Ganita Prakash I · Chapter 6 — Number Play

Six friends stand in a line, one behind the other, all facing the front. Each one calls out a number using this rule: count the people in front of you who are taller than you. Nobody says their height, yet the numbers tell us a lot about the line.

An example. Suppose the heights from front to back are 150, 142, 155, 138, 160 and 147 cm.

The second person (142) has only the first person (150) taller in front, so says 1. The fourth person (138) is shorter than all three in front, so says 3. The sixth person (147) has 150, 155 and 160 taller in front, so says 3.

What the numbers tell us. The first person always says 0, because nobody is in front. A person who says 0 is taller than everyone ahead of them, but not necessarily the tallest in the whole line: in our example the first person says 0 even though two people behind are taller. The tallest person, however, always says 0.

Always, sometimes or never? Good mathematical thinking asks whether a statement is always true, only sometimes true, or never true.

• ‘The tallest person says 0.’ Always true: nobody in front can be taller.
• ‘A person who says 0 is the tallest.’ Sometimes true: see the first person in the example.
• ‘The person with the largest number is the shortest.’ Sometimes true: a short person standing near the front has few people in front to count.
• ‘In a line of 8, the largest possible number is 7.’ True: the last person has 7 people in front, and all 7 could be taller.

Building a line from the numbers. Can you make a line whose numbers read 0, 1, 2, 3, 4, 5? Yes: put people in order from tallest to shortest, so each person has everyone in front taller. For 0, 0, 0, 0, 0, 0, put them from shortest to tallest. For 0, 1, 0, 1, 0, 1, alternate: each even-position person is shorter than the person just ahead but taller than everyone before that, as in heights 140, 135, 150, 145, 160, 155.

Why this matters. These numbers describe an arrangement without the actual measurements. Mathematicians and computer scientists use such counts to see how ‘mixed up’ a list is, for example when sorting names or scores.

Kishor must fill 5 boxes with 5 cards so that they add to 30, but every card shows an odd number. He tries again and again and fails. Is there a reason? Yes, and the reason is parity.

Parity means whether a number is even or odd. An even number can be arranged in pairs with nothing left over. An odd number is a set of pairs with one extra left over.

Even + even. Two sets of pairs together are still pairs. So the sum of any number of even numbers is even.

Odd + odd. Each odd number has one left-over. Put the two left-overs together and they form a new pair. So odd + odd is even.

Even + odd. The pairs stay pairs and the one left-over stays alone. So even + odd is odd.

Many odd numbers. Left-overs pair up two at a time. With an even count of odd numbers, all left-overs pair up and the sum is even. With an odd count, one left-over remains and the sum is odd.

Back to Kishor. Five odd numbers is an odd count, so their sum is odd. 30 is even. So the puzzle has no solution, and we know this without trying every choice of cards.

Consecutive numbers. Two siblings born exactly a year apart say their ages add up to 112. Impossible! Their ages are consecutive numbers, and of any two consecutive numbers one is even and one is odd. Their sum is odd, but 112 is even.

Coins. Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an even number of ₹10 coins, and claims a total of ₹205. An odd number of ₹1 coins gives an odd amount; an odd number of ₹5 coins gives an odd amount (odd count of odd numbers); any number of ₹10 coins gives an even amount. Odd + odd + even is even, but 205 is odd. He has made a mistake.

A light switch. A bulb is on. Each press of the switch changes it. After an even number of presses it is back on; after an odd number it is off. After 77 presses it is off.

A 3 × 3 grid has 9 small squares, an odd number. A 3 × 4 grid has 12, an even number. Can we tell the parity of the number of squares in a 27 × 13 grid without multiplying?

Products. A grid of r rows and c columns has r × c squares, which is r added to itself c times. If r is even, every row has pairs, so the total is even. If r is odd but c is even, we add an odd number an even count of times, which is even. Only when both are odd is the product odd.

So 27 × 13 is odd (both odd), 42 × 78 is even and 135 × 654 is even (654 is even).

Parity of expressions. The value of 3n + 4 depends on n. For n = 3 it is 13 (odd); for n = 8 it is 28 (even). Since 3n has the same parity as n and 4 is even, 3n + 4 is odd when n is odd and even when n is even.

Some expressions are always even, such as 100p or 2k + 6, because they are 2 times a whole number. Some are always odd, such as 2m + 1 or 4n + 3, because they are an even number plus 1.

The nth even and odd numbers. The even numbers are 2, 4, 6, 8, … The first is 2 × 1, the second 2 × 2, so the nth even number is 2n. Line up the odd numbers 1, 3, 5, 7, … under them: each odd number is 1 less than the even number above it. So the nth odd number is 2n − 1. The 100th even number is 200 and the 100th odd number is 199.

Checking claims.

• ‘4m − 1 always gives odd numbers.’ True: 4m is even, minus 1 is odd.
• ‘6j − 4 gives every even number.’ False: it gives 2, 8, 14, … and misses 4 and 6.
• ‘2f + 3 gives both even and odd numbers.’ False: it is always odd.
• ‘2p + 1 gives every odd number when p = 1, 2, 3, ….’ False: it starts at 3 and misses 1, while 2q − 1 starts at 1.

