How to Win a Number Game Before It Starts
Learn to solve arrangement puzzles from clues, use odd and even to prove something is impossible, fill number grids that always work, and find a winning strategy.
Can you know the answer to a puzzle before solving it?
Often, yes. Reasoning about whether numbers are odd or even — their parity — can tell you an outcome is impossible without any calculation at all. The same kind of thinking cracks arrangement puzzles and decides who wins a two-player game.
This page covers everything in the CBSE Class 7 Mathematics chapter on number play: working out arrangements from clues, parity arguments, number grids, and winning strategies.
This page covers everything in the CBSE Class 7 Mathematics chapter on number play: working out arrangements from clues, parity arguments, number grids, and winning strategies.
How do you work out an arrangement from numerical clues?
Start with the clue that leaves no choice, and let it fix one position; the rest usually follows.
Five children stand in a line, and each announces how many taller children stand ahead of them. Suppose the announcements, front to back, are 0, 0, 1, 2, 1.
The first child says 0 — with nobody ahead, that tells you nothing yet. The second also says 0, so nobody ahead is taller, meaning the second is taller than the first. The third says 1, so exactly one of the two ahead is taller. The fourth says 2, so both of some pair ahead are taller.
Work through the line assigning relative heights one child at a time, checking each new claim against everyone already placed. If a claim contradicts what you have built, an earlier choice was wrong — go back rather than forcing it.
For example, in a school assembly line you could reconstruct the whole height order from such statements without measuring anyone.
The method is always the same: find the most restrictive clue, act on it, then test every later clue against what is already fixed.
Five children stand in a line, and each announces how many taller children stand ahead of them. Suppose the announcements, front to back, are 0, 0, 1, 2, 1.
The first child says 0 — with nobody ahead, that tells you nothing yet. The second also says 0, so nobody ahead is taller, meaning the second is taller than the first. The third says 1, so exactly one of the two ahead is taller. The fourth says 2, so both of some pair ahead are taller.
Work through the line assigning relative heights one child at a time, checking each new claim against everyone already placed. If a claim contradicts what you have built, an earlier choice was wrong — go back rather than forcing it.
For example, in a school assembly line you could reconstruct the whole height order from such statements without measuring anyone.
The method is always the same: find the most restrictive clue, act on it, then test every later clue against what is already fixed.
What are the rules for odd and even results?
Parity simply means whether a number is odd or even, and it follows fixed rules.
For addition and subtraction:
- even + even = even
- odd + odd = even
- even + odd = odd
Subtraction behaves identically, so odd − odd is even.
For multiplication:
- even × anything = even
- odd × odd = odd
So a product is odd only when every factor is odd; a single even factor makes the whole product even.
For example, adding up the page numbers 1 to 10 gives an odd count of odd numbers, and the parity rules let you predict whether the total is odd or even before adding a thing.
The most useful consequence is about sums of odd numbers: adding two odds gives an even, so adding an even number of odds always gives an even total, and an odd number of odds gives an odd total.
For addition and subtraction:
- even + even = even
- odd + odd = even
- even + odd = odd
Subtraction behaves identically, so odd − odd is even.
For multiplication:
- even × anything = even
- odd × odd = odd
So a product is odd only when every factor is odd; a single even factor makes the whole product even.
For example, adding up the page numbers 1 to 10 gives an odd count of odd numbers, and the parity rules let you predict whether the total is odd or even before adding a thing.
The most useful consequence is about sums of odd numbers: adding two odds gives an even, so adding an even number of odds always gives an even total, and an odd number of odds gives an odd total.
How does parity prove something is impossible?
If a task requires an odd result but the arithmetic can only produce an even one, the task cannot be done — and parity proves it without trying a single case.
Can you pick three odd numbers that add up to 30? Three odds added together give an odd total, because odd + odd = even and even + odd = odd. But 30 is even, so it is impossible, no matter which odd numbers you choose.
Another: can the numbers 1 to 9 be split into two groups with equal sums? Their total is 45, which is odd, and an odd total cannot split into two equal whole-number halves. Impossible again.
This is a genuinely different kind of answer from "I tried and could not find one". Trying examples can never prove impossibility — there is always another combination untested — whereas a parity argument settles every case at once.
That is what makes it powerful, and why questions ask you to explain why, not merely to answer yes or no.
Can you pick three odd numbers that add up to 30? Three odds added together give an odd total, because odd + odd = even and even + odd = odd. But 30 is even, so it is impossible, no matter which odd numbers you choose.
Another: can the numbers 1 to 9 be split into two groups with equal sums? Their total is 45, which is odd, and an odd total cannot split into two equal whole-number halves. Impossible again.
This is a genuinely different kind of answer from "I tried and could not find one". Trying examples can never prove impossibility — there is always another combination untested — whereas a parity argument settles every case at once.
That is what makes it powerful, and why questions ask you to explain why, not merely to answer yes or no.
How do you fill a grid so the sums always work?
Use letter-numbers to show the arrangement works for any starting value, rather than checking one example.
In a simple grid task, place the numbers so that each row, column and diagonal adds to the same total. With the numbers 1 to 9, that total must be
because the three rows together use every number once and must share the total equally. Knowing the answer is 15 before placing anything makes the filling far easier, and 5 must sit in the centre since it appears in four of the lines.
