Class 8 Mathematics Chapter 13: Algebra Play

Algebra for fun

Algebra has been used over the last two years to model situations and to solve equations for an unknown. It has another use, and this chapter is about that one: explaining why a puzzle or a trick works, and then inventing new ones.

A trick to start with

Here is a familiar one. Think of a number. Double it. Add four. Divide by two. Now take away the number you first thought of. The answer is 2. Try it with 3, with 11, with 40. It is 2 every time.

Why guessing will not do

Trying more starting numbers can never settle it, because there are always more left to try. What is needed is an argument that covers every possible start at once, and that is exactly what a letter-number gives.

Following the steps with a letter

Call the starting number x. Doubling it gives 2x. Adding four gives 2x plus 4. Dividing by two gives x plus 2. Taking away the original x leaves x plus 2 minus x, which is simply 2.

What the working shows

The letter has vanished. Not because of anything special about x, but because the final step removed exactly as much x as was left. Whatever number you started from, the answer is 2.

Changing the answer

Now the trick can be redesigned. The 4 was added and then halved, so what survives is half of it. To land on 3, tell people to add six. To land on 5, tell them to add ten. The rest of the chain is untouched.

Inventing a longer chain

Any chain works so long as the x cancels at the end. Think of a number. Add five. Multiply by four. Take away eight. Divide by four. Take away the original number. Following it through: 4x plus 12, then x plus 3, then 3.

A trick of a different kind

Here is one that does not cancel its letters. Instead it hides two of them, and the point is to get them back. Mukta thinks of a date. Shubham tells her the steps, hears one number, and names the date.

The steps

Take the month and multiply it by five. Add six. Multiply by four. Add nine. Multiply by five. Finally add the day, and report the result. For the 26th of January that comes to 291.

Following it with letters

Write M for the month, D for the day. Five times M is 5M. Adding six gives 5M plus 6. Multiplying by four gives 20M plus 24. Adding nine gives 20M plus 33. Multiplying by five gives 100M plus 165. Adding the day gives 100M plus 165 plus D.

Where the 100 comes from

The month was multiplied by five, then by four, then by five again, and 5 times 4 times 5 is 100. That is the whole purpose of those three multiplications: they push the month up into the hundreds and leave the lower columns free.

Where the 165 comes from

The six added early was multiplied by four and then by five, which makes 120. The nine added later was multiplied by five, which makes 45. Together they come to 165, a fixed number that is in the answer whatever the date was.

Undoing it

So Shubham subtracts 165 from whatever he is told, and what is left is 100M plus D. Mukta said 291, and 291 minus 165 is 126. The 1 is the month and the 26 is the day, which is the 26th of January.

Why the reading is safe

This only works because a day is at most 31. Two digits are always enough for it, so it never carries over into the hundreds where the month sits. The two quantities stay in separate columns and can be read apart.

Two more

Mukta next reports 1390. Subtracting 165 leaves 1225, so the month is 12 and the day is 25: the 25th of December. A report of 868 leaves 703, which is the 3rd of July. Notice that a single-digit day shows as 03.

Making it your own

The constants can all be changed. Add a different number at the start, or in the middle, and the fixed part will no longer be 165. Work out what it becomes, and subtract that instead. The structure of the trick is untouched: three multiplications still lift the month into the hundreds, and the day is still read off the last two digits. What changes is only the one number you take away at the end. That is what inventing a trick amounts to. Decide what you want to happen, write the chain with letters, and let the algebra tell you which constants will deliver it.

Scenario:

A school fete has a stall where visitors are given number tricks. Asha is told to think of a number, double it and add four. She then halves the result and takes away her starting number. Bikram wants the same chain of steps to land on 9 instead. Chhavi is taken through the date trick and reports the single number 473. The stallholder names her birthday at once. A fourth visitor is told only to think of a number, double it and add four. That visitor is then asked what the answer will be, before saying a word.

Analysis:

Three of the four can be answered without knowing any starting number. The fourth cannot, and the reason matters.

Title:

Run the date trick on your class

Safety:

Ask before writing down anybody’s birthday, and do not share the list afterwards.

What You Notice:

Nearly every failure is an arithmetic slip in one of the three multiplications. Ask the classmate to redo that step rather than starting again.

You can design a trick to land on any answer you like. In the double-add-halve-subtract chain, the answer is always half of whatever you told people to add.

The 165 in the date trick is not chosen at random. The 6 added at the start is multiplied by 4 and then by 5, and the 9 added in the middle is multiplied by 5, which comes to 24 plus 9, then 165.

If the letter goes, the answer is fixed — Follow the trick with a letter. If the letter cancels at the last step, every starting number gives one answer.

What a number pyramid is

A number pyramid is a stack of rows, each one shorter than the row below it by a single entry. One rule governs the whole thing: every entry is the sum of the two entries directly beneath it.

