Class 8 Mathematics Chapter 5: Number Play

Sums of a run of numbers

Anshu notices that 7 is 3 plus 4, and that 12 is 3 plus 4 plus 5. He asks whether every number can be written as a sum of consecutive numbers, and whether some can be written that way in more than one manner.

The easy half of the answer

Every odd number can. An odd number is one more than twice something, so split it into two halves that differ by one. Nine is 4 plus 5. Twenty-one is 10 plus 11. That works for every odd number, however large.

And some have several forms

Twenty-one also comes out as 6 plus 7 plus 8, and again as 1 up to 6. Forty-five has four different forms, including 5 up to 10. So the number of ways is not fixed, and finding them all is a puzzle in itself.

The numbers that refuse

Try 32 and you will fail. Try 8 or 16 and you will fail too. The numbers with no consecutive-sum form at all are exactly the powers of two, which is a surprising answer to a question that looked like simple arithmetic.

Four numbers and eight expressions

Now a different game. Take four consecutive numbers, say 6, 7, 8 and 9. Put a plus or a minus between each neighbouring pair. The first sign is fixed as a plus, so there are three free choices, which gives eight expressions in all.

Working them out

The eight values come to 30, 16, 14, 12, 0, minus 2, minus 4 and minus 18. Look at them. Every one is even. Zero is in the list, and zero is even. So is minus 18, since negative numbers have parity too.

Try it again

Start from 11, 12, 13 and 14 instead. The values come to 50, 26, 24, 22, 0, minus 2, minus 4 and minus 28. Even again, every one. This is beginning to look like a rule rather than an accident.

Why guessing is not enough

There are infinitely many sets of four numbers. Checking them one by one can never finish, so a few agreeing cases only give a conjecture. What is wanted is a reason that covers every set at once.

Switching one sign

Here is the reason. Take any expression, say a plus b minus c minus d, and change the plus before b into a minus. The new value is a minus b minus c minus d. Subtract one from the other and almost everything cancels.

What is left

The difference comes to twice b. That is an even number whatever b happens to be. So switching one sign moves the answer by an even amount.

And an even move keeps the parity

Two numbers that differ by an even amount must have the same parity. If one were odd and the other even, their difference would be odd. So switching a sign cannot turn an even answer into an odd one.

Finishing the argument

Any one of the eight expressions can be reached from any other by switching one sign at a time, at most three times. Each switch keeps the parity, so all eight share it. No particular numbers were used, so this holds for every set of four.

It does not need consecutive numbers

Look back at the argument. It never used the fact that the four numbers follow one another. Take any four numbers at all and the eight signed sums still share one parity.

The parity rules behind it

Three short facts carry a lot of weight. An odd plus or minus an odd is even. An even plus or minus an even is even. An odd plus or minus an even is odd. Notice that plus and minus behave identically throughout.

Reading an expression for parity

Now use this on algebra. Which expressions must come out even, whatever whole numbers their letters stand for? Take 6m plus 2n. Both parts are even for every m and n, so the sum is even.

The quicker way to see it

Rewrite it as 2 times the quantity 3m plus n. Written that way, 2 is a factor of the whole expression, so it cannot be anything but even. With m equal to 5 and n equal to minus 7, it comes to 16.

One counterexample sinks a claim

Try x times x, plus 4. With x equal to 4 it gives 20, which is even. But with x equal to 5 it gives 29, which is odd. So the expression is not always even, and that single case settles it.

Collecting terms first

Some expressions hide their factor. Take 9k minus 3k. Collect the terms and it is 6k, which has 2 as a factor, so it is always even. By contrast 6m minus 3n is 3 times the quantity 2m minus n, and with m and n both 1 it comes to 3.

Scenario:

A class is set the following task. Take the four numbers 8, 11, 2 and 7, put a plus or a minus between each neighbouring pair, and write down all the answers. One pupil hands in eight values, and seven of them are even. The eighth is written as 13. The teacher marks the eighth wrong without working it out, and the pupil protests that the numbers were not consecutive, so the rule about parity should not apply. A second pupil adds that one of the four numbers is odd, which must surely matter.