For a pattern that always works, algebra proves it. If three consecutive numbers are , and , their sum is
So the total is always three times the middle number — true for every starting value, which no list of examples could establish.
Work out the required total first. Placing numbers by trial without knowing what they must add to is what turns a two-minute question into a ten-minute one.
In a simple grid task, place the numbers so that each row, column and diagonal adds to the same total. With the numbers 1 to 9, that total must be
because the three rows together use every number once and must share the total equally. Knowing the answer is 15 before placing anything makes the filling far easier, and 5 must sit in the centre since it appears in four of the lines.
For a pattern that always works, algebra proves it. If three consecutive numbers are , and , their sum is
So the total is always three times the middle number — true for every starting value, which no list of examples could establish.
Work out the required total first. Placing numbers by trial without knowing what they must add to is what turns a two-minute question into a ten-minute one.
How do you find a winning strategy in a number game?
Work backwards from the winning position, and look for multiples.
A common game: players take turns removing 1, 2 or 3 matchsticks from a pile of 20, and whoever takes the last one wins.
Work back. If you leave your opponent 4 sticks, they cannot win — whatever they take (1, 2 or 3), you take the rest. The same holds at 8, 12, 16 and 20: every multiple of 4 is a losing position for whoever must move.
Since the pile starts at 20, a multiple of 4, the second player wins by always restoring a multiple of 4. If the first player takes 1, the second takes 3; if the first takes 2, the second takes 2.
The key number comes from the rules: you may take 1 to 3, so is the magic step.
Parity is the same idea in its simplest form — a game where players alternate and only the count's oddness matters is a game decided before the first move.
A common game: players take turns removing 1, 2 or 3 matchsticks from a pile of 20, and whoever takes the last one wins.
Work back. If you leave your opponent 4 sticks, they cannot win — whatever they take (1, 2 or 3), you take the rest. The same holds at 8, 12, 16 and 20: every multiple of 4 is a losing position for whoever must move.
Since the pile starts at 20, a multiple of 4, the second player wins by always restoring a multiple of 4. If the first player takes 1, the second takes 3; if the first takes 2, the second takes 2.
The key number comes from the rules: you may take 1 to 3, so is the magic step.
Parity is the same idea in its simplest form — a game where players alternate and only the count's oddness matters is a game decided before the first move.
Exam tip
Exam tip: explaining why, not just answering yes or no
Most marks in this chapter sit in the justification. "No, it is not possible" earns almost nothing on its own.
For an impossibility question, give the parity chain: three odd numbers always add to an odd total, but 30 is even, so it cannot be done. Three short clauses — what the arithmetic forces, what the target is, why they conflict.
For a strategy question, name the position you aim to leave and say why it wins: always leave a multiple of 4, because whatever your opponent takes from 1 to 3, you can take the rest and restore the next multiple of 4.
And never answer an impossibility question by listing failed attempts. Examples show you tried; only the argument shows it cannot be done.
For an impossibility question, give the parity chain: three odd numbers always add to an odd total, but 30 is even, so it cannot be done. Three short clauses — what the arithmetic forces, what the target is, why they conflict.
For a strategy question, name the position you aim to leave and say why it wins: always leave a multiple of 4, because whatever your opponent takes from 1 to 3, you can take the rest and restore the next multiple of 4.
And never answer an impossibility question by listing failed attempts. Examples show you tried; only the argument shows it cannot be done.
Did you know
Why can three odd numbers never add to an even total?
Pair them up and watch the parity. The first two odds add to an even number, and adding the third odd to an even gives an odd.
So three odds always end odd, whatever the numbers are. The same reasoning generalises: an odd count of odd numbers gives an odd total, and an even count gives an even one — which is why the number of odd numbers matters more than their size.
So three odds always end odd, whatever the numbers are. The same reasoning generalises: an odd count of odd numbers gives an odd total, and an even count gives an even one — which is why the number of odd numbers matters more than their size.
Key takeaways
Number play: quick revision
- Solve arrangement puzzles by acting on the most restrictive clue first, then testing every later clue against what is already fixed.
- Parity rules: even+even and odd+odd are even, even+odd is odd; a product is odd only if every factor is odd.
- A parity argument proves impossibility for all cases at once — three odd numbers can never total 30, because three odds always give an odd sum.
- In grid tasks, calculate the required total first ( for 1 to 9), and use letter-numbers to show a pattern holds for every value.
- Find winning strategies by working backwards to the losing positions, which in a take-1-to-3 game are the multiples of 4.
You will remember all of this far better after answering five questions on it than after reading it twice.
- Parity rules: even+even and odd+odd are even, even+odd is odd; a product is odd only if every factor is odd.
- A parity argument proves impossibility for all cases at once — three odd numbers can never total 30, because three odds always give an odd sum.
- In grid tasks, calculate the required total first ( for 1 to 9), and use letter-numbers to show a pattern holds for every value.
- Find winning strategies by working backwards to the losing positions, which in a take-1-to-3 game are the multiples of 4.
You will remember all of this far better after answering five questions on it than after reading it twice.