PrepShorts · Study sheet · Class 7 Mathematics · Chapter 6, Number Play
Chapter 6 · Number Play
The parity of a sum or product is fixed before you compute it
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A table with seven rows, three filled in and four blank. The blank ones are the subtraction rules, and the explanation refuses to just hand them over.
The idea
Parity survives arithmetic. Add, subtract or multiply, and the parity of the answer is already decided by the parities of the inputs — so you can work in a world with only two numbers in it, and you can call a 135 × 654 grid odd or even without ever performing the multiplication. The sharp part is the case that is not decided: 3n + 4 is sometimes odd and sometimes even, and telling those expressions apart from the ones with a fixed parity is what the section is really teaching. An even coefficient hides its variable from parity altogether, and then the constant alone decides; an odd coefficient hands the decision back to n.
What you should be able to do
- Complete the parity table for addition and for subtraction, and justify each row from the dot picture rather than from examples
- State the parity of a sum from nothing but how many odd numbers it contains
- Say whether an m × n grid holds an odd or an even count of unit squares, from m and n alone and without multiplying
- Explain why one even side is enough to make a rectangular count even
- Evaluate an expression at several values of the letter-number and read its parity off the table
- Sort expressions into those with a fixed parity for every value and those whose parity changes, and say which feature of the expression decides that
- Write the nth even number and the nth odd number as expressions, and say which numbers each expression reaches
- Judge a claim of the form "this expression always gives odd numbers" as a claim about every value, and refute it with one counterexample where it is false
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| parity | whether a number is even or odd | printed in §6.2, p.131 and used throughout pp.131–133 |
| grid | a rectangle ruled into unit squares, described by its two dimensions | printed in "Small Squares in Grids", §6.2, p.131 |
| small squares | the unit cells a grid is ruled into; their count is the product of the two dimensions | printed in "Small Squares in Grids", §6.2, p.131 |
| expression | a rule such as 3n + 4 that turns a letter-number into a value | printed in "Parity of Expressions", §6.2, p.132 |
| letter-number | a letter standing for a number, here usually a position in a sequence | printed in §6.2, p.132 |
| nth term | the value an expression takes at position n | printed in §6.2, p.132 |
| formula | an expression written to produce a term at any position | printed in §6.2, p.133 |
| value of an expression | what the expression evaluates to once the letter-number is fixed | printed in Part I, §4.2, p.83 and §4.3, p.87. Not printed in this chapter: the §6.2 table on p.132 heads its middle column for the one expression in front of it, not for expressions in general — checked against p.132. An explanation may use the general term, but should not show it as this chapter's wording. |
| fixed parity | the explanation's phrase for an expression that is odd for every value, or even for every value | the explanation's phrasing; not printed in this chapter |
Where people slip up
- "Subtraction will need its own set of rules." It will not. Taking away a block of pairs never touches a leftover, so subtraction copies addition row for row. Deriving that, instead of memorising four more lines, is section 2.
- "An odd number times an even number could be odd." Never: an even factor means the whole array pairs up along one direction. The grid picture makes this visible in a way the times table does not — one even side, and the cells fall into columns of two.
- "3n + 4 is even, because 4 is even." It is even only when 3n is. The coefficient 3 is odd, so 3n takes the parity of n and the expression flips with it. The printed table is built to catch exactly this.
- "An expression that always gives even values gives all the even values." 6k + 2 is the counterexample the chapter prints. Always-even is a statement about the outputs it produces; listing all the evens is a much stronger claim about the ones it does not miss.
- "2p + 1 and 2q − 1 are the same thing, so both list every odd number." As expressions they differ only by where the count starts, and in this chapter a letter-number counts positions beginning at one. Under that convention 2q − 1 starts at 1 while 2p + 1 starts at 3 and never reaches 1. Say the convention out loud; without it the claim is neither true nor false.
- "Parity checks out, so the answer is yes." Matching parity only means the parity argument has nothing to say. The loose-sheets question on p.144 is exactly this trap — see Notes, where the finer argument is set out.
- "77 is odd, so the bulb ends up odd." A bulb is not a number. What the parity of 77 controls is whether the state has been flipped an even or an odd number of times, and only that.
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Worked answers to this chapter’s exercises · this video explains Figure it Out · 6.2 Q1, Figure it Out · 6.2 Q2, Figure it Out · 6.2 Q3, Figure it Out · 6.5 Q3, Figure it Out · 6.5 Q5, Figure it Out · 6.5 Q6, Figure it Out · 6.5 Q10
Transcript1,346 words
Here is a table with seven rows, and three of them we already know. Even plus even is even. Odd plus odd is even. Even plus odd is odd. Every one of those came from a picture rather than from examples. But the other four rows are about subtraction, and they are empty. Even take away even. Odd take away odd. Even take away odd. Odd take away even. Now, you could go and work out four more rules and memorise those too.
Do not. There is nothing new down there, and I can show you why in about thirty seconds. Because subtraction cannot do anything to a leftover dot that addition has not already done. Picture a number as dots, paired off, two to a row. Now take some away. Not any old dots. Take away a whole block of complete rows. Watch what happens to the leftover at the top. Nothing. It is still sitting there.
You removed pairs, and pairs are exactly the part that was never in question. So an odd number minus an even number is still odd. The stray dot survived. And an odd minus an odd? Each one has a stray. Take one away and the other goes with it. That leaves whole rows only. So the answer is even, and that is the same as the addition row. Fill all four in, and they match the addition table line for line. Subtraction brought nothing new.
