Problem-solving techniques for olympiad maths

An olympiad problem comes with no ready-made method. What it has instead are techniques that work across many problems at once. Here are the 11 techniques that come up most often, each with a problem, a drawing and a solution.

Try the problem with your child before you open the solution. What matters most is noticing which technique worked, so as to recognise it next time.

Where to begin any problem

Before you choose a technique, it helps to go through four steps. The mathematician George Pólya set them out in his book How to Solve It (1945).

Understand the problem
Say it in your own words. What do you know? What do you have to find?
Make a plan
What does it remind you of? Which technique on this page could you try?
Carry it out
Step by step, writing down what each step gives.
Look back
Put the answer back into the problem. Does everything fit? Could there be another answer?

See the problem

First the problem has to be understood. These techniques turn words into a picture, a table or a small example.

Draw the problem

A picture or a diagram turns words into something you can look at. Lines, dots and arrows are enough; it does not have to be pretty.

When it helps: “so many more than”, ages, journeys, parts of a whole.

Problem · Grades 1–2

A brother is 4 years older than his sister. Together they are 16. How old is the brother?

Show the solution

Draw the sister's age as a bar, and the brother's as the same bar and 4 years more.
Take away the extra 4 years: 16 − 4 = 12. That is two equal bars.
One bar is 12 ÷ 2 = 6 years: that is the sister. The brother is 6 + 4 = 10.

The diagram shows what is hard to hold in your head: without the extra four years, brother and sister have the same.

Answer The brother is 10.

In the topics:Calendar and age

Make a table

When there are many clues, they are hard to keep in your head. A table keeps them for you: every clue is a cross, and where a row has one free cell left, that is the answer.

When it helps: several people, several things, and clues with “not”.

Problem · Grades 1–2

Anya, Borya and Vera have a cat, a dog and a parrot, one pet each. Anya does not keep the cat. Borya is afraid of dogs. The cat does not live with Borya. Who has which pet?

Show the solution

Put a cross for each clue: Anya, not the cat; Borya, not the dog; the cat, not with Borya.
Borya has one cell left: the parrot. So neither Anya nor Vera has the parrot.
Anya is left with the dog, and Vera with the cat.

The small number in a cell is the step of the solution at which it was marked.

Answer Anya has the dog, Borya the parrot, Vera the cat.

In the topics:Ordering

Start with small numbers

A special case

When the numbers are big and you cannot see where to start, solve the same problem with small numbers. Spot the rule, then apply it to the big ones.

When it helps: big numbers, “a hundred”, “a thousand”, long rows.

Problem · Grades 3–4

Posts are put along a road 100 metres long, one every 5 metres, at both ends too. How many posts are needed?

Show the solution

You cannot draw a hundred metres, so take a road of 10 m. Posts at 0, 5 and 10 m: 3 posts for 2 gaps.
For 15 and 20 m it is the same: there is always one post more than there are gaps.
On 100 m there are 100 ÷ 5 = 20 gaps, so there are 21 posts.

The rule shows on a short road already; then it is enough to apply it to a hundred metres.

Answer 21 posts.

In the topics:Gaps and boundaries

Find the answer

Once it is clear what is going on, these techniques lead to the answer without losing anything on the way.

List systematically

Listing at random is bound to miss something or count it twice. Listing systematically goes in order: first by one feature, then by the next. A tree is handy for it.

When it helps: “how many ways”, “how many numbers”, “write out every option”.

Problem · Grades 1–2

How many different two-digit numbers can be made from the digits 2, 4, 6 and 8, if no digit is used twice in a number?

Show the solution

Choose an order: first the tens digit, then the units.
The tens digit is 2, 4, 6 or 8. Each leaves three others for the units.
Write them out branch by branch: 24, 26, 28, 42, 46, 48, 62, 64, 68, 82, 84, 86. None is missed, and none is repeated.

The tree lets nothing slip: each first digit has branches of its own.

Answer 12 numbers.

In the topics:Enumeration

Work from the end

Working backwards

When you know how the story ends and are asked how it began, go backwards. Turn every action into its opposite: “ate” into “gave back”, “multiplied” into “divided”.

When it helps: you know what came out at the end, and have to find what there was at the start.

Problem · Grades 3–4

There were some apples in a bowl. Tanya ate half of them. Then Petya ate half of what was left, and one apple more. 2 apples were left. How many apples were there at the start?

Show the solution

Start from what you know: at the end there were 2 apples.
Before Petya: give back the one apple, 2 + 1 = 3. That was half, so there were 6.
Before Tanya: 6 was half, so there were 12. Check: 12 → 6 → 6 − 3 − 1 = 2.

Go from right to left and do everything the other way round: “ate half” becomes “times 2”, “ate one more” becomes “plus 1”.

Answer 12 apples.

In the topics:Arithmetic with a trick Weighing and pouring Games with a winning strategy

Suppose they are all the same

The assumption method

When a problem has things of two kinds, imagine they are all of one kind. Count how far the answer is off, and you will see how many have to be swapped.

When it helps: two kinds of things and a known total: heads and legs, coins of two values.

Problem · Grades 3–4

Hens and rabbits are walking in a yard. Together they have 10 heads and 28 legs. How many rabbits are in the yard?

Show the solution

Suppose all 10 are hens. Then there are 10 × 2 = 20 legs.
In fact there are 28 legs: 8 more.
Each rabbit in place of a hen adds 2 legs: 8 ÷ 2 = 4 rabbits. Check: 4 × 4 + 6 × 2 = 28.

