Algebra Play

Here is a trick you can try on anyone. Think of a number, and keep it to yourself. Double it. Add 4. Halve the answer. Now take away the number you first thought of. Your answer is 2. It is always 2, and the person doing the trick never had to know what you started with.

Test it once. Start with 9: doubling gives 18, adding 4 gives 22, halving gives 11, and 11 − 9 = 2. Test it again with 25: doubling gives 50, adding 4 gives 54, halving gives 27, and 27 − 25 = 2. Two successes are encouraging, but they are not a reason. There are endlessly many numbers you could have started with, and you cannot try them all.

So do the steps once more, this time with a letter in place of the number. Let the number be x. Doubling gives 2x. Adding 4 gives 2x + 4. Halving gives (2x + 4) ÷ 2 = x + 2. Taking away the number you started with gives x + 2 − x = 2. The x has vanished, and that is the whole secret. The instructions were built so that the starting number cancels itself out, so what is left cannot depend on it. One line of algebra has settled every case at once.

Once you can see inside a trick you can build your own. Keep the same steps but add k instead of 4. Halving now gives x + k ÷ 2, and the last step leaves k ÷ 2. So the answer is always half of whatever you told the person to add. Tell them to add 6 and the answer will be 3; tell them to add 10 and the answer will be 5. Nothing else in the trick has to change.

A longer trick. Think of a number, add 7, multiply by 3, subtract 6, divide by 3, and then subtract the number you started with. Follow it with x. Adding 7 gives x + 7. Multiplying by 3 gives 3(x + 7) = 3x + 21. Subtracting 6 gives 3x + 15. Dividing by 3 gives x + 5. Subtracting the start leaves 5. Try it with 4: 11, then 33, then 27, then 9, and 9 − 4 = 5. Try it with 12: 19, then 57, then 51, then 17, and 17 − 12 = 5.

Generalise this one too. If the second instruction is add k, the working becomes (3(x + k) − 6) ÷ 3 − x = x + k − 2 − x = k − 2. The answer is always 2 less than the number you asked for. So if you want the trick to land on 8, tell your friend to add 10. Check with a start of 4: 14, then 42, then 36, then 12, and 12 − 4 = 8, exactly as promised.

A trick where the letter does not cancel. Not every trick is built to lose the starting number. Here is one that keeps it. Think of the number of the month you were born in, multiply it by 5, add 6, multiply by 4, add 9, multiply by 5, and finally add the day of the month. Tell me the answer and I will tell you your birthday. Let the month be M and the day be D. Then 5M becomes 5M + 6, which becomes 4(5M + 6) = 20M + 24, which becomes 20M + 33, which becomes 5(20M + 33) = 100M + 165, and adding the day gives 100M + 165 + D.

All the trickster does is subtract 165, which leaves 100M + D. Since the day is never more than 31, the last two digits are the day and whatever is in front is the month. Someone born on the 15th of the eighth month works out 40, then 46, then 184, then 193, then 965, then 980; and 980 − 165 = 815, which reads as month 8 and day 15. Someone born on the 3rd of the first month gets 5, then 11, then 44, then 53, then 265, then 268; and 268 − 165 = 103, which is month 1 and day 3.

Two lessons come out of all this. Testing a trick on a few numbers only tells you about those few numbers. Writing the steps with a letter tells you about every number there is. And when the expression you end up with contains no letter at all, that is a promise that the answer can never depend on what anybody chose.

A number pyramid starts with a bottom row of numbers. Every cell above is the sum of the two cells directly below it, and you keep going until one cell is left at the top. Take the bottom row 3, 5, 4. The row above is 3 + 5 = 8 and 5 + 4 = 9, and the top is 8 + 9 = 17.

Now build the same pyramid with letters. Let the bottom row be a, b, c. The row above is a + b and b + c. The top is (a + b) + (b + c) = a + 2b + c. Test the rule on the numbers above: 3 + 2 × 5 + 4 = 3 + 10 + 4 = 17, which is the top we already found.

