← All Learn lessons · Start adaptive practice · Practise on paper

You will learn: Use odd and even numbers to rule out impossible outcomes and build a short proof.

Before you start: Whole-number arithmetic; multiplying brackets is explained below.

Start by pairing things

Put counters into pairs. If none is left over, the number is even. If one is left over, it is odd. This odd-or-even property is called parity. It often answers an Olympiad question without a long calculation.

An odd number has one unpaired counter.

For any integer k, 2k is even and 2k + 1 is odd. These descriptions include negative integers: −3 = 2(−2) + 1 is odd. Parity is a property of integers; it does not classify a fraction such as 1/2.

What happens when numbers combine?

Adding an even number preserves parity. Adding an odd number changes it. In particular, odd + odd is even: the two leftover counters make a pair. A product is odd only when every factor is odd; one even factor makes the product even.

Here is the algebra behind odd + odd. If the numbers are 2a + 1 and 2b + 1, their sum is 2a + 2b + 2 = 2(a + b + 1). It is twice an integer, so it is even. The letters a and b stand for arbitrary integers, making this a proof for every pair.

Worked example 1: a long sum

Is 3 + 5 + 7 + ⋯ + 21 odd or even? First count the terms: they are 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, so there are ten. Pair the ten odd terms into five pairs. Each pair has an even sum, and a sum of even numbers is even. We did not need the total.

Worked example 2: an impossible finish

A number starts at 8. Every move adds 4 or subtracts 2. Can it become 15? Both moves change the number by an even amount, so its parity never changes. It starts even and stays even. Since 15 is odd, reaching it is impossible. A property that remains unchanged during every allowed move is called an invariant.

A common trap

Even is not the same as divisible by 4. The number 6 is even but not divisible by 4. Parity distinguishes only two classes; use remainders when you need finer information. Also, a few numerical examples can suggest a pattern, but a statement about every integer needs a general argument.

Practise at your next step

A six-question session chooses from nine questions. Two correct answers in a row without hints move you up a level; an incorrect answer brings a simpler next question where one is available. Hints keep you at the same level. This is a practice suggestion, not an exam score or proof of mastery.

Interactive practice loads here. You can also use the complete question set below.

Write a proof of your own

Prove that n(n + 1) is even for every integer n.

Write your reasoning on paper before comparing. The practice checker does not grade a written proof.

Compare your proof with a full solution

The integers n and n + 1 are consecutive. If n is even, the product has an even factor. If n is odd, n + 1 is even, so the product again has an even factor. These two cases cover every integer n. Therefore n(n + 1) is always even.

Check: did you state the assumptions, explain the key step, and reach the requested conclusion?

All nine practice questions, hints and solutions

This complete set works without the interactive practice. Hide each solution until you have made an attempt.

1. Which integer is odd?

Foundation

  1. 0
  2. −8
  3. −5
  4. 12
Hint

Try writing each option as twice an integer, or twice an integer plus 1.

Answer and explanation

−5. −5 = 2(−3) + 1, so −5 is odd. The other options are multiples of 2. Negative integers can be odd or even.

2. What is the parity of 27 + 35?

Foundation

  1. Odd
  2. Even
  3. It depends
  4. Neither
Hint

Both terms have one leftover counter after pairing.

Answer and explanation

Even. 27 and 35 are odd. Their two leftovers form a pair, so their sum is even. Indeed, 27 + 35 = 62.

3. What is the parity of 13 × 18?

Foundation

  1. Odd
  2. Even
  3. It depends on the order
  4. Neither
Hint

One factor is a multiple of 2.

Answer and explanation

Even. 18 = 2 × 9, so 13 × 18 = 2 × (13 × 9). The product is even.

4. What is the parity of the sum of seven odd integers?

Core

  1. Always even
  2. Always odd
  3. Sometimes even, sometimes odd
  4. Impossible to decide without their values
Hint

Pair six of the terms. What is left?

Answer and explanation

Always odd. Six odd terms can be grouped into three pairs, each with an even sum. Adding the seventh odd term gives an odd total. This also works for negative odd terms.

5. For an integer n, which expression is always even?

Core

  1. n + 1
  2. n² + 1
  3. n(n + 1)
  4. 2n + 1
Hint

Two consecutive integers have opposite parity.

Answer and explanation

n(n + 1). Among n and n + 1, one is even. Their product is therefore even. The other expressions are not always even: n = 2 disproves the first two, and 2n + 1 is always odd.

6. Start at 9. Each move adds 6 or subtracts 4. Which target is impossible?

Core

  1. 5
  2. 11
  3. 15
  4. 12
Hint

Each allowed change is even.

Answer and explanation

12. Starting from the odd number 9, adding or subtracting an even number keeps the result odd. Thus 12 is impossible. The other targets are reachable: 5 = 9 − 4, 11 = 9 + 6 − 4, and 15 = 9 + 6.

7. If n² is odd for an integer n, what must be true?

Stretch

  1. n is even
  2. n is odd
  3. n is positive
  4. n is prime
Hint

Test what would happen if n were even.

Answer and explanation

n is odd. If n were even, n = 2k for an integer k, giving n² = 4k², which is even. This contradicts the given odd square. Therefore n is odd. Negative odd integers are allowed.

8. Five integers have an odd sum. Which count of odd integers among them is possible?

Stretch

  1. 0
  2. 2
  3. 3
  4. 4
Hint

Even terms do not change the parity of a sum.

Answer and explanation

3. The sum has the same parity as the number of odd terms. An odd total requires 1, 3 or 5 odd terms. Of the options, only 3 works; for example 1 + 1 + 1 + 2 + 2 = 7.

9. Start at 0 and make exactly seven moves, each adding 1 or subtracting 1. Can the final value be 0?

Stretch

  1. Yes, always
  2. Yes, with the right order
  3. No, the final value must be odd
  4. No, the final value must exceed 7
Hint

Each move changes parity.

Answer and explanation

No, the final value must be odd. Each of the seven moves changes even to odd or odd to even. After an odd number of changes, the final value is odd. Since 0 is even, it cannot be the final value.

Where to go next

Move on to the mixed proof-practice plan · Explore the wider Olympiad topic map

Teaching examples prepared for this guide, not attributed to a contest paper. Familiar elementary problems may appear in other learning materials. Report an unclear step or an error.