Olympiad maths is a different sport from school and entrance-exam maths: fewer questions, deeper thinking, and an emphasis on proof and insight over speed. The only way to get better is to wrestle with hard problems and then study clean solutions. Maths Daily Helper generates fresh olympiad-difficulty problems across the classic areas, each with a full worked solution, so you always have a new challenge and a clear model answer to learn from.
Areas covered
- Number theory: divisibility, modular arithmetic, primes, Diophantine equations.
- Combinatorics: counting, pigeonhole, graph theory, invariants.
- Geometry: angle chasing, similar triangles, cyclic quadrilaterals, proofs.
- Inequalities: AM–GM, Cauchy–Schwarz, functional inequalities.
Try a few olympiad problems
Show that among any 5 integers, there are two whose difference is divisible by 4.
1. Every integer leaves a remainder of 0, 1, 2 or 3 when divided by 4. That is only 4 possible remainders.
2. With 5 integers and 4 remainder classes, by the pigeonhole principle at least two integers share the same remainder.
3. If two integers a and b have the same remainder mod 4, then a − b is divisible by 4.
Hence two of the five always differ by a multiple of 4.
For positive reals a and b, show that a/b + b/a ≥ 2.
1. By the AM–GM inequality, for any two positive numbers x and y, (x + y)/2 ≥ √(xy).
2. Let x = a/b and y = b/a. Then xy = (a/b)(b/a) = 1, so √(xy) = 1.
3. Therefore (a/b + b/a)/2 ≥ 1, which gives a/b + b/a ≥ 2, with equality when a = b.
In how many ways can 8 people be seated around a round table? (Rotations counted as the same arrangement.)
1. For n people in a line, there are n! arrangements. Around a circle, rotations are equivalent, so we fix one person's seat.
2. That leaves (n − 1)! ways to arrange the rest. Here n = 8.
3. (8 − 1)! = 7! = 5040.
Answer: 5040.
Find the last digit of 7¹⁰⁰.
1. The last digit of powers of 7 cycles: 7¹ = 7, 7² = 49 (9), 7³ = 343 (3), 7⁴ = 2401 (1), then repeats every 4.
2. Find 100 mod 4 = 0, which corresponds to the 4th position in the cycle (last digit 1).
Answer: 1.
Show that in any group of 13 people, at least two share a birth month.
1. There are 12 possible birth months (the pigeonholes).
2. With 13 people (the pigeons) and only 12 months, by the pigeonhole principle at least one month must contain two or more people.
Hence at least two people share a birth month.
In a triangle, two angles measure 50° and 60°. A point lies such that the third angle is bisected. Find each half of the bisected angle.
1. Angles of a triangle sum to 180°, so the third angle = 180° − 50° − 60° = 70°.
2. The bisector splits 70° into two equal parts: 70° ÷ 2 = 35°.
Answer: 35° each.
Prove that the product of any three consecutive integers is divisible by 6.
1. Among any three consecutive integers, at least one is even, so the product is divisible by 2.
2. Among any three consecutive integers, exactly one is divisible by 3, so the product is divisible by 3.
3. Divisible by both 2 and 3, and since gcd(2,3) = 1, the product is divisible by 2 × 3 = 6.
Hence the product is always divisible by 6.
For positive reals a, b, show that (a + b)(1/a + 1/b) ≥ 4.
1. Expand: (a + b)(1/a + 1/b) = 1 + a/b + b/a + 1 = 2 + (a/b + b/a).
2. By AM–GM, a/b + b/a ≥ 2 for positive reals.
3. So the expression ≥ 2 + 2 = 4, with equality when a = b.
Hence (a + b)(1/a + 1/b) ≥ 4.
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