The 2 in front of b is worth staring at. The middle number is used twice on the way up, once by each of the two cells sitting above it, while the numbers at the ends are used only once. So the middle cell has twice the pull of either end. Raise the middle from 5 to 6 and the top goes from 17 to 19, a rise of 2; check it directly with 3 + 6 = 9 and 6 + 4 = 10, giving 19. Raise the left end from 3 to 4 instead and the top only goes to 18, a rise of 1. If you want the top as large as possible and you may arrange the same three numbers as you like, put the biggest one in the middle.

A wider pyramid. With four numbers a, b, c, d at the bottom, the second row is a + b, b + c and c + d. The third row is (a + b) + (b + c) = a + 2b + c and (b + c) + (c + d) = b + 2c + d. Adding those two gives the top: a + 3b + 3c + d. Now the two inner cells count three times each. Test it with the bottom row 2, 1, 4, 3. The second row is 3, 5, 7; the third row is 8, 12; the top is 20. The rule gives 2 + 3 × 1 + 3 × 4 + 3 = 2 + 3 + 12 + 3 = 20, which agrees.

Working backwards. Suppose the bottom row is 5, then a gap, then 8, and you are told the top is 27. Call the missing number x. The top is 5 + 2x + 8 = 2x + 13, so 2x + 13 = 27, which gives 2x = 14 and x = 7. Check by building it: 5, 7, 8 gives 12 and 15, and 12 + 15 = 27. Here is the second use of algebra in this chapter. First you generalise the pyramid into an expression; then you set that expression equal to a given number and solve.

When it cannot be done. Keep the ends 5 and 8 but ask for a top of 20. Then 2x + 13 = 20, so 2x = 7 and x = 3.5. If the pyramid is meant to hold whole numbers, there is no answer at all. The expression even explains why. When x is a whole number, 2x is even, so 2x + 13 is always odd; with ends of 5 and 8 the top can only ever be an odd number, and 20 is even. No amount of searching would have found a filling, and now you know not to look.

One more for practice. The bottom row is 6, then a gap, then 9, and the top is 31. The top is 6 + 2x + 9 = 2x + 15, so 2x = 16 and x = 8. Building 6, 8, 9 gives 14 and 17, and 14 + 17 = 31. If instead the top were 30, you would need 2x = 15 and x = 7.5, so that pyramid has no whole-number filling either.

The point is not the pyramids. It is that one short expression answers every question you could ask about them: what the top will be, which cell to change, what a missing entry must be, and whether the puzzle you have been handed is possible in the first place.

A calendar month is a grid in which each row is a week. That gives it two simple rules: moving one step to the right adds 1 to the date, and moving one step down adds 7. Those two rules are all the algebra you need here.

A block of four dates. Draw a square around any four dates that form a 2 by 2 block. Call the date in the top left corner a. The date to its right is a + 1, the date below it is a + 7, and the date diagonally across is a + 8. Add the four: a + (a + 1) + (a + 7) + (a + 8) = 4a + 16. Try it with a = 9. The block is 9, 10, 16, 17, and 9 + 10 + 16 + 17 = 52, while the rule gives 4 × 9 + 16 = 52.

That expression is a trick waiting to happen. Ask a friend to ring any 2 by 2 block of dates and tell you only the total. Suppose they say 72. Then 4a + 16 = 72, so 4a = 56 and a = 14, and you can announce the dates 14, 15, 21 and 22 without seeing the calendar. Check the total: 14 + 15 + 21 + 22 = 72.

Some totals are impossible, and the same expression shows it. Since 4a + 16 = 4(a + 4), the total of a block is always a multiple of 4. If a friend claims a total of 62, you can say straight away that they have made a mistake, because 62 is not a multiple of 4. Solving would give 4a = 46 and a = 11.5, which is not a date at all.