First count as if they were all the same, then change exactly as many as it takes to fit.

Answer 4 rabbits.

Look for a pattern

Look at the first few steps and find what repeats. The repeating piece lets you jump far ahead without drawing everything in between.

When it helps: patterns, rows, the calendar: “which will be the twentieth”, “which will be the hundredth”.

Problem · Grades 1–2

Flags hang in order: red, yellow, green, blue, then red, yellow, green, blue again, and so on. What colour is the 23rd flag?

Show the solution

Find the piece that repeats: red, yellow, green, blue, four flags.
23 flags are 5 whole fours, which is 20 flags, and 3 more.
The 23rd is the third of its four, and the third is always green.

There is no need to draw twenty-three flags: it is enough to see which place in its four the flag takes.

Answer Green.

In the topics:Parity and alternation Calendar and age

Prove it

At an olympiad the question is often not “how many” but “can it happen”. Then counting is not enough: you have to explain why it cannot be otherwise.

Suppose the opposite

Proof by contradiction

To prove that something cannot be, suppose it can, and follow the reasoning until you run into a contradiction. Almost every knights-and-liars problem is solved this way.

When it helps: “can it happen”, “prove it is impossible”, who is telling the truth.

Problem · Grades 3–4

On an island live knights, who always tell the truth, and liars, who always lie. Anya said: “Borya is a liar.” Borya said: “Anya and I are both knights.” Who is who?

Show the solution

Suppose Borya is a knight. Then his words are true, and Anya is a knight too. But Anya, a knight, said that Borya is a liar. A contradiction.
So Borya is a liar, and his words are false: Anya and he are not both knights. That fits.
Anya said “Borya is a liar”, and that is true. Only a knight says true things, so Anya is a knight.

One path leads to a dead end, so the other one is right.

Answer Anya is a knight, and Borya is a liar.

In the topics:Knights and liars

Watch the parity

Sometimes you need not know everything: it is enough to know whether a number is even or odd. If something switches between even and odd at every step, the number of steps tells you at once where you can end up and where you cannot.

When it helps: jumps, moves, swaps, “every other one”, “can it turn out”.

Problem · Grades 3–4

A grasshopper sits at zero on a number line. Every jump takes it exactly 1 to the left or to the right. Can it be back at zero after 5 jumps?

Show the solution

Colour the cells in turn: even and odd.
Every jump changes the colour: from an even cell to an odd one, and back.
After 5 jumps, an odd number, the grasshopper is sure to be on an odd cell, and zero is even.

There is no need to follow where exactly the grasshopper went: it is enough to follow the colour of its cell.

Answer No, it cannot.

In the topics:Parity and alternation

Pigeons and pigeonholes

The pigeonhole principle

If there are more pigeons than holes, some hole holds at least two pigeons. It sounds obvious, yet it proves things that seem impossible to check.

When it helps: “prove there must be”, “at least two”, “for certain”.

Problem · Grades 3–4

How many times must you throw a die to be certain that some number comes up at least twice?

Show the solution

A die has 6 numbers: those are 6 “holes”. The throws are the “pigeons”.
After 6 throws the numbers can still all be different: 1, 2, 3, 4, 5, 6.
The seventh throw has no free hole left: some number comes up a second time. So 7 throws.

Even if the first six throws all differ, the seventh has to repeat one of them.

Answer 7 throws.

In the topics:Pigeonhole principle

Mirror the moves

A symmetric strategy

In two-player games the winner is often the one who can mirror the other's moves. Then you always have a reply, as long as your opponent has a move.

When it helps: two-player games: “who wins with the right play”.

Problem · Grades 5–6

A strip has 9 cells. Two players take turns to put a counter in an empty cell that is not next to a filled one. Whoever cannot put a counter loses. Where should the first player put the first counter to win, however the other plays?

Show the solution

With the first move, put a counter in the middle, the fifth cell.
Answer every move of the opponent with its mirror: the cell opposite, across the middle.
If the opponent's cell was free and not next to a counter, so is its mirror. So the opponent runs out of moves first.

A strip has one middle, and every other cell has a mirror. That is why the first player never runs out of moves.

Answer In the middle, the fifth cell, and then mirror every move.

In the topics:Games with a winning strategy

Which technique to try

A hint from the words of the problem. Not a rule: many problems give way to two techniques at once.

If the problem has……try

“so many more than”, ages, parts of a whole: 01Draw the problem
many people, things and clues with “not”: 02Make a table
big numbers, and no idea where to start: 03Start with small numbers
“how many ways”, “how many numbers”: 04List systematically
the end is known, and the question is the start: 05Work from the end
two kinds of things and a total: 06Suppose they are all the same
something repeats: “which will be the twentieth”: 07Look for a pattern
“can it happen”, “prove it is impossible”: 08Suppose the opposite 09Watch the parity
“for certain”, “at least two”: 10Pigeons and pigeonholes
a game for two: “who wins”: 11Mirror the moves

Try the techniques on problems

They come in handy in the problems of every MathTrail topic.

Add to Claude All topics

Problem-solving techniques for olympiad maths

Where to begin any problem

Understand the problem

Make a plan

Carry it out

Look back

See the problem

Draw the problem

Make a table

Start with small numbers

Find the answer

List systematically

Work from the end

Suppose they are all the same

Look for a pattern

Prove it

Suppose the opposite

Watch the parity

Pigeons and pigeonholes

Mirror the moves

Which technique to try

Try the techniques on problems