Modular Arithmetic
Modular arithmetic is the maths of remainders, and it quietly powers a big slice of CAT Number System. The idea is simple: instead of caring about the whole number, you only care about what is left over after dividing by a fixed number m, called the modulus. We write a ≡ b (mod m) to say a and b leave the same remainder when divided by m. That single idea collapses enormous computations — the units digit of 7^100, the remainder of 2^500 divided by 7, the day of the week 1,000 days from now — into a few clean steps. CAT loves these because they look terrifying but reward a student who knows the rules. This chapter builds the toolkit in three layers: first the language of congruences and why a ≡ b (mod m) behaves like equality; then the operations — how congruences add, multiply, raise to powers, and how to invert and solve linear congruences such as 3x ≡ 1 (mod 7); and finally the high-yield applications, especially day-of-week and clock problems that turn calendar trivia into a 30-second calculation. Master this and remainder questions stop being guesswork.
Topics
⚡ CAT shortcuts & speed methods
The fastest ways to crack this chapter under time pressure — the techniques that separate a 95+ percentiler from the rest.
- Always reduce the base to its smallest residue first; prefer a −1 residue (e.g. 6 ≡ −1 mod 7) so the power becomes a sign.
- Find a small power that is ≡ 1 (the cycle length), divide the exponent by it, and only the remaining power matters.
- Fermat: for prime p and gcd(a,p)=1, reduce the exponent modulo (p−1) before computing a^k mod p.
- Euler: for any m with gcd(a,m)=1, reduce the exponent modulo φ(m); for m = 10, φ(10)=4 explains units-digit cycles of length 4.
- For a units digit, work mod 10 — every base cycles with period 1, 2 or 4; use exponent mod 4 (treat remainder 0 as the 4th term).
- Calendar/clock questions are just "N mod 7" (weekday) or "N mod 12" (clock hour); count odd days = 1 per ordinary year, 2 per leap year.
⚠️ Common mistakes & traps
CAT is designed so that careless errors here cost you marks. Internalise each trap before the exam.
- Cancelling a common factor across a congruence without dividing the modulus too — you may only cancel a when gcd(a, m) = 1.
- Assuming a modular inverse always exists; it exists only when gcd(a, m) = 1, so 2 has no inverse mod 4.
- Reducing the exponent modulo m instead of modulo (p−1) or φ(m) — the base is reduced mod m, the exponent is not.
- Forgetting leap years in calendar problems: a leap year contributes 2 odd days, not 1, and century years are leap only if divisible by 400.
- Misreading the remainder 0 in a cyclicity problem as "the first term"; a remainder of 0 means the last term of the cycle.
📈 CAT exam insight & PYQ analysis
🎴 Flashcards — instant recall
Tap a card to reveal the answer. Drill these until they are automatic.
📌 Quick revision
Chapter test
🏆 Vidaara CAT success checklist
You have truly mastered Modular Arithmetic when you can tick every box below.
- Recall every formula in this chapter without looking them up
- Solve each topic’s practice set with at least 80% accuracy
- Use the chapter shortcuts to cut your solving time in half
- Spot and avoid every common trap listed above
- Score 80%+ on the timed chapter test
📋 Chapter mastery scorecard
Track where you stand. Aim for the target before moving to the next chapter.
| Skill checkpoint | Target |
|---|---|
| Concept theory & formulas understood | 100% |
| Topic practice sets attempted (3 topics) | 3/3 |
| Best topic-test score | — → 80%+ |
| Chapter test score | — → 80%+ |
| Flashcards drilled to instant recall | 12 cards |
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, … |