The diagonals. Multiply the two dates on one diagonal and then the two on the other. One diagonal gives a(a + 8) = a² + 8a. The other gives (a + 1)(a + 7) = a² + 8a + 7. The two answers differ by exactly 7, whatever a is, and the a² + 8a part never has to be worked out. With the block starting at 9: 9 × 17 = 153 and 10 × 16 = 160, and 160 − 153 = 7. With the block starting at 14: 14 × 22 = 308 and 15 × 21 = 315, and again the gap is 7.

A bigger block. For a 3 by 3 block it pays to name the middle date rather than the corner. Call the middle date c. The three dates in its own row are c − 1, c and c + 1. The row above is c − 8, c − 7 and c − 6, and the row below is c + 6, c + 7 and c + 8. Add all nine and every offset pairs off with its opposite: −8 with +8, −7 with +7, −6 with +6, −1 with +1. What is left is 9c. So the total of a 3 by 3 block of dates is always nine times the middle date. A block centred on 16 contains 8, 9, 10, 15, 16, 17, 22, 23 and 24, and those add to 144, which is 9 × 16. If a friend gives you a total of 99, the middle date is 99 ÷ 9 = 11, and the block runs 3, 4, 5, 10, 11, 12, 17, 18, 19, which does indeed total 99.

Notice what naming the middle cell did. Had you called the top left date a instead, the nine dates would have been a, a + 1, a + 2, a + 7, a + 8, a + 9, a + 14, a + 15 and a + 16, and the total would have come to 9a + 72. That is correct, but 9c is far easier to use. Choosing where to put your letter is part of the skill.

The same idea on another grid. On a hundred square with ten numbers in each row, moving down adds 10 instead of 7. A 2 by 2 block with n in the top left holds n, n + 1, n + 10 and n + 11, so the total is 4n + 22, and the diagonal products differ by 10 because (n + 1)(n + 10) − n(n + 11) = 10. Take n = 34: the block is 34, 35, 44, 45, which adds to 158, and 4 × 34 + 22 = 158. The products are 34 × 45 = 1530 and 35 × 44 = 1540, a gap of 10 as promised. Change the shape of the grid and only one number in the algebra changes.

Take three different digits, say 2, 3 and 5. Use each of them once to make a two-digit number and a one-digit number, then multiply the two together. Which arrangement gives the largest product? You could guess, but with only six arrangements you can simply list them all.

• 32 × 5 = 160
• 52 × 3 = 156
• 23 × 5 = 115
• 53 × 2 = 106
• 25 × 3 = 75
• 35 × 2 = 70

The winner is 32 × 5 = 160. Look at what the winning arrangement did: the biggest digit, 5, was left standing alone as the multiplier; the next biggest, 3, took the tens place; and the smallest, 2, was pushed into the units place where it counts for least. That is a rule worth testing. With the digits 1, 4 and 7 it predicts 41 × 7, which is 287, and the full list agrees: 71 × 4 = 284, 14 × 7 = 98, 74 × 1 = 74, 17 × 4 = 68 and 47 × 1 = 47.

Why the rule works. Testing two sets of digits is not a proof, so write the digits as letters. Let them be a, b and c with a the largest and c the smallest, so a > b > c. The smallest digit clearly belongs in the units place, because that place is worth only 1 while the tens place is worth 10. That leaves two serious candidates: put a alone and b in the tens place, giving (10b + c) × a, or put b alone and a in the tens place, giving (10a + c) × b.

Now expand both. The first is 10ab + ac. The second is 10ab + bc. The 10ab is exactly the same in the two products, so it cannot decide anything, and the whole contest comes down to ac against bc. The difference is ac − bc = c(a − b). Since a is bigger than b, the bracket a − b is positive, so as long as c is not 0 the first arrangement wins. That settles it for every set of digits at once: the biggest digit stands alone.

The expression c(a − b) also tells you the size of the gap before you multiply anything. For 2, 3 and 5 it gives 2 × (5 − 3) = 4, and indeed 160 − 156 = 4. For 1, 4 and 7 it gives 1 × (7 − 4) = 3, and 287 − 284 = 3. For 3, 6 and 8 it gives 3 × (8 − 6) = 6, and the two products are 63 × 8 = 504 and 83 × 6 = 498, which do differ by 6.

