NUMERICAL — TIER 1

The seven topics that carry the section · Days 1–5

694 of the 1,214 numerical questions in my corpus came from these seven topics. If you only had five days, you would spend all five here.

§1 · NUMBER SYSTEM — 175 questions (the largest topic)

Sub-type breakdown from the real bank:

1.1 Remainders — the highest-value 90 minutes of your prep

This is the single biggest sub-type in the exam and the biggest hole in your material. Everything below is written from scratch.

The one idea everything rests on

When you only care about the remainder, you can replace any number by its remainder at any point. Remainders travel through addition, subtraction and multiplication untouched.

(a + b) mod n  =  (a mod n  +  b mod n) mod n
(a − b) mod n  =  (a mod n  −  b mod n) mod n
(a × b) mod n  =  (a mod n  ×  b mod n) mod n

Division is the exception — it does NOT work this way. Never divide inside a modulus.

Worked example. Remainder of 123 × 124 × 125 divided by 7.

Reduce each first: 123 = 17×7 + 4, 124 = 17×7 + 5, 125 = 17×7 + 6. So the answer is (4 × 5 × 6) mod 7 = 120 mod 7 = 1.

You never multiplied the big numbers. That is the whole technique.

Negative remainders — the shortcut most candidates never learn

A remainder of n − 1 can be written as −1. This turns ugly problems into trivial ones.

Worked example. Remainder of 6^100 divided by 7.

6 ≡ −1 (mod 7), so 6^100 ≡ (−1)^100 = 1. Answer: 1. No calculation at all.

Worked example. Remainder of 15^50 divided by 16. 15 ≡ −1 (mod 16) → (−1)^50 = 1. Answer: 1.

Rule of thumb: whenever the base is one less than the divisor, the answer is 1 for even powers and n−1 for odd powers. Check for this before doing anything else.

Cyclicity of remainders

Powers of a number mod n eventually cycle. Find the cycle length, take the exponent mod that length, read off the answer.

Worked example. Remainder of 2^100 divided by 7.

2^1 mod 7 = 2
2^2 mod 7 = 4
2^3 mod 7 = 1   ← cycle closes, length 3

100 mod 3 = 1, so 2^100 ≡ 2^1 = 2. Answer: 2.

Watch the edge case: if the exponent divides evenly by the cycle length, the answer is the last element of the cycle (here 2^3 ≡ 1), not the first. 2^99 would give 1, not 2.

Fermat's little theorem — for prime divisors

If p is prime and a is not divisible by p, then a^(p−1) ≡ 1 (mod p).

This hands you the cycle length for free when the divisor is prime.

Worked example. Remainder of 3^100 divided by 7. 7 is prime, so 3^6 ≡ 1 (mod 7). Now 100 = 6×16 + 4, so 3^100 ≡ 3^4 = 81 ≡ 4 (mod 7). Answer: 4.

Worked example. Remainder of 2^1000 divided by 13. 2^12 ≡ 1 (mod 13). 1000 mod 12 = 4. So 2^1000 ≡ 2^4 = 16 ≡ 3. Answer: 3.

Euler's theorem — for composite divisors

When the divisor is not prime, Fermat does not apply. Use Euler:

If a and n are coprime, then a^φ(n) ≡ 1 (mod n), where φ(n) counts the numbers below n that are coprime to it.

Computing φ(n): prime factorise, then

φ(n) = n × (1 − 1/p₁) × (1 − 1/p₂) × …

Worked example. Remainder of 7^100 divided by 100. φ(100) = 40, and 7 is coprime to 100. 100 mod 40 = 20, so 7^100 ≡ 7^20. Then 7^4 = 2401 ≡ 1 (mod 100), so 7^20 = (7^4)^5 ≡ 1. Answer: 1.

Note this also gives you the last two digits of 7^100, which is how TCS often phrases it.