Ganita Prakash II · Chapter 6 — Algebra Play

‘Think of a number. Double it. Add 4. Divide by 2. Take away the number you first thought of. Your answer is 2!’ Try it with 5, with 13, with 100. It always works. Why?

Let algebra do the thinking. Call the secret number x and follow each step.

The x disappears in the last step, so the answer is 2 whatever number you start with. A single example shows that the trick works for that number; the algebra proves it works for every number.

Changing the answer. The ‘add 4’ step decides the final answer: after halving, it becomes + 2. To make the answer 3, add 6 instead; for 5, add 10. In general, adding 2k gives a final answer of k.

Designing your own trick. The recipe: do some steps that build an expression in x, then undo the x part so only a number remains. Example:

• Think of a number: x.
• Multiply by 3: 3x.
• Add 12: 3x + 12.
• Divide by 3: x + 4.
• Subtract your number: 4.

A trick that reveals the number. Instead of cancelling x, keep it and then work backwards. ‘Think of a number, add 7, multiply by 2, subtract 4, tell me the answer.’ The expression is 2(x + 7) − 4 = 2x + 10. If your friend says 36, then 2x + 10 = 36, so x = 13. To find the number quickly, subtract 10 and halve.

Why tricks fail. If a step like ‘divide by 2’ comes when the expression is odd for some starting numbers, your friend gets a fraction and may stop. Check that every step works for whole numbers. Also, if your friend makes an arithmetic slip, the trick ‘fails’, which is a good reason to ask them to check each step.

The big idea. A letter-number lets us follow every possible starting number at once. Simplifying the expression after each step shows exactly what the trick does.

Mukta thinks of a date and follows Shubham’s steps: multiply the month by 5, add 6, multiply by 4, add 9, multiply by 5, then add the day. She says ‘291’. Shubham instantly says ‘26 January, Republic Day!’ How?

Algebra. Let the month be M and the day D.

So the answer is always 100M + D + 165. Shubham just subtracts 165. That leaves 100M + D.

Place value does the rest. A day is at most 31, so D fits in the last two digits, and 100M sits in the hundreds and above. For 291: 291 − 165 = 126, so M = 1 and D = 26: 26 January.

• Answer 1390: 1390 − 165 = 1225, so 25 December.
• Answer 1269: 1104, so 4 November.
• Answer 394: 229, so 29 February.
• Answer 296: 131, so 31 January.

Try it on a birthday. 14 November: 5 × 11 = 55, + 6 = 61, × 4 = 244, + 9 = 253, × 5 = 1265, + 14 = 1279. Subtract 165: 1114, which is 11 and 14, the 14th of November.

Making your own version. The aim is to end with 100M + D plus some fixed number. Any steps that multiply M by 100 overall will do. For example: multiply the month by 4, add 3, multiply by 25, add the day. This gives 100M + 75 + D, so you subtract 75. Check with 15 August: 4 × 8 = 32, + 3 = 35, × 25 = 875, + 15 = 890, and 890 − 75 = 815: month 8, day 15.

Why must the multiplier be 100? If M were multiplied by only 10, the month and the day would overlap: 10M + D for 1 and 23 gives 33, and so does 2 and 13. Multiplying by 100 keeps the two-digit day in its own place, so the answer can be split up uniquely.

The same idea lies behind tricks that find someone’s age and shoe size, or two dice throws: build 100 × first + second (or 10 × first + second when the second is a single digit) plus a constant.

In a number pyramid, each number is the sum of the two numbers just below it. With 1, 9, 4 in the bottom row, the next row is 10 and 13, and the top is 23.

Filling gaps by working backwards. If the top is 10, the left middle is 4 and the left bottom is 1, then the right middle is 10 − 4 = 6, the middle bottom is 4 − 1 = 3, and the right bottom is 6 − 3 = 3.

When no simple start exists, use letters. The top is 60 and the bottom row is 12, c, 8. Call the middle row a and b.