Here is a puzzle. Fill a 3 × 3 grid with the numbers 1 to 9, each used once. Beside each row and below each column a circle shows the sum of that row or column. Given only the sums, can you fill the grid?

An example.

A surprising fact. Add the three row sums: 14 + 16 + 15 = 45. Add the three column sums: 14 + 15 + 16 = 45. This is not a coincidence. Every number from 1 to 9 is in exactly one row, so the row sums together add up all of them: 1 + 2 + … + 9 = 45. The same is true for the columns. So all six circles together always total 90.

Spotting impossible puzzles. The smallest possible sum of three different numbers from 1 to 9 is 1 + 2 + 3 = 6. The largest is 7 + 8 + 9 = 24. So if a puzzle shows a circle with 5 or 26, it cannot be solved. Also, if the three row sums do not add to 45, the puzzle is impossible. These checks save time before you start trying.

Strategies for solving.

• Start with the extreme sums. A row sum of 24 must be 7, 8 and 9; a row sum of 6 must be 1, 2 and 3.
• Look where a row and a column cross: the number there belongs to both sums.
• Use parity: a row sum that is odd needs one or three odd numbers in that row.
• Check every row and column at the end.

Many ways to fill. There are 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,62,880 ways to place 1 to 9 in a 3 × 3 grid. Trying them one by one is hopeless, so careful reasoning is the only sensible way.

Make your own. Fill a grid with 1 to 9 in any order, write down only the six sums, and give it to a friend. Check that your sums pass the tests: each between 6 and 24, rows adding to 45, columns adding to 45.

This kind of reasoning, using a total to check the parts, is used whenever accountants check that row and column totals of a table agree.

A magic square is a square grid of numbers in which every row, every column and both diagonals add up to the same number, called the magic sum. Let us build a 3 × 3 magic square with 1 to 9 by reasoning, not guessing.

Step 1: the magic sum. The three rows together contain 1 to 9, so they add to 45. In a magic square the three row sums are equal, so each is 45 ÷ 3 = 15.

Step 2: the centre. The centre number lies on the middle row, middle column and both diagonals. Could it be 9? Then 8 is somewhere else, and the line through 8 and the centre would need 8 + 9 + something = 15, so the third number would be −2. Impossible. Could it be 1? Then the line through 1 and 2 would need 12 more, which is too big. Checking the same way rules out 2, 3, 4, 6, 7 and 8. Only 5 works.

Step 3: where do 1 and 9 go? A corner number lies on three lines (a row, a column, a diagonal), so it must be part of three different triples that sum to 15. For 1, the only triples are 1 + 5 + 9 and 1 + 6 + 8, just two. So 1 cannot be in a corner; it goes in the middle of a side, with 9 opposite it (1 + 5 + 9 = 15). The same reasoning keeps 9 out of the corners.

Step 4: complete it. Put 1 at the top middle and 9 at the bottom middle. The top row needs two numbers adding to 14 with 1: these are 8 and 6. The rest follows:

Check: rows 15, 15, 15; columns 8 + 3 + 4 = 15, 1 + 5 + 9 = 15, 6 + 7 + 2 = 15; diagonals 8 + 5 + 2 = 15 and 6 + 5 + 4 = 15. Every other 1–9 magic square is this one turned or flipped.

The general pattern. If the centre is m, the square is:

Every line adds to 3m. With m = 5 we get the square above. With m = 20 we get a square of 16 to 24 with magic sum 60. Adding the same number to every entry, or multiplying every entry by the same number, gives another magic square: adding 1 to each entry raises the magic sum by 3, and doubling each entry doubles it.

Magic squares have fascinated people for more than two thousand years, and India has a special place in their story.

The Lo Shu square. The oldest recorded magic square comes from ancient China. A legend says a turtle came out of a flooded river with a 3 × 3 pattern of dots on its shell. Written as numbers it is the square 4 9 2 / 3 5 7 / 8 1 6, the same 1–9 square we built, turned over.

The Chautīsā Yantra. The first recorded 4 × 4 magic square is carved in a 10th-century inscription at the Pārśvanātha Jain temple in Khajuraho:

‘Chautīs’ means 34, and every row, column and diagonal adds to 34. For example, 7 + 12 + 1 + 14 = 34 and 7 + 13 + 10 + 4 = 34. Many other groups of four also make 34: the four corners, 7 + 14 + 9 + 4 = 34; the central 2 × 2 block, 13 + 8 + 3 + 10 = 34; and each 2 × 2 corner block, such as 7 + 12 + 2 + 13 = 34. Why 34? The numbers 1 to 16 add to 136, and 136 ÷ 4 = 34.

