classes

Suppose $x \equiv 4 \pmod 6$ and $y \equiv 5 \pmod 6$ .

What remainder does $x + y$ leave on division by $6$ ?

What remainder does $xy$ leave on division by $6$ ?

✅ Correct!

4 + 5 = 9 = 6 \cdot 1 + 3

and

4 \cdot 5 = 20 = 6 \cdot 3 + 2

, whichever

x

and

y

were chosen.

❌ Not quite.

9

is not a remainder on division by

6

: it must lie between

0

and

5

. Reduce

9 = 6 \cdot 1 + 3

❌ Not quite.

20

4 \cdot 5

, but a remainder on division by

6

lies between

0

and

5

❌ Check the sum. By congruence arithmetic

x + y \equiv 4 + 5 \pmod 6

; reduce

9

❌ Check the product. By congruence arithmetic

xy \equiv 4 \cdot 5 \pmod 6

; reduce

20

Show solution

Since $x \equiv 4$ and $y \equiv 5 \pmod 6$ , congruence arithmetic gives

x + y \equiv 4 + 5 = 9 = 6 \cdot 1 + 3 \equiv 3 \pmod 6, \qquad xy \equiv 4 \cdot 5 = 20 = 6 \cdot 3 + 2 \equiv 2 \pmod 6.

Check with members: $x = 10$ , $y = -1$ give $x + y = 9$ (remainder $3$ ) and $xy = -10 = 6 \cdot (-2) + 2$ (remainder $2$ ).

Problem 2 · The Remainder of a Negative Integer

What is the remainder when $-23$ is divided by $6$ ?

✅ Correct!

-23 = 6 \cdot (-4) + 1

with

0 \le 1 < 6

❌ Not quite.

-23 = 6 \cdot (-3) - 5

is true, but a remainder must satisfy

0 \le r < 6

; go one more multiple of

6

down.

❌ Not quite.

5

is the remainder of

23

. Check:

-23 - 5 = -28

is not a multiple of

6

❌ Not quite. Find the integer

q

with

6q \le -23 < 6q + 6

Show solution

The Division Algorithm needs $-23 = 6q + r$ with $0 \le r < 6$ . The largest multiple of $6$ not above $-23$ is $6 \cdot (-4) = -24$ , so

-23 = 6 \cdot (-4) + 1, \qquad r = 1.

Check: $-23 - 1 = -24 = 6 \cdot (-4)$ , so $-23 \equiv 1 \pmod 6$ and $-23 \in [1]$ .

Problem 3 · Two Remainders Modulo 9

Let $n = 8642$ . Using $r = n - 9 \cdot \operatorname{int}(n/9)$ , find the remainder of $n$ on division by $9$ .

✅ Correct!

8642 = 9 \cdot 960 + 2

, and the digit sum

20 = 9 \cdot 2 + 2

leaves the same remainder.

❌ Not quite.

20

is the digit sum

s(8642)

, and it is larger than

8

, so it is not a remainder. What matches is the remainder it leaves.

❌ Not quite.

960 = \operatorname{int}(8642/9)

is the quotient

q

, not the remainder.

❌ Not quite. Compute

9 \cdot 960

and subtract it from

8642

Show solution

$8642 / 9 \approx 960.2$ , so $\operatorname{int}(8642/9) = 960$ and

r = 8642 - 9 \cdot 960 = 8642 - 8640 = 2.

The digit sum is $s(8642) = 8 + 6 + 4 + 2 = 20 = 9 \cdot 2 + 2$ , which also leaves remainder $2$ .

Problem 4 · A Polynomial in x Modulo 6

Suppose $x \equiv 5 \pmod 6$ . What remainder does $x^2 + x$ leave on division by $6$ ?

✅ Correct!

x^2 + x \equiv 25 + 5 = 30 = 6 \cdot 5 \pmod 6

❌ Not quite.

30

5^2 + 5

, but a remainder on division by

6

lies between

0

and

5

❌ Not quite.

x^2 \equiv 25 \equiv 1

, but the expression also has the term

x \equiv 5

; add them:

1 + 5 = 6

❌ Not quite. Replace

x

5

x^2 + x

, then reduce modulo

6

Show solution

From $x \equiv 5 \pmod 6$ , congruence arithmetic gives $x^2 \equiv 5^2 = 25 \pmod 6$ and then $x^2 + x \equiv 25 + 5 = 30 \pmod 6$ . Since $30 = 6 \cdot 5 + 0$ ,

x^2 + x \equiv 0 \pmod 6.

Check with $x = -1 \in [5]$ : $(-1)^2 + (-1) = 0$ . In fact $x^2 + x = x(x + 1)$ is always even, and here it is a multiple of $3$ as well.

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