The Lumenary · Grades 6–8

Clock arithmetic: remainders and modular arithmetic

Name: ______________________

Date: ______________________

You will learn to

Compute sums and products modulo $n$ using remainders, on 12- and 24-hour clocks and in a 7-day week
Build addition and multiplication tables modulo $n$ and find which remainders have a multiplicative partner
Prove that the remainder of a sum or a product depends only on the remainders of its parts
Prove that a number sharing a factor greater than 1 with $n$ has no partner modulo $n$ , and explain why every non-zero remainder has a partner modulo a prime
Use repeating cycles of remainders to answer questions about huge powers and calendars

1.
A long-distance bus leaves Nairobi at 17:00 and the journey takes 30 hours. On the 24-hour clock, at what hour does it arrive? (Give a number from 0 to 23.)
2.
Today is Tuesday. What day of the week will it be in 45 days?
- A.Friday
- B.Saturday
- C.Wednesday
- D.Tuesday
3.
Without multiplying the numbers out, find $(53 \cdot 86) \bmod 9$ .
4.
What is $-23 \bmod 6$ ? (Remember that a remainder modulo $6$ is one of $0, 1, \dots, 5$ .)
5.
Which of these remainders have a multiplicative partner modulo $10$ ? (Choose all that apply.)
- A. $1$
- B. $2$
- C. $3$
- D. $5$
- E. $7$
- F. $9$
6.
Find the partner of $3$ modulo $11$ : the number $b$ from $0$ to $10$ with $3b \bmod 11 = 1$ .
7.
Amara claims: “If $a \cdot b \bmod n = 0$ , then $a \bmod n = 0$ or $b \bmod n = 0$ .” For which modulus does a counterexample exist?
- A. $n = 7$
- B. $n = 10$ , since $2 \cdot 5 = 10 \equiv 0$
- C. $n = 11$
- D.None — the claim is always true
8.
What is the last digit of $3^{2026}$ ?
9.
In Lean, a % n is the remainder of a on division by n. Which statement says that the remainder of a product depends only on the remainders of the factors (Theorem 2)?
- A.(a * b) % n = (a % n * (b % n)) % n
- B.(a * b) % n = (a % n) * (b % n)
- C.(a + b) % n = (a % n + b % n) % n
- D.(a * b) % n = 0 → a % n = 0 ∨ b % n = 0
10.
Write a function partners(n) that returns the list of remainders a in range(n) that have a multiplicative partner modulo n — some b in range(n) with a * b % n == 1. Then use it to compare partners(12) with partners(13).
Computational — Practice

Answer key — Clock arithmetic: remainders and modular arithmetic

1.
$17 + 30 = 47 = 1 \cdot 24 + 23$ , so $47 \bmod 24 = 23$ . The bus arrives at 23:00 on the next day (the quotient $1$ tells us it is one day later).
2.
A week has 7 days, so we need $45 \bmod 7$ . Since $45 = 6 \cdot 7 + 3$ , the remainder is $3$ . Three days after Tuesday is Friday.
3.
$53 = 5 \cdot 9 + 8$ and $86 = 9 \cdot 9 + 5$ . By Theorem 2, $(53 \cdot 86) \bmod 9 = (8 \cdot 5) \bmod 9 = 40 \bmod 9 = 4$ . Check: $53 \cdot 86 = 4558 = 506 \cdot 9 + 4$ . ✓
4.
$-23 = (-4) \cdot 6 + 1$ with $0 \le 1 \le 5$ , so $-23 \bmod 6 = 1$ . On a 6-hour clock, walking 23 steps backwards from $0$ is the same as walking 24 steps back (four full turns) and then 1 step forwards.
5.
Partners exist for 1, 3, 7, 9: $1 \cdot 1 = 1$ , $3 \cdot 7 = 21$ , $9 \cdot 9 = 81$ all end in $1$ . The others share a factor with $10$ ( $2$ or $5$ ), so by Theorem 4 they have no partner. These four are exactly the remainders with no common factor with $10$ .
6.
$3 \cdot 4 = 12 = 1 \cdot 11 + 1$ , so $3 \cdot 4 \equiv 1 \pmod{11}$ and the partner is $b = 4$ . Theorem 5 promised one, because $11$ is prime.
7.
Modulo $10$ : $2 \cdot 5 = 10 \equiv 0$ although $2 \not\equiv 0$ and $5 \not\equiv 0$ . In general, if $n = r \cdot s$ with $1 < r, s < n$ , then $r \cdot s \equiv 0 \pmod n$ . For a prime modulus the claim is true — that is Euclid's lemma.
8.
$3^4 = 81 \equiv 1 \pmod{10}$ . Since $2026 = 4 \cdot 506 + 2$ , Theorem 2 gives $3^{2026} = (3^4)^{506} \cdot 3^2 \equiv 1^{506} \cdot 9 = 9 \pmod{10}$ . The last digit is 9.
9.
Theorem 2 is (a * b) % n = (a % n * (b % n)) % n. Leaving out the last % n gives a false statement (try $a = b = 5$ , $n = 6$ ), and the other two options are different statements.
10.
```
def partners(n):
    return [a for a in range(n) if any(a * b % n == 1 for b in range(n))]
```
partners(12) is [1, 5, 7, 11]: only the remainders with no common factor with $12$ . partners(13) is all of 1, …, 12, because $13$ is prime (Theorem 5). For $n = 1$ the only remainder is $0$ , and $0 \ne 1$ , so the list is empty. The program checks particular moduli; Theorems 4 and 5 explain the pattern for all of them.