• a = 12 + c and b = c + 8.
• a + b = 60, so (12 + c) + (c + 8) = 60.
• 20 + 2c = 60, so 2c = 40 and c = 20.
• Then a = 32 and b = 28. Check: 32 + 28 = 60.

A formula for the top. With two rows, the top of a and b is a + b. With three rows and bottom a, b, c, the middle is a + b and b + c, so the top is

(a + b) + (b + c) = a + 2b + c.

The middle number counts twice, because it feeds both numbers above it. For 4, 13, 8 the top is 4 + 26 + 8 = 38, without building the pyramid. For 7, 11, 3 it is 7 + 22 + 3 = 32.

With four rows and bottom a, b, c, d:

For 8, 19, 21, 13 the top is 8 + 57 + 63 + 13 = 141. The multipliers 1, 2, 1 and 1, 3, 3, 1 are the rows of a famous triangle of numbers (Pascal’s triangle, known in India as Meru Prastāra from Piṅgala’s work).

Pyramids from Virahāṅka-Fibonacci numbers. In the sequence 1, 2, 3, 5, 8, 13, 21, ... each number is the sum of the two before it. Put 1, 2, 3 in the bottom row: the next row is 3, 5 and the top is 8, all numbers from the sequence. With 1, 2, 3, 5 in a four-row pyramid you get 3, 5, 8, then 8, 13, then 21. The reason: adding two neighbouring terms always gives the next term, so every row is part of the sequence, shifted along by 2 each time. The top of an n-row pyramid is the term 2(n − 1) places after the first one.

A friend picks a 2 × 2 square of dates on a calendar page, adds the four dates and tells you only the sum, say 36. You announce: ‘Your dates are 5, 6, 12 and 13!’

The secret. On a calendar, the next date to the right is 1 more, and the date just below is 7 more (one week later). If the top-left date is a, the square is

The sum is a + (a + 1) + (a + 7) + (a + 8) = 4a + 16. If the sum is 36, then 4a + 16 = 36, 4a = 20 and a = 5. So the dates are 5, 6, 12, 13. Quick rule: subtract 16 and divide by 4.

Check with another square. 6, 7, 13, 14 add to 40, and (40 − 16)/4 = 6. Correct.

Other shapes.

• A 3 × 3 square with centre m: the dates are m − 8, m − 7, m − 6, m − 1, m, m + 1, m + 6, m + 7, m + 8. The sum is 9m, so divide the sum by 9 to get the centre.
• Three dates in a column: a, a + 7, a + 14 add to 3a + 21.
• A plus shape around centre m: m − 7, m − 1, m, m + 1, m + 7 add to 5m.

Algebra grids. In a grid puzzle, each shape stands for a number, and the last column (or row) gives the sum of the shapes in that row. For example:

• Row 1: three circles = 27, so a circle is 9.
• Row 2: two triangles and a circle = 19, so 2 × triangle + 9 = 19, and a triangle is 5.
• Row 3: a circle, a triangle and a square = 20, so 9 + 5 + square = 20, and a square is 6.

The plan is the same as in a pyramid: start with a row that has only one unknown, find it, and then use it in the other rows. Each row is really an equation such as 3c = 27 or 2t + c = 19.

Create your own. Choose values for the shapes first, build the rows, and work out the sums. Then hide the values and give the puzzle to a friend. Make sure some row lets them start.

Put the digits 2, 3 and 5 into the boxes □□ × □ (a two-digit number times a one-digit number), using each digit once. Which arrangement gives the largest product?

List the cases. The first box can take 3 digits and then the next box 2, leaving 1, so there are 3 × 2 = 6 arrangements: 23 × 5, 25 × 3, 32 × 5, 35 × 2, 52 × 3, 53 × 2.

Compare cleverly, not by brute force. Group by the multiplier.

