Applications
Two CAT favourites fall straight out of modular arithmetic: inverses/linear congruences and calendar–clock problems. A modular inverse of a (mod m) is the number a⁻¹ with a·a⁻¹ ≡ 1 (mod m); it exists only when gcd(a, m) = 1. To solve 3x ≡ 1 (mod 7), test residues: 3×5 = 15 ≡ 1, so x ≡ 5. More generally ax ≡ b (mod m) is solvable exactly when gcd(a,m) divides b. For calendars, every 7 days repeats the weekday, so you work mod 7. The count of odd days (the remainder after dividing total days by 7) tells you the shift: an ordinary year has 1 odd day, a leap year 2. For clocks, hours work mod 12 and minutes mod 60. The exam shortcut is to map the question to "what is N mod 7?" or "what is N mod 12?" and add that shift to the known starting point.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Congruence basics & operations
| Definition | a ≡ b (mod m) ⇔ m divides (a − b) |
|---|---|
| Addition | a ≡ b, c ≡ d ⇒ a + c ≡ b + d (mod m) |
| Multiplication | a ≡ b, c ≡ d ⇒ ac ≡ bd (mod m) |
| Exponentiation | a ≡ b (mod m) ⇒ a^k ≡ b^k (mod m) |
| Reduce the base first | a^k mod m = (a mod m)^k mod m |
Inverses, theorems & calendars
| Modular inverse | a·a⁻¹ ≡ 1 (mod m); exists ⇔ gcd(a, m) = 1 |
|---|---|
| Linear congruence ax ≡ b | solvable ⇔ gcd(a,m) divides b |
| Fermat’s little theorem | p prime, gcd(a,p)=1 ⇒ a^(p−1) ≡ 1 (mod p) |
| Euler’s theorem | gcd(a,m)=1 ⇒ a^φ(m) ≡ 1 (mod m) |
| Day shift | days mod 7 → 0=same weekday, 1=next day, … |