Modular Arithmetic for JEE Main: Number Theory Guide

Modular arithmetic — the mathematics of remainders — underpins JEE Main questions on units digits, large power remainders, and divisibility. These questions appear embedded in binomial theorem, sequences, and permutation-combination problems. Thinking in congruences converts seemingly impossible computations on huge numbers into quick mental arithmetic.

Congruences: The Language of Remainders

We write a ≡ b (mod m) if a and b leave the same remainder when divided by m — equivalently, if m divides (a − b). Key properties: if a ≡ b and c ≡ d (mod m), then a + c ≡ b + d and a × c ≡ b × d. These allow reducing large numbers before computing: 123 ≡ 3 (mod 10), so 123⁴ ≡ 3⁴ = 81 ≡ 1 (mod 10). The units digit of 123⁴ is 1. This is the congruence approach to units-digit problems. For the number theory foundation see our number systems guide.

Units-Digit Cyclicity

The units digit of powers cycles with a short period. Powers of 2: 2, 4, 8, 6 (period 4). Powers of 3: 3, 9, 7, 1 (period 4). Powers of 7: 7, 9, 3, 1 (period 4). Powers of 8: 8, 4, 2, 6 (period 4). Powers of 4: 4, 6 (period 2). Powers of 9: 9, 1 (period 2). Powers of 0, 1, 5, 6: always end in 0, 1, 5, 6 (period 1). To find the units digit of a power: find the remainder of the exponent modulo the cycle length, then read off the corresponding term. If the remainder is 0, take the last term of the cycle.

Remainders of Large Powers

To find the remainder of a^n divided by m: reduce a modulo m first. Then find the cycle length of the reduced base. Find n modulo the cycle length. Compute the corresponding power. Example: 7^100 mod 8. 7 ≡ −1 (mod 8). (−1)^100 = 1. So remainder is 1. Alternative for mod prime p: Fermat's little theorem: a^(p−1) ≡ 1 (mod p) for a not divisible by p. So a^n ≡ a^(n mod (p−1)) (mod p).

Binomial Theorem for Remainders

Write the base as (multiple of modulus ± 1): a^n = (qm ± 1)^n. Expanding: all terms except the last (±1)^n are divisible by m. So a^n ≡ (±1)^n (mod m). Example: 99^50 mod 100 = (100−1)^50 ≡ (−1)^50 = 1 (mod 100). So the last two digits of 99^50 are 01. This expansion trick works whenever the base can be written as (multiple of modulus ± small number). For the binomial theorem connection see our binomial theorem guide.

Divisibility Rules

Divisibility by 2: last digit even. By 3 or 9: digit sum divisible. By 4: last two digits divisible. By 8: last three digits divisible. By 11: alternating digit sum divisible. By 7: more complex (double the last digit, subtract from rest; repeat). Modular arithmetic gives a unified framework: a number N ≡ (its digit sum) (mod 9), because 10 ≡ 1 (mod 9). This is why the digit sum divisibility rule for 3 and 9 works. After mastering modular arithmetic, practise ten remainder problems spanning all the above types, then take a free mock test.