• × 2: 35 × 2 and 53 × 2. The bigger number 53 wins.
• × 3: 25 × 3 and 52 × 3. 52 wins.
• × 5: 23 × 5 and 32 × 5. 32 wins.

Now compare 53 × 2, 52 × 3 and 32 × 5. Clearly 52 × 3 is more than 53 × 2. For the last two, expand:

32 × 5 = (3 × 10 × 5) + (2 × 5) and 52 × 3 = (5 × 10 × 3) + (2 × 3).

The first parts are equal (150), and the second parts are 10 and 6. So 32 × 5 = 160 is the largest. (The others: 52 × 3 = 156, 53 × 2 = 106.)

Is there a rule? It seems we should use the largest digit as the multiplier and arrange the other two in decreasing order. Let us prove it for any three different digits p, q, r with p < q < r. Within each pair with the same multiplier, the larger tens digit wins, leaving rq × p, rp × q and qp × r.

• rp × q beats rq × p: rp × q = 10rq + pq and rq × p = 10rp + pq, and 10rq is bigger than 10rp because q is bigger than p.
• Now compare qp × r = 10qr + pr with rp × q = 10rq + pq. The first parts are equal, and pr is bigger than pq because r is bigger than q.

So qp × r is always the largest: the largest digit is the multiplier, and the other two digits form the multiplicand in decreasing order.

Check. Digits 1, 3, 7: the rule gives 31 × 7 = 217. The nearest rival 71 × 3 = 213 is smaller. Digits 3, 5, 9: 53 × 9 = 477 beats 93 × 5 = 465.

Why algebra helps. Trying all six cases works for one set of digits. The algebra shows the rule works for every set of three different digits, and also explains why: the tens parts tie, and the units parts decide.

Mukta says: ‘Choose a two-digit number with different digits, reverse the digits, subtract the smaller from the larger and divide by 9. There will be no remainder!’ With 47: 74 − 47 = 27, and 27 ÷ 9 = 3.

Why? A two-digit number with tens digit a and units digit b is 10a + b. Reversed it is 10b + a. If b is larger:

(10b + a) − (10a + b) = 9b − 9a = 9(b − a).

So the difference is always a multiple of 9, and the quotient is b − a, the difference of the digits. For 47 the quotient is 7 − 4 = 3. (If a is larger, the difference is 9(a − b).)

Add instead of subtract. 31 + 13 = 44, 28 + 82 = 110, 12 + 21 = 33. All are multiples of 11, because (10a + b) + (10b + a) = 11a + 11b = 11(a + b).

Cycling three digits. From abc make bca and cab, and add: for 123, 123 + 231 + 312 = 666. In general

(100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111(a + b + c).

Since 111 = 3 × 37, the sum is always divisible by 37 and by 3. For 123: 666 = 111 × 6 = 37 × 18.

The abcabc trick. Write a three-digit number twice, like 425425. Divide by 7, then 11, then 13: you get back 425! Because abcabc = abc × 1000 + abc = abc × 1001, and 1001 = 7 × 11 × 13.

Even and odd averages. Algebra also settles quick claims. Is the average of two even numbers always even? 2m and 2n have average m + n, which can be odd (2 and 4 average 3). So no. Is the average of five multiples of 5 a multiple of 5? 5a + 5b + 5c + 5d + 5e over 5 is a + b + c + d + e, a whole number, but not necessarily a multiple of 5 (5, 10, 15, 20, 30 average 16). So no again: one counter-example settles it.

The lesson. Examples suggest a pattern; algebra with place value (10a + b, 100a + 10b + c) proves it for every case.

Many old puzzles become easy once we name the unknown with a letter and write an equation.

Horses and hens. A farm has 55 heads and 150 legs. How many horses and hens? Let the horses be h; then the hens are 55 − h. Legs: 4h + 2(55 − h) = 150, so 2h + 110 = 150, 2h = 40 and h = 20. There are 20 horses and 35 hens.

