These questions are written as interview-style practice rather than as a prediction of any university's admissions material. Ask another person to reveal the hints one at a time if possible. A useful attempt includes false starts, provided they are explained and reassessed.

How to use the problems Spend a few minutes exploring aloud before opening a hint. State what is known, try small cases, describe any pattern and say what would count as a proof. After finishing, explain the solution again in a shorter form.

1. Two positive roots

For which real values of t does the equation

x² − 2tx + t + 2 = 0

have two distinct positive real roots?

Hint 1

Separate “real and distinct” from “both positive”. What does the discriminant tell you?

Hint 2

If the roots are r and s, use r + s and rs rather than solving explicitly.

Solution outline

The discriminant is 4(t² − t − 2), so distinct real roots require t < −1 or t > 2. By Vieta, r + s = 2t and rs = t + 2. Two positive roots require a positive sum and product. Combining all conditions gives t > 2.

2. A divisibility claim

Prove that n³ − n is divisible by 6 for every integer n.

Hint 1

Factorise the expression completely.

Hint 2

What must occur among any three consecutive integers?

Solution outline

Write n³ − n = n(n − 1)(n + 1). These are three consecutive integers. At least one is even and exactly one is divisible by 3, so their product is divisible by both 2 and 3 and therefore by 6.

3. The largest rectangle

A rectangle has fixed perimeter P. Which dimensions give the greatest area, and how do you know the maximum is global?

Hint 1

If one side is x, express the other side and the area in terms of x.

Hint 2

Complete the square. This also answers the question about a global maximum.

Solution outline

The second side is P/2 − x, so A = x(P/2 − x) = P²/16 − (x − P/4)². The squared term is never negative, so A ≤ P²/16, with equality when both sides are P/4. The rectangle of greatest area is therefore a square.

4. A sequence approaching √2

Let x₁ > √2 and define

xn+1 = ½(xn + 2/xn).

Show that the sequence converges and find its limit.

Hint 1

Compare xn+1 first with √2 and then with xn.

Hint 2

Factor the numerators of xn+1 − √2 and xn+1 − xn.

Solution outline

If xn > √2, then xn+1 − √2 = (xn − √2)²/(2xn) > 0, so the sequence stays above √2. Also xn+1 − xn = (2 − xn²)/(2xn) < 0, so it decreases. It is decreasing and bounded below, hence convergent. If its limit is L > 0, then L = ½(L + 2/L), giving L² = 2 and therefore L = √2.

5. A consecutive sum

Given any n integers a₁, a₂, …, an, prove that some non-empty consecutive block has sum divisible by n.

Hint 1

Consider the n partial sums sk = a₁ + ··· + ak.

Hint 2

Look at the remainders of these partial sums when divided by n. What happens if none has remainder 0?

Solution outline

If some sk is divisible by n, the block a₁ + ··· + ak works. Otherwise the n partial sums occupy only the n − 1 non-zero remainder classes, so two have the same remainder. If si and sj have the same remainder with i < j, then sj − si = ai+1 + ··· + aj is divisible by n.

Questions to ask after each solution

  • What was the first observation that genuinely reduced the problem?
  • Which assumption or example was useful, and which was a distraction?
  • Where did the argument move from evidence to proof?
  • Could the solution be explained coherently without repeating every calculation?
  • What natural extension would you investigate next?

These independent practice problems are not Cambridge admissions questions and are not endorsed by the University of Cambridge or any college. They do not use confidential interview material.