Filling one upwards

With a full bottom row there is nothing to it. Add neighbouring pairs to get the row above, then add pairs of those, and keep going until one entry is left. A bottom row of 1, 3 and 3 gives 4 and 6, and then 10.

Filling one downwards

Now suppose the top is given instead. If an entry is known and one of the two beneath it is known, the other is their difference. With 10 on top and 4 below it, the entry beside the 4 must be 10 minus 4, which is 6.

Working on down

That 6 and that 4 now have their own pairs beneath them. If the leftmost bottom entry is 1, then the one beside it is 4 minus 1, which is 3. And the last bottom entry is 6 minus 3, which is 3. The pyramid closes.

When neither direction starts

Some pyramids give too little for either method. Take a top of 60, with 12 and 8 at the two ends of the bottom row and nothing else. No sum can be formed and no difference either, so the working cannot begin.

Letters get it moving

Name the gaps. Let the middle bottom entry be c and the two entries above be a and b. The rule now says three things: a equals 12 plus c, b equals c plus 8, and a plus b equals 60.

Solving the equations

Put the first two into the third. That gives 12 plus c plus c plus 8 equal to 60, which is 20 plus 2c equal to 60. So 2c is 40 and c is 20. Then a is 32 and b is 28, and the pyramid is complete.

From bottom row to top

There is a better question than how to fill a pyramid: what does the top depend on? Start small. Two rows with a and b at the bottom give a plus b at the top. That much is just the rule itself.

Three rows

With a, b and c along the bottom, the row above holds a plus b and b plus c. Adding those gives a plus 2b plus c at the top. The middle entry counts twice, and the two ends only once each.

Four rows

Carry on in the same way with a, b, c and d. The top comes to a plus 3b plus 3c plus d. The pattern of coefficients is 1, 3, 3, 1, so the two middle entries now count for three times as much as the ends.

Why the middles pull harder

A coefficient counts the routes from that bottom entry up to the top. An end entry has only one route, straight up the side. A middle entry has several, and each of them carries its value to the top once more.

Using the formula

Now the top can be had without building anything. A bottom row of 8, 13 and 4 gives 8 plus 26 plus 4, which is 38. A bottom row of 21, 8, 19 and 13 gives 21 plus 24 plus 57 plus 13, which is 115.

A pyramid of Virahanka-Fibonacci numbers

The Virahanka-Fibonacci sequence runs 1, 2, 3, 5, 8 and so on, each term the sum of the two before it. Put 1, 2 and 3 along the bottom of a three-row pyramid. The next row is 3 and 5, and the top is 8.

What that pyramid shows

Every entry that appeared is itself in the sequence. That is no accident: the pyramid rule and the sequence rule are the same rule. With the first n terms along the bottom, the top is the term numbered 2n minus 1.

A calendar grid

Grids yield to the same treatment. Cut a two-by-two block from a calendar page and name the top left number a. The one beside it is a plus 1. The one below is a plus 7, since a week is seven days. The fourth is a plus 8.

Recovering the block

Those four add to 4a plus 16. So a friend who tells you only the total has told you everything. A total of 40 gives 4a equal to 24 and a equal to 6, so the block is 6, 7, 13 and 14.

And in general

A total of 88 gives 4a equal to 72 and a equal to 18, so the block is 18, 19, 25 and 26. Any block of any shape can be handled the same way: name one cell, write the others in terms of it, and add.

Scenario:

A puzzle club pins up three challenges for the week. The first is a four-row pyramid with 19, 7, 18 and 6 along the bottom. Its top entry is wanted, without building the rest. The second is a pyramid with 40 at the top. It has 7 at the left of the bottom row and 5 beside that 7, and it has to be completed. The third gives only the total of a two-by-two calendar block, which is 72. It asks for the four numbers in that block.

Analysis:

One puzzle wants a formula, one wants careful subtraction, and one wants a single letter and an equation.

Title:

Test the calendar trick

Safety:

Use a calendar nobody needs. Do not write on a wall calendar that belongs to someone else.

What You Notice:

The three-by-three block totals nine times the middle number, which is a neater result than the two-by-two one and worth proving for yourself.

The middle entries of a bottom row pull far harder than the ends. In a four-row pyramid each middle entry is counted three times over, while each end entry is counted only once.

Put the first few Virahanka-Fibonacci numbers along the bottom of a pyramid and every single entry comes out a Virahanka-Fibonacci number, the one at the top included.

One, three, three, one — A four-row pyramid counts its bottom entries once, three times, three times and once.

A placing problem

Here is the question. Take the digits 2, 3 and 5 and fill in a two-digit number multiplied by a one-digit number, using each digit exactly once. Which arrangement gives the largest product?

Counting the arrangements

Be systematic. The first box can take 2, 3 or 5, and for each of those there are two ways to place the remaining pair. That makes six arrangements in all: 23 times 5, 25 times 3, 32 times 5, 35 times 2, 52 times 3 and 53 times 2.