Without letters. If all 55 were hens there would be 110 legs. The real count is 40 more. Each horse adds 2 extra legs, so there are 40 ÷ 2 = 20 horses. The algebra and the reasoning say the same thing.

Ages. A mother is 5 times her daughter’s age. In 6 years she will be 3 times her daughter’s age. Let the daughter be d; the mother is 5d. In 6 years: 5d + 6 = 3(d + 6) = 3d + 18. So 2d = 12, d = 6. The daughter is 6 and the mother 30. Check: in 6 years, 36 = 3 × 12.

Cows. Naina has twice as many cows as Gauri. If Naina gave Gauri 3 cows, they would be equal. Let Gauri have g; Naina has 2g. Then 2g − 3 = g + 3, so g = 6. Gauri has 6 cows and Naina 12. Check: 9 and 9.

A dosa cart. Rent is ₹5000 a day and each dosa costs ₹10 to make.

• To earn a profit of ₹2000 by selling 100 dosas at a price s: 100s − (5000 + 10 × 100) = 2000, so 100s = 8000 and s = ₹80.
• If customers pay only ₹50: 50n − (5000 + 10n) = 2000, so 40n = 7000 and n = 175 dosas.

A pattern with odd numbers. Look at 1/3, (1 + 3)/(5 + 7), (1 + 3 + 5)/(7 + 9 + 11). They are 1/3, 4/12 and 9/27, all equal to 1/3! The first n odd numbers add to n², and the first 2n odd numbers add to (2n)² = 4n². So the next n odd numbers add to 4n² − n² = 3n², and the fraction is n²/3n² = 1/3 every time.

A method for word problems.

• Choose a letter for one unknown and write the others in terms of it.
• Turn each sentence into an equation.
• Solve, and answer in words.
• Check the answer in the original story.

Some puzzles repeat the same action again and again. Algebra keeps track of every round.

Karim and the genie. A genie tells Karim: ‘Each time you walk round that tree, the coins in your pocket will double, but you must then pay me 8 coins.’ Karim walks round three times and after the third payment he has nothing left. How many coins did he start with?

Let him start with x coins.

He ends with nothing, so 8x − 56 = 0 and x = 7. Check: 7 → 14 → 6 → 12 → 4 → 8 → 0.

When is the deal good for Karim? After one round he has 2x − c if the fee is c. This is more than x only when c is less than x. With 7 coins he gains only if the fee is less than 7 coins per round; at 7 he stays at 7 for ever.

How could the genie take everything? With fee c, after three rounds Karim has 8x − 7c. Setting this to 0 gives c = 8x/7. For x = 7 that is 8 coins, exactly what the genie asked. The genie chose the fee knowing Karim’s coins!

The magic ponds. A pilgrim dips his flowers into a magic pond that doubles them, then leaves p flowers at the first shrine. He repeats this at a second pond and shrine, and at a third, where he leaves all the remaining flowers, again p. Let him start with f flowers.

• After pond 1 and shrine 1: 2f − p.
• After pond 2 and shrine 2: 2(2f − p) − p = 4f − 3p.
• After pond 3: 8f − 6p, and this is all placed at shrine 3, so 8f − 6p = p.

So 8f = 7p. The smallest whole numbers are f = 7 and p = 8. He starts with 7 flowers and places 8 at each shrine. Check: 7 → 14 → 6 → 12 → 4 → 8 → 0. (Notice it is the same chain as Karim’s coins!) Any multiple also works: 14 flowers and 16 per shrine.

Working backwards is another method. Before the last shrine he had 8 flowers, so before the last pond 4; before the second shrine 4 + 8 = 12, so before the second pond 6; before the first shrine 6 + 8 = 14, so he started with 7. Reverse each step: halve instead of double, add back instead of taking away.

Why algebra wins. Working backwards needs the final number. Algebra also handles questions like ‘for what fee does Karim gain?’, which have no single final number.