IOQM-level practice

IOQM algebra practice set

8 original problems at the level of the first olympiad stage: roots and Vieta, identities, AM–GM, functional equations and telescoping sums. Each answer is a whole number. Try each one before opening the hints; the full solution explains why the method works and where the idea leads.

The first 3 are free with full solutions. Hints, answer checking and full solutions for problems marked ‘With a plan’ are included with every A Level, IB, IGCSE and CBSE plan. See plans. These are our own problems, written for this site; none is a past IOQM or RMO question or a reworded one.

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Problem S157

AlgebraShort answer

The roots of x² − 7x + 3 = 0 are p and q. Find p³ + q³.

Hint

You do not need the roots: use p + q and pq.

Second hint

p³ + q³ = (p + q)³ − 3pq(p + q).

Full worked solution

Answer: 280

Vieta: p + q = 7 and pq = 3.
Expand (p + q)³ = p³ + q³ + 3pq(p + q).
So p³ + q³ = 7³ − 3 × 3 × 7 = 343 − 63.
p³ + q³ = 280.

Why this works: Every symmetric expression in the roots can be written in terms of their sum and product, so the messy roots never appear.

Where it leads: Newton’s identities generalise this to power sums of the roots of any polynomial.

Strategy: Symmetry

Problem S158

AlgebraShort answer

A real number x satisfies x + 1/x = 5. Find x⁴ + 1/x⁴.

Hint

Square both sides.

Second hint

x² + 1/x² = 23. Square again.

Full worked solution

Answer: 527

Square: x² + 2 + 1/x² = 25, so x² + 1/x² = 23.
Square again: x⁴ + 2 + 1/x⁴ = 529.
So x⁴ + 1/x⁴ = 529 − 2.
x⁴ + 1/x⁴ = 527.

Why this works: Squaring x + 1/x always produces a constant middle term 2, so the powers climb without ever solving for x.

Where it leads: The values xⁿ + 1/xⁿ satisfy a linear recurrence; with x = e^iθ they become 2cos nθ and Chebyshev polynomials.

Strategy: Spot the pattern and generalise

Problem S159

AlgebraShort answer

An arithmetic progression has first term 5 and common difference 6. How many terms must be added to give a sum of 1240?

Hint

S_n = n/2 × (2a + (n − 1)d).

Second hint

Here S_n simplifies to n(3n + 2). Solve n(3n + 2) = 1240 for a positive integer n.

Full worked solution

Answer: 20

S_n = n/2 × (10 + 6(n − 1)) = n(3n + 2).
Solve 3n² + 2n − 1240 = 0: (n − 20)(3n + 62) = 0.
The positive root is n = 20; check 20 × 62 = 1240.
So 20 terms.

Why this works: The sum of an AP is a quadratic in n, so ‘how many terms’ is a quadratic equation with one meaningful root.

Where it leads: Sums of polynomial sequences are polynomials of one higher degree: the discrete version of integration.

Strategy: Working backwards