Yantras in daily life. 3 × 3 magic squares appear on a pillar of an 8th-century temple in Palani, Tamil Nadu, and in homes and shops as yantras. The Navagraha Yantras give a different magic sum for each graha. A Kubera Yantra uses the numbers 20 to 28:

Its centre is 24 and every line adds to 72 = 3 × 24, just as our general rule predicts. It is the 1–9 square with 19 added to every entry.

Indian methods. Indian mathematicians did not only find magic squares; they wrote down general methods for building them, including 4 × 4, 5 × 5 and larger squares. You will meet some of these methods in later classes.

Try it. Make a 3 × 3 magic square with magic sum 0 using some negative numbers. Using the general pattern with m = 0 gives 3, −4, 1 / −2, 0, 2 / −1, 4, −3. Check that every line adds to 0.

Magic squares show that mathematics has always been part of art, architecture and culture, not only of textbooks.

In the poetry of Sanskrit, Prakrit and many Indian languages, each syllable is short (1 beat) or long (2 beats). Poets asked: how many different rhythms fill exactly 8 beats? In maths language: in how many ways can we write 8 as a sum of 1s and 2s, where order matters?

Small cases.

A clever rule. Every rhythm of 5 beats starts with either a short syllable (1) or a long one (2). If it starts with 1, the rest is a 4-beat rhythm: there are 5 of those. If it starts with 2, the rest is a 3-beat rhythm: there are 3. So there are 5 + 3 = 8 rhythms of 5 beats. In the same way, 6 beats give 8 + 5 = 13 rhythms.

The sequence. Each number is the sum of the two before it: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, … So 8 beats give 34 rhythms.

History. The Prakrit scholar Virahāṅka wrote this rule around 700 CE, inspired by the earlier work of Piṅgala (around 300 BCE). Gopāla (around 1135 CE) and Hemacandra (around 1150 CE) also wrote about it. The Italian mathematician Fibonacci described the numbers in 1202 CE, about 500 years after Virahāṅka. That is why we call them the Virahāṅka–Fibonacci numbers.

Same problem, different story. Angaan climbs an 8-step staircase taking 1 or 2 steps at a time. The number of ways is again 34, because each way to climb is a sum of 1s and 2s making 8.

Parity pattern. The parities go odd, even, odd, odd, even, odd, odd, even, … The block ‘odd, even, odd’ repeats. Why? Odd + even = odd, even + odd = odd, odd + odd = even, and then it starts again. So every third term from the 2nd is even: the 2nd, 5th, 8th, …, 20th terms are even.

Going backwards. If two neighbours are 987 and 1597, the next ones are 987 + 1597 = 2584 and 1597 + 2584 = 4181. The one before 987 is 1597 − 987 = 610, and before that 987 − 610 = 377.

In nature. The number of petals on many daisies is a number from this sequence, such as 13, 21 or 34. These numbers link poetry, music, nature and mathematics.

In a cryptarithm (also called an alphametic), the digits of a sum are hidden behind letters. Each letter stands for one digit, different letters stand for different digits, and a number does not begin with 0. Your job is to be a detective and find the digits.

Puzzle 1: T + T + T = UT. Three copies of a digit T make a two-digit number ending in T. So 3 × T ends in T. Try the digits: 3 × 5 = 15 ends in 5. Other digits fail (3 × 2 = 6, 3 × 4 = 12, 3 × 6 = 18 …) or give a one-digit answer (3 × 0 = 0). So T = 5 and U = 1.

Puzzle 2: K2 + K2 = HMM. The units: 2 + 2 = 4, so M = 4. The tens: K + K must end in 4 and carry 1 so that the answer has three digits. K + K = 14 gives K = 7. So 72 + 72 = 144, with H = 1, M = 4.

Puzzle 3: AB + B = BA. Adding a single digit B to AB changes the tens digit from A to B, so there must be a carry and B = A + 1. In the units, B + B ends in A, with a carry: B + B = 10 + A. Put B = A + 1: 2A + 2 = 10 + A, so A = 8 and B = 9. Check: 89 + 9 = 98.

Puzzle 4: AA + B = BCC. Two digits plus one digit can be at most 99 + 9 = 108, so the answer is between 100 and 108 and B = 1. Then AA + 1 is at least 100, so AA = 99. The answer is 100, and C = 0. Check: 99 + 1 = 100.

Puzzle 5: UT + TA = TAT. Two two-digit numbers add to less than 200, so T = 1. Then U1 + 1A = 1A1. The units: 1 + A ends in 1, so A = 0. Then U1 + 10 = 101, so U1 = 91 and U = 9.

Detective tips.

• The first digit of a longer answer is usually a carry, so it is often 1.
• Look at the units column first; it has no carry coming in.
• Use limits: the largest two-digit sum is 99 + 99 = 198.
• Use parity: a digit added to itself always gives an even units digit.
• Always check your answer by doing the sum.

Why bother? Cryptarithms train the same careful thinking that this whole chapter uses: look for what must be true, rule out what cannot be, and prove your answer works. Make one for a friend: write a true sum, replace the digits by letters, and check that the puzzle has only one answer.