Analysis:

The teacher is relying on a result whose proof never mentions either objection, so the two protests can be answered from the argument itself.

Title:

Hunt the consecutive sums

Safety:

No equipment is involved, so no safety step is needed.

What You Notice:

The ringed numbers come out as 1, 2, 4, 8, 16 and 32, which are the powers of two. Nothing about the hunt suggested that answer in advance, which is what makes it worth finding.

The numbers that cannot be written as a sum of consecutive whole numbers are exactly the powers of two: 2, 4, 8, 16, 32 and so on.

The eight signed sums of 6, 7, 8 and 9 come to 30, 16, 14, 12, 0, minus 2, minus 4 and minus 18. Every one of them is even.

Flip a sign, move by double — Changing one plus to a minus changes the answer by twice that number. A change of an even amount cannot change the parity, so all eight answers agree.

A question about pairs

Take two even numbers and add them. Is the answer a multiple of 4? Try 20 and 36 and you get 56, which is. Try 20 and 14 and you get 34, which is not. So the answer is sometimes yes and sometimes no, and the question is what decides it.

Two kinds of even number

Divide an even number by 4 and the remainder is either 0 or 2. It cannot be 1 or 3, since the number is even. So every even number is either 4p, a multiple of 4, or 4p plus 2, which is not.

First case: both are multiples of 4

Add 4p and 4q. Taking the 4 out gives 4 times the quantity p plus q, which is a multiple of 4. Try it: 20 plus 36 is 56, which is 4 times 14. So this case always works.

Second case: neither is

Add 4p plus 2 to 4q plus 2. That is 4p plus 4q plus 4, which is 4 times the quantity p plus q plus 1. A multiple of 4 again. The two remainders of 2 made a fresh 4 between them. Try it: 14 plus 22 is 36, which is 4 times 9.

Third case: one of each

Add 4p to 4q plus 2. That is 4 times the quantity p plus q, plus 2. So the sum leaves a remainder of 2, and is never a multiple of 4. Try it: 20 plus 14 is 34, which is 4 times 8 with 2 over.

The pattern underneath

Notice how closely this mirrors parity. There, two odds or two evens gave an even number, while one of each gave an odd one. Here, two of a kind give a multiple of 4 while one of each does not. The remainders are doing the same work.

A general question

Now ask it with any divisor. If 8 divides two numbers separately, must it divide their sum? The two numbers are 8a and 8b. Their sum is 8 times the quantity a plus b, so yes, always.

And the difference too

The same move works for subtraction. The difference is 8 times the quantity a minus b, so 8 divides it as well. Try it: 8 divides 24 and 56, and it divides both 80 and 32.

The general statement

Put generally: when a goes into M and also into N, it goes into M plus N and into M minus N. One line of algebra does the whole job, for every divisor and every pair of multiples.

Now turn it round

Here is a different claim. If a number is divisible by 8, must any two numbers adding to it also be divisible by 8? Take 72. It is 48 plus 24, and both of those are multiples of 8. So far so good.

Where it fails

But 72 is also 50 plus 22, and neither of those is a multiple of 8. So the claim is sometimes true and sometimes false. It gets the verdict sometimes true, and the two examples are what earn that verdict.

Three verdicts

This is worth naming. A statement about numbers can be always true, sometimes true, or never true. Always true needs a proof. Never true needs a proof as well. Sometimes true needs one case of each kind, and nothing more.

Multiples of a multiple

Another claim: if a number is divisible by 7, is every multiple of it divisible by 7? Write the number as 7j. A multiple of it is 7j times m, which is 7 times the quantity j times m. So 7 is still a factor, and the claim is always true.

Why that one feels obvious

It is obvious once written down, but it does real work. It means a divisor can never be lost by multiplying. Only adding and subtracting put a divisor at risk, and even then only when the parts do not share it.