So here is a long sum. Six numbers, added up. What is the parity of the answer? You could add them. You do not have to. Cross out every even number in the list, because none of them can change anything. Every even number is complete rows. It brings no stray dot to the party at all. Now count what is left. Just count them. Do not add them. Four odd numbers. Four strays, which pair off, leaving nothing. So the total is even.
That is the whole calculation. How many odd numbers are in the list, and is that count odd or even? And once you can do that, some questions become almost embarrassingly quick. A piggy bank. Inside it are coins worth one, coins worth five, and coins worth ten. You are not told how many of each. You are told something much thinner. There is an odd number of the one-coins. An odd number of the five-coins.
And an even number of the ten-coins. Somebody claims the whole lot comes to two hundred and five. Can that be right? The one-coins: an odd count of them, each worth one. That is an odd amount. The five-coins: odd times odd is odd, so that is an odd amount too. And the ten-coins are even, whatever their count. Odd, odd, even. Two strays, which pair up. The total is even. And two hundred and five is not.
Now something that looks like a different subject, and is not. Rule a rectangle into small squares. Three rows by three columns. Count them. Nine. Which is odd. Now add a column. Three by four. Twelve small squares. Even. One extra column, and the answer changed parity. So the count is a product, and we are asking about the parity of a product. And that is a question we have never actually answered.
But the grid is about to answer it for us, and rather more convincingly than the times table would. Take the three by four grid and try to cover it in dominoes. Two squares each. Because four is even, every row splits into exactly two dominoes. Nothing is left over. It did not matter that the other side was three. It would work if it were three hundred. One even side, and the whole grid falls into pairs.
Now the three by three. Try the same thing. You get four dominoes down, and one square that will not pair with anything. It is the stranded dot again, wearing a different hat. Two odd sides leaves exactly one over. Any even side leaves none. That is the entire rule. Which means we can now do something that ought to feel like cheating. Twenty-seven by thirteen. Is the number of small squares odd or even?
Both sides odd. So it is odd. I have not multiplied anything. Forty-two by seventy-eight. Forty-two is even, so the grid is even. Done. A hundred and thirty-five by six hundred and fifty-four. Six hundred and fifty-four is even. So it is even. That is the whole working. For the record, that last grid has eighty-eight thousand two hundred and ninety small squares. But we knew its parity without going anywhere near that number, and that is the point.
Now the interesting bit, where this stops being tidy. Here is an expression. Three n plus four. Feed it a number and it gives you back a number. What is its parity? You might say even, because four is even. Let us just try it. When n is three, it gives thirteen. Odd. When n is eight, it gives twenty-eight. Even. When n is ten, thirty-four. Even again. So it has no answer. Sometimes odd, sometimes even, and which one depends on what you feed it.
Compare that with a hundred p. Whatever p is, a hundred p is even. Or forty-eight w minus two. Always even. Every value, no exceptions. So what is different? Look at the number in front of the letter. A hundred is even. Forty-eight is even. And three is not. An even number times anything is even, so an even coefficient wipes the letter out completely. Whatever n is doing, it cannot get through. The constant on the end decides everything on its own.
But three is odd. Odd times n keeps n's parity, so the letter is back in charge. Even coefficient, the parity is fixed. Odd coefficient, the parity follows the letter, and it moves. One more thing about these expressions, and it is the one people slide past. Take six k plus two. Six is even, so this one has a fixed parity. Always even. Its values start at eight, then fourteen, then twenty.
So here are the even numbers, and let us light up the ones this expression reaches. Eight. Fourteen. Twenty. Twenty-six. Look at everything still dark. Two, four, six, ten, twelve, sixteen, eighteen. Ten of the first fourteen even numbers never come out of it. It only lands on every third one. Always even is a statement about what it gives you. Reaching all the evens is a completely different claim.
So which expression does reach every even number? Write them out. Two, four, six, eight, ten. The first is two. The second is four. The third is six. Every one of them is twice its own position. So the nth even number is two n. Now write the odd numbers underneath, lined up position by position. One, three, five, seven, nine. And look at the gap between the rows. One, all the way along. Every odd number sits exactly one below the even number above it.
So the nth odd number is two n minus one. The hundredth even is two hundred, and the hundredth odd is a hundred and ninety-nine. Last thing. Here are four claims, and exactly one of them is true. Four m minus one always gives odd numbers. Four m is even, take one, always odd. True. Every even number can be written as six j minus four. We have just seen that trick. False.
Two f plus three gives both odd and even values. Two f is even, plus three. Always odd. False. And the fourth: two p plus one and two q minus one both list all the odd numbers. Two q minus one starts at one, then three, five. That reaches all of them. But two p plus one starts at three. It never produces one, so it misses the very first odd number.
And notice what settled each of those. One value, chosen to break the promise. Always means always.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Odd and even as "can this be arranged in pairs"Class 7 · Ch 6, Number Play
- Brackets decide which operation happens firstClass 7 · Ch 2, Arithmetic Expressions
Comes up again in
- Virahāṅka–Fibonacci numbers, and the poetry-counting problem that produced themClass 7 · Ch 6, Number Play
- Cryptarithms: recovering digits from constraints aloneClass 7 · Ch 6, Number Play
Either side of this one
- Using row and column sums to prove a grid cannot be filledClass 7 · Ch 6, Number Play