Cutting the work down

Six products could simply be worked out, but there is a neater route. Group them in pairs that share a multiplier: 35 and 53 both times 2, 25 and 52 both times 3, 23 and 32 both times 5.

Three survivors, then two

The multiplier matches within each pair, so whichever arrangement has the bigger two-digit number must win. That leaves only three to compare: 53 times 2, 52 times 3, and 32 times 5. Of those, 52 times 3 plainly beats 53 times 2, since 52 and 53 are so close while 3 is half again as much as 2. So the contest is between 52 times 3 and 32 times 5.

Expanding to compare

Split each two-digit factor into tens and units. Then 32 times 5 is 3 times 10 times 5, plus 2 times 5. And 52 times 3 is 5 times 10 times 3, plus 2 times 3. The first terms are both 150.

The decision

Since the first terms match, everything rests on the second. Two times 5 is 10 and 2 times 3 is only 6, so 32 times 5 is the larger. Its value is 160, and that is the largest of all six.

Was that a coincidence?

In the winning arrangement the largest digit became the multiplier and the other two went in decreasing order. Is that always how it works, or did the particular digits 2, 3 and 5 make it happen? Algebra can settle it.

The general case

Let the three digits be p, q and r with p smaller than q and q smaller than r. The six arrangements group by multiplier exactly as before, and whichever member of a pair has the bigger tens digit wins it. Three remain: rq times p, rp times q, and qp times r.

Expanding again

Compare the first two of those. They share the same tens digit r, but the second has the larger multiplier, since q is bigger than p. So rp times q beats rq times p, and rp times q now faces qp times r.

Comparing the last two

Written out, qp times r is 10 times q times r, plus p times r. And rp times q is 10 times r times q, plus p times q. The first terms are identical. The second terms are p times r and p times q.

The rule, proved

Since r is larger than q, the term p times r is larger than p times q. So qp times r wins. The largest digit is the multiplier and the other two go in decreasing order, for any three digits at all.

Using it

For the digits 1, 3 and 7 the rule gives 31 times 7, which is 217. For 3, 5 and 9 it gives 53 times 9, which is 477. Neither needed any of the other five arrangements to be worked out.

A trick about reversing

Now a different kind of puzzle. Choose a two-digit number with unequal digits and keep it to yourself. Reverse the digits. Subtract the smaller from the larger. Divide by 9. There is never a remainder.

Why there is no remainder

Write the number as 10a plus b. Reversed, it is 10b plus a. Suppose b is the larger digit. The difference is 10b plus a minus 10a minus b, which collects to 9b minus 9a, that is 9 times the quantity b minus a.

What the quotient tells you

So the difference is nine times something, and that something is the gap between the two digits. With 47 and 74 the difference is 27, and 27 divided by 9 is 3, which is exactly 7 minus 4. If a is the larger digit the working is the same the other way round.

Adding instead of subtracting

Try the sum. Ten a plus b, added to 10b plus a, gives 11a plus 11b, which is 11 times the quantity a plus b. So the sum is always a multiple of 11, and the quotient is the digit sum. For 31 and 13 the sum is 44, and 44 over 11 is 4.

Three digits, cycled

Take a three-digit number abc, which is 100a plus 10b plus c. Cycle the digits round twice to get bca and cab, and add all three. Every digit appears once in each column, so the total is 111 times the quantity a plus b plus c. Now factorise. Since 111 is 3 times 37, that sum is always divisible by 37, and always by 3 as well. One factorisation answers both questions at once.

Repeating the digits

One more. Take abc and repeat it to make the six-digit number abcabc. That is abc times 1001, because the first copy has been shifted three places. And 1001 is 7 times 11 times 13, so dividing by those three in turn hands abc straight back.

Scenario:

Four challenges are set out on a table. The first asks for the largest product obtainable from the digits 1, 3 and 7 in a two-digit times a one-digit shape. The second asks why reversing 62 and subtracting leaves a multiple of 9. The third is about three shrines with magical ponds. Each pond doubles whatever flowers are dipped in it, a person leaves an equal number at each shrine and ends with none. The last is a farm with 55 heads and 150 legs, holding only horses and hens.

Analysis:

Two challenges use the rules just proved. The other two are ordinary equations wearing a disguise.

Title:

Test the reversing trick

Safety:

No hazards. Work in pen so your own earlier answers cannot be quietly corrected.

What You Notice:

The quotient column should match the digit gap every time, and in the second round it should match the digit sum. A mismatch means an arithmetic slip somewhere.

The three-digit cycling trick works because 111 is 3 times 37. That single factorisation explains both why the sum divides by 37 and why it divides by 3.

Repeating a three-digit number, so that 428 becomes 428428, multiplies it by 1001. And 1001 is 7 times 11 times 13, which is why dividing by those three in turn hands the number straight back.

Biggest multiplies, the rest come down — Give the largest digit the one-digit place, then write the other two in decreasing order.