The exception hiding in the algebra. The gap is c(a − b), and that is 0 when c is 0. So if the smallest of the three digits is zero the two arrangements are not merely close, they are exactly equal. Try 0, 3 and 5: 30 × 5 = 150 and 50 × 3 = 150. There are two best answers, not one. It is worth being fussy about the word always, because the algebra tells you precisely when the rule holds and when it does not.

A zero brings one more caution. A two-digit number cannot start with 0, so 05 is not allowed. With the digits 0, 3 and 5 only four arrangements are legal at all: 30 × 5, 50 × 3, 35 × 0 and 53 × 0, and the last two are worth nothing.

Listing six cases was easy enough here. The reason for doing the algebra is that it works for every set of three digits without any listing, it tells you how much you win by, and it warns you about the one case where the rule breaks down.

Every trick in this section comes from one idea: write a number in place value form using letters. A two-digit number whose tens digit is a and whose units digit is b is worth 10a + b. It is not a times b, and the letters written side by side are not the digits side by side. The number with the digits swapped round is 10b + a.

Reverse and subtract. Say the tens digit is the bigger of the two, so a is more than b and 10a + b is the larger number. Take the smaller away from the larger: (10a + b) − (10b + a) = 10a + b − 10b − a = 9a − 9b = 9(a − b). Whatever digits you started with, the answer is 9 times something, so it is always a multiple of 9. Test it with 72 and 27: the difference is 45, which is 9 × 5, and the digits differ by 7 − 2 = 5. Test it with 63 and 36: the difference is 27, which is 9 × 3, and 6 − 3 = 3. The trick even gives you a bonus, because dividing the answer by 9 hands back the difference of the two digits.

Reverse and add. This time (10a + b) + (10b + a) = 11a + 11b = 11(a + b), so the total is always a multiple of 11, and dividing it by 11 gives the sum of the digits. With 72 and 27 the total is 99, which is 11 × 9, and the digits do add to 9.

Three digits. A three-digit number is 100a + 10b + c and its reverse is 100c + 10b + a. Subtracting, the middle digit disappears completely: (100a + 10b + c) − (100c + 10b + a) = 99a − 99c = 99(a − c). So the answer is always a multiple of 99, and since 99 = 9 × 11 it is a multiple of both 9 and 11. Try 852 and 258: the difference is 594, which is 99 × 6, and the outer digits differ by 8 − 2 = 6. The middle digit 5 played no part at all, which is why 731 − 137 gives the same 594.

The 37 trick. Start with three digits a, b and c and write the three numbers you get by moving the leading digit to the end each time: 100a + 10b + c, then 100b + 10c + a, then 100c + 10a + b. Add all three. Each digit gets multiplied once by 100, once by 10 and once by 1, so each contributes 111 of itself, and the total is 111(a + b + c). Because 111 = 3 × 37, that total is always a multiple of 3 and always a multiple of 37, however the digits were chosen. With the digits 1, 5 and 3: 153 + 531 + 315 = 999, which is 111 × 9 and also 37 × 27. With 4, 7 and 2: 472 + 724 + 247 = 1443, which is 111 × 13 and also 37 × 39.

Why the digit-sum test for 9 works. You already know that a number is divisible by 9 when its digits add to a multiple of 9. Place value explains it. Split a two-digit number as 10a + b = 9a + (a + b). The 9a part is a multiple of 9 no matter what, so the whole number is a multiple of 9 exactly when a + b is. The same split works for three digits: 100a + 10b + c = 99a + 9b + (a + b + c), and since 99a + 9b is already a multiple of 9, everything depends on the digit sum a + b + c.

