Olympiad Questions - Preparing for Maths Competitions

Mathematics competitions test the deepest kind of thinking: not speed at routine calculations, but genuine problem-solving creativity, elegance, and insight. Olympiad problems reward those who have learned to see patterns others miss, apply ideas from one area of mathematics to another, and construct clear, complete solutions. This page introduces the key ideas and gives you a taste of competition mathematics.

Major Mathematics Competitions

Competition	Country / Level	Format
UKMT Junior / Intermediate / Senior	United Kingdom	Multiple choice, then Olympiad papers
AMC 8 / 10 / 12	United States	Multiple choice, 30–40 minutes
AIME	United States (invitation)	Short answer, 3 hours
IMO	International	6 problems over 2 days, proof required
Kangaroo	Worldwide	Multiple choice, age groups

Key Olympiad Techniques

1. The Pigeonhole Principle

If n+1 objects are placed into n boxes, at least one box contains 2 or more objects.

Prove that among any 5 integers, two must have the same remainder when divided by 4.

There are only 4 possible remainders when dividing by 4: 0, 1, 2, 3 (four “boxes”). With 5 integers (“pigeons”), by the pigeonhole principle at least two must share a remainder. □

2. Modular Arithmetic

We write a ≡ b (mod n) to mean that a − b is divisible by n. Modular arithmetic lets you find remainders of large powers and prove divisibility results without full computation.

Find the remainder when 7¹⁰⁰ is divided by 5.

7 ≡ 2 (mod 5). Powers of 2 mod 5 cycle: 2, 4, 3, 1, 2, 4, 3, 1, ... (period 4).
100 = 4 × 25, so 2¹⁰⁰ ≡ 1 (mod 5).
Therefore 7¹⁰⁰ ≡ 1 (mod 5). The remainder is 1.

3. Parity Arguments

Parity (odd vs even) is a powerful tool. Many competition problems are solved simply by showing that one side of an equation is odd and the other is even – a contradiction.

Prove that the sum of three consecutive integers is always divisible by 3.

Let the integers be n, n+1, n+2. Sum = 3n + 3 = 3(n+1). Divisible by 3 for any integer n. □

4. Extremal Principle

Consider the largest or smallest element in a set. Its extremal position often forces properties that simplify the problem dramatically.

In a group of people where every pair of people are either friends or strangers, show there must exist a person who is friends with everyone, or a person who is strangers with everyone (for n = 3).

With 3 people A, B, C: A has 2 relationships. Either A knows both others, A knows neither, or A knows exactly one.
If A knows both B and C: if B and C know each other, all 3 are mutual friends. If not, B and C are strangers – three strangers’ case applies for that pair.
The Ramsey theorem generalises this: R(3,3) = 6, meaning any 6 people must contain 3 mutual friends or 3 mutual strangers.

5. AM-GM Inequality

For any non-negative numbers a and b:
(a + b)/2 ≥ √(ab) (Arithmetic Mean ≥ Geometric Mean)
Equality holds only when a = b. Used widely in optimisation problems.

Among all rectangles with a fixed perimeter of 40 cm, which has the greatest area?

Let sides be l and w. Then l + w = 20. By AM-GM: (l+w)/2 ≥ √(lw), so 10 ≥ √(Area). Area ≤ 100 cm².
Maximum area = 100 cm², achieved when l = w = 10 cm (a square).

Sample Olympiad Problems

[Number Theory] Find all positive integers n such that n² + 7n + 1 is a perfect square.

Let n² + 7n + 1 = k² for some positive integer k. Then k² − n² = 7n + 1, so (k−n)(k+n) = 7n+1.
Since k > n, try k = n + d for small d. For d = 3: (3)(2n+3) = 7n+1 → 6n+9 = 7n+1 → n = 8.
Check: 64 + 56 + 1 = 121 = 11². ✓ Answer: n = 8.

[Combinatorics] In how many ways can you place 8 non-attacking rooks on an 8×8 chessboard?

Each rook must be in a different row and a different column. The number of ways = the number of permutations of 8 columns assigned to 8 rows = 8! = 40 320.

[Geometry] Prove that the angles of any triangle sum to 180°.

Draw a line through vertex A parallel to BC. Let angles at B and C be β and γ. The alternate interior angles with the parallel line equal β and γ. The three angles at A (along the straight line) are α, β, γ. Since they form a straight line: α + β + γ = 180°. □

Competition Tips

Read every problem before starting – easier ones may appear later in the paper.
For multiple-choice, eliminate impossible answers first.
Try small cases to identify patterns before generalising.
Draw diagrams for every geometry problem.
Show all working clearly – partial credit is awarded in proof-based competitions.
If stuck, change representation: algebra → geometry or vice versa.
Practise regularly with past papers from UKMT, AMC, and Kangaroo.

Key Takeaways

Pigeonhole: n+1 objects in n categories guarantees a repeated category.
Modular arithmetic: find the cycle length of remainders to handle large powers.
AM-GM: arithmetic mean ≥ geometric mean, with equality when values are equal.
Extremal principle: analyse the largest or smallest element to force a useful structure.

Practice Olympiad Questions

Prove that among any 6 integers, at least two have the same remainder when divided by 5.
Find the last two digits of 3²⁰²⁶ (i.e. find 3²⁰²⁶ mod 100).
The product of two positive integers is 48. Their sum is as small as possible. Find both integers.
A 4×4 grid is to be coloured using 2 colours. Prove that there must exist a rectangle whose four corners are all the same colour.
Find all integer solutions to x² − y² = 100.

You have completed the Competitive Mathematics section. Continue your maths journey with the topics below.

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