Two multiples of 3

Here is a harder one. When do two multiples of 3 add to a multiple of 6? Write them as 3a and 3b. Their sum is 3 times the quantity a plus b, so 3 is a factor already. For 6, we need a factor of 2 as well.

Finishing it with parity

The sum is a multiple of 6 exactly when a plus b is even, which happens when a and b share a parity. In other words, when the two multiples of 3 are both even or both odd. Check: 9 plus 15 is 24, but 9 plus 6 is 15.

Examining a claim

Snehal says this. Take a number leaving 8 on division by 12, and another that sits 4 below a multiple of 12. The total is always a multiple of 8. Notice first that a number sitting 4 below a multiple of 12 leaves 8, so both numbers are of one kind.

The verdict on it

Their sum is 12 times something plus 16, so it always leaves 4 on division by 12. But a multiple of 8? Eight plus 8 is 16, which is. Twenty plus 8 is 28, which is not. So Snehal is wrong, and one counterexample is all it took.

Scenario:

A packer fills boxes with exactly 4 jars each. At the end of a shift she has two part-filled boxes left over. On Monday each of them holds 2 jars. On Tuesday one holds 2 jars and the other is a full box of 4 that she has not sealed. Her supervisor says that since every leftover count is even, combining the two part-filled boxes will always give whole boxes with nothing over. The packer is not convinced and wants to know which days work and why.

Analysis:

Both leftover counts are even, but what decides the outcome is the remainder each leaves on 4, not the fact that they are even.

Title:

Hunt the runs of three

Safety:

No equipment is involved, so no safety step is needed.

What You Notice:

The runs start at 2, 14, 26, 38 and 50, so the gap is always twelve. Twelve is the smallest number that 2, 3 and 4 all divide, which is why the pattern repeats on that cycle.

Three consecutive numbers where the first is a multiple of 2, the second of 3 and the third of 4 start at 2, then 14, then 26, then 38. They recur every twelve numbers, for ever.

Two multiples of 3 add to a multiple of 6 exactly when both are odd or both are even, which ties this topic straight back to parity.

Forwards safe, backwards not — A common divisor always carries into the sum and the difference. Going backwards from a sum to its parts is not safe, and one counterexample shows it.

Writing a number in general

A four-digit number with digits d, c, b and a, reading from the left, is 1000d plus 100c plus 10b plus a. In 4075 that makes d equal to 4, c equal to 0, b equal to 7 and a equal to 5. Every shortcut in this topic comes out of that one line.

Divisible by ten

Look at the place values. All of them except the units are multiples of 10. So the whole number is a multiple of 10 plus the units digit. That makes it divisible by 10 exactly when the units digit is 0, and nothing else matters.

Five and two come free

The same argument works for 5 and 2, because every place value above the units is a multiple of both. So only the units digit decides. For 4 you need the last two digits, since 100 and every higher place value is a multiple of 4, and for 8 the last three.

Nine is harder

Which of 999, 909, 900, 90 and 990 are divisible by 9? All of them. Each digit is 0 or 9, so each term in the expanded form is 9 times a place value or 0 times one, and both are multiples of 9.

Most multiples of 9 use other digits, so this is no use as a general test. Nor can the units digit decide: 99 and 109 both end in 9, and only the first is a multiple of 9.

The key observation

Divide 10 by 9 and the remainder is 1. Divide 100 by 9 and the remainder is 1 again. The reason is plain once written down: 10 is 9 plus 1, 100 is 99 plus 1, 1000 is 999 plus 1, and 9, 99 and 999 are all multiples of 9.

Using it on a number

Take 7309. Written out, it is 7 times 999 plus 1, plus 3 times 99 plus 1, plus 0 times 9 plus 1, plus 9. Separate the nines from the ones. The nines part is a multiple of 9, and the ones part is 7 plus 3 plus 0 plus 9, which is the digit sum.

So the rule explains itself

Every number is a multiple of 9 plus its digit sum. A number is therefore divisible by 9 exactly when its digit sum is. Take 7461: the digits add to 18, so it is a multiple of 9.