A six-digit surprise. Write any three-digit number twice over, such as 314314. That number is 314 × 1000 + 314 = 314 × 1001, and 1001 = 7 × 11 × 13. So dividing by 7, then by 11, then by 13 must bring you back to where you began: 314314 ÷ 7 = 44902, then 44902 ÷ 11 = 4082, then 4082 ÷ 13 = 314. It looks like magic and it is really just factorising 1001.

None of these results needed a single new fact about numbers. Each one is place value written with letters and then tidied up.

In the tricks at the start of this chapter the letter always cancelled, and that was the point. Sometimes it does not cancel, and that is just as useful, because then the expression can be set equal to a known number and solved. The unknown that would not go away turns out to be the thing you were asked for.

The doubling toll. A traveller sets out with some coins and has to cross three bridges. At each bridge a genie doubles the coins in her purse, and she then pays a toll of 8 coins. Let her start with x coins. After the first bridge she has 2x − 8. After the second she has 2(2x − 8) − 8 = 4x − 16 − 8 = 4x − 24. After the third she has 2(4x − 24) − 8 = 8x − 48 − 8 = 8x − 56.

Suppose her purse is empty at the end. Then 8x − 56 = 0, so 8x = 56 and x = 7. Always check against the story rather than only against the equation: 7 coins double to 14, and paying 8 leaves 6; 6 double to 12, and paying 8 leaves 4; 4 double to 8, and paying 8 leaves 0. She started with 7 coins.

Change the ending and the same expression answers again. If she finishes with 8 coins, then 8x − 56 = 8, so 8x = 64 and x = 8. There is something pleasing about that answer: 8 coins double to 16, and the toll of 8 brings the purse straight back to 8, so nothing changes at any bridge. Algebra says the same thing in one line, because 2x − 8 = x gives x = 8.

The rules for solving. An equation is a balance, and the two sides must stay equal, so whatever you do you must do to both sides: add the same number, subtract the same number, multiply by the same non-zero number, or divide by the same non-zero number. Transposing is only a quicker wording of the same moves. A term that crosses the equals sign changes its sign, and a number that multiplies on one side crosses over as a divisor.

An equation with a bracket. Somebody following the earlier trick was told to think of a number, add 7 and multiply by 3, then subtract 6, and they stopped there with 45. What was their number? The instructions give 3(x + 7) − 6 = 45. Open the bracket: 3x + 21 − 6 = 45, which tidies to 3x + 15 = 45. Then 3x = 30 and x = 10. Check it through the story: 10 + 7 = 17, and 17 × 3 = 51, and 51 − 6 = 45.

Letters on both sides. Solve 5x − 3 = 3x + 9. Collect the letters on one side and the plain numbers on the other: 5x − 3x = 9 + 3, so 2x = 12 and x = 6. Check each side of the original separately. The left side is 5 × 6 − 3 = 27 and the right side is 3 × 6 + 9 = 27, so the two agree.

Clearing fractions. Solve x ÷ 2 + x ÷ 3 = 15. The denominators are 2 and 3, whose LCM is 6, so multiply every term on both sides by 6. That gives 3x + 2x = 90, so 5x = 90 and x = 18. Check: 18 ÷ 2 = 9 and 18 ÷ 3 = 6, and 9 + 6 = 15. The word every is doing real work in that instruction, because the 15 has no fraction in it but must still be multiplied by 6.

When there is no answer. Could the traveller finish with exactly 3 coins? That needs 8x − 56 = 3, so 8x = 59 and x = 7.375. Coins come in whole numbers, so no starting amount can leave her with 3. This is the same situation as a pyramid whose top would need a bottom entry of 3.5, or a block of dates whose total is not a multiple of 4. In each case the algebra does not just fail to find an answer, it proves that no answer exists, which is a much stronger statement than not having found one yet.

Three habits will carry you through this whole chapter. Name the unknown clearly and say what it counts. Write every other quantity in terms of that one letter. And when you have an answer, put it back into the original story, not merely into the line of working just above it, because an arithmetic slip early on will otherwise survive all the way to the end.