And it gives the remainder too

Add the digits repeatedly until one digit is left, and that digit is the remainder on division by 9. For 61274 the digits give 20, and 2 plus 0 gives 2. So 61274 leaves a remainder of 2.

One caveat worth stating

The repeated digit sum of 7461 comes out as 9, not 0, yet 7461 is a multiple of 9 and leaves no remainder. So the rule needs a rider: when the final digit is 9, the remainder is zero.

Three for free

Now look again at 9, 99 and 999. They are multiples of 3 as well. So exactly the same argument shows that a number is a multiple of 3 when its digit sum is. The digit sum of 61274 is 20, which is not a multiple of 3, so neither is 61274.

Eleven, and a near miss

Eleven needs a different look. Ten is one LESS than 11. A hundred is 99 plus 1, and 99 is a multiple of 11, so a hundred is one MORE. A thousand is 1001 less 1, and 1001 is 11 times 91, so a thousand is one less again.

So the place values take turns. The units, hundreds and ten-thousands sit one OVER a multiple of 11. The tens, thousands and lakhs sit one UNDER. That alternation is what the rule for 11 is made of.

Working an example

Take 3,28,105. Add the digits in the over places: 5 plus 1 plus 2, which is 8. Add the digits in the under places: 0 plus 8 plus 3, which is 11. Subtract to get 8 less 11, or minus 3.

Reading the answer

So 3,28,105 is 3 short of a multiple of 11, which is the same as being 8 over one. When the difference is 0 or a multiple of 11, the number is divisible by 11. Check 6248: the sums are 8 plus 2, and 4 plus 6, so the difference is 0.

Building a test for a bigger divisor

What about 24? Since 24 is 3 times 8, a number divisible by both 3 and 8 is divisible by 24. We already have tests for 3 and for 8, so the work is done. For 1848 the digit sum is 21 and the last three digits make 848, which is 8 times 106.

Why the factors must be chosen carefully

This only works when the two factors share no common factor. Testing 24 by checking 4 and 6 looks reasonable and fails: 12 passes both tests and is not a multiple of 24. The trouble is that 4 and 6 share a factor of 2, so between them they only guarantee 12.

Digits in disguise

Last of all, a puzzle. In a cryptarithm each letter stands for a digit, no two letters share a digit, and no number starts with 0. Take PQ times 8 equals RS, a two-digit number whose product with 8 is still two digits.

Reasoning it out

Thirteen times 8 is 104, so PQ cannot be 13 or more. Ten is out, because 80 repeats the units digit. Eleven is out, because P and Q would be the same digit. That leaves 12, and 12 times 8 is 96. One answer, found without trying every case.

Scenario:

A shop assistant has to check a long till roll at the end of the day. The total reads 72,633 rupees, and the batteries in the calculator have died. She needs to know three things before she can file the sheet. Whether the total is divisible by 9, because the day’s sales were all in multiples of 9 rupees. Whether it is divisible by 11, because a second column should match. And whether it is divisible by 99, since the manager insists on that check too. She remembers the digit-sum rule and the alternating rule.

Analysis:

All three answers can be had from the digits alone, and the third one depends on whether the first two can safely be combined.

Title:

Solve a cryptarithm

Safety:

No equipment is involved, so no safety step is needed.

What You Notice:

The survivor is 24 times 4, which is 96. Candidates like 16 times 6 give 96 as well, but then K and H would both be 6, which the rules of a cryptarithm forbid.

The repeated digit sum of a multiple of 9 comes out as 9, not as zero. So the rule needs one small caveat: a final 9 means the remainder is zero.

Testing for 24 by checking 4 and 6 fails. Twelve is divisible by both and not by 24, because 4 and 6 share a factor of 2.

Every ten is a nine and a one — Ten is 9 plus 1, a hundred is 99 plus 1, a thousand is 999 plus 1. The nines go into a multiple of 9 and the ones add up to the digit sum.