Ganita Prakash · Chapter 3 — Number Play

Numbers do more than count. They can carry information. Five children of different heights stand in a line. Each child says a number: how many of the children standing right next to them are taller than them. A child with both neighbours taller says 2. A child with one taller neighbour says 1. A child with no taller neighbour says 0.

Say the heights from left to right are 120, 135, 128, 140 and 132 cm. The first child has one neighbour, 135, who is taller: says 1. The second (135) has neighbours 120 and 128, both shorter: says 0. The third (128) has 135 and 140 beside them, both taller: says 2. The fourth (140) is the tallest: says 0. The fifth (132) has only the 140 beside them: says 1. So the line says 1, 0, 2, 0, 1.

Reasoning with the rule. Now we can answer questions without trying every arrangement.

• Can a child at the end say 2? No. An end child has only one neighbour, so the most they can say is 1.
• Can everyone say 0? No. Look at any two neighbours: one of them is shorter, and that child has a taller neighbour, so says at least 1.
• Can everyone say 1? No. The tallest child has no taller neighbour, so the tallest always says 0.
• Can four say 1 and one say 0? Yes. Stand in order from shortest to tallest: the line says 1, 1, 1, 1, 0.
• Is 0, 1, 2, 1, 0 possible? Yes. Put the shortest child in the middle and taller children towards the ends. Heights 150, 130, 120, 140, 160 give exactly 0, 1, 2, 1, 0.

How many can say 2? A child who says 2 cannot be at an end, and two such children cannot stand side by side (the taller of the two would not have both neighbours taller). With 5 children, the middle positions 2 and 4 can both say 2, so the most is 2. Example heights: 150, 120, 160, 125, 155 give 0, 2, 0, 2, 0.

The bigger lesson. We did not need to test all 120 ways of lining up five children. A short argument, ‘the tallest always says 0’, settled the question for every arrangement at once. That is the power of reasoning in mathematics.

Numbers are written in a row of cells. A cell is a supercell if its number is larger than the numbers in all the cells next to it. A cell at the end has only one neighbour, so it needs to beat only that one.

Here 626 is a supercell, as it is larger than both 577 and 345. 200 is not, because 577 beside it is larger. The end cell 198 has only one neighbour, 109, and 198 is larger, so 198 is a supercell too. 345 and 109 are not. Check each cell this way, one neighbour at a time.

Facts about supercells.

• The cell with the largest number is always a supercell, because no neighbour can be bigger.
• The cell with the smallest number is never a supercell, because every neighbour is bigger.
• Two supercells can never sit side by side: of two neighbours, one is smaller than the other.
• So with numbers all different, there is always at least one supercell, and you cannot fill a table with none.

How many supercells at most? Since supercells cannot touch, at most every second cell can be one. In a row of 9 cells, put large numbers in cells 1, 3, 5, 7 and 9 and small numbers in cells 2, 4, 6 and 8. That gives 5 supercells. For a row of 10 cells, the most is also 5.

Can the second-largest number fail to be a supercell? Yes: place it right next to the largest. Can the second-smallest be a supercell? Yes, if it is placed at the end next to the smallest, so its only neighbour is smaller.

Grids. In a grid, the neighbours of a cell are the cells directly left, right, above and below. The rule stays the same: a supercell must be larger than all of its neighbours. In a grid, 8632 is a supercell if it is greater than 4580, 8280, 4795 and 1944 around it, and it is.

Using place value. Supercell puzzles train you to compare big numbers quickly. To compare 60,319 and 61,039, look at the first place where they differ: the thousands digit, 0 against 1, so 61,039 is bigger. Commas after the thousands digit make 5-digit numbers easier to read.

How many numbers have each number of digits? There are 9 one-digit counting numbers (1 to 9). Two-digit numbers run from 10 to 99, so there are 99 − 10 + 1 = 90. Three-digit numbers run from 100 to 999: 900. Four-digit: 9,000. Five-digit: 90,000. Each time the count is multiplied by 10.

Placing numbers on a number line. On a line from 0 to 10,000, 1050 sits just past 1000 and 9950 sits just before 10,000. 5030 and 5300 look alike, but 5300 is further right because its hundreds digit is larger. Numbers with the same digits can be far apart: 9590 is to the left of 9950.

Digit sums. The digit sum of a number is what you get by adding its digits. 68 has digit sum 6 + 8 = 14. So do 176 (1 + 7 + 6) and 545 (5 + 4 + 5). Many numbers share a digit sum.

• Smallest number with digit sum 14. Use as few digits as possible, with the biggest digit last: two digits, 5 and 9, give 59.
• Largest 5-digit number with digit sum 14. Put the biggest digits first: 9, then 5, then zeros: 95,000.
• Even bigger numbers. With no limit on the number of digits, a number made of fourteen 1s has digit sum 14, and putting zeros at its end makes it larger and larger without changing the digit sum. So there is no largest such number.

A pattern. 3-digit numbers with consecutive digits: 123 has digit sum 6, 234 has 9, 345 has 12, 456 has 15. The digit sum is always 3 times the middle digit, because the three digits are (middle − 1), middle and (middle + 1).

Counting a digit. Dinesh writes the numbers from 1 to 100. How many times does he write 7? In the units place: 7, 17, 27, …, 97, which is 10 times. In the tens place: 70 to 79, another 10 times (77 counts in both places, since it has two 7s). Total 20. From 1 to 1000, each of the three places has 7 a hundred times: 100 + 100 + 100 = 300.

Why digit sums are useful. They help in checking calculations and, in a later chapter, in divisibility tests. For now, notice how much can be found by looking at digits alone.

Look at 66, 848, 575, 797 and 1111. Each reads the same from left to right and from right to left. Such numbers are called palindromes or palindromic numbers. Words can be palindromes too, like ‘madam’ and ‘level’.

Palindromes from the digits 1, 2, 3. A 3-digit palindrome has the form aba: the first and last digits are the same. With digits 1, 2 and 3 (repeats allowed), there are 3 choices for a and 3 for b, so 3 × 3 = 9 palindromes: 111, 121, 131, 212, 222, 232, 313, 323, 333.

Reverse and add. Start with a 2-digit number. Add it to its reverse. If the answer is a palindrome, stop; if not, repeat with the answer.

Some numbers need many steps. 89 needs 24 steps and reaches the large palindrome 8,813,200,023,188. Yet every 2-digit number does reach a palindrome in the end. For 3-digit numbers nobody knows for sure: starting with 196, no one has ever reached a palindrome, even with computers, and it is suspected that it never will.

A palindrome riddle. ‘I am a 5-digit palindrome. I am odd. My tens digit is double my units digit, and my hundreds digit is double my tens digit.’ A 5-digit palindrome has the form abcba, so the units digit is a, the tens digit b and the hundreds digit c. Then b = 2a and c = 2b = 4a. Since c is a single digit, 4a is at most 9, so a is 1 or 2. Odd means a is odd, so a = 1, b = 2, c = 4. The number is 12421, twelve thousand four hundred twenty-one.

Smallest and largest 5-digit palindromes. The smallest is 10001 and the largest is 99999. Their sum is 110000 and their difference is 89998.

Palindromes in time. On a clock, 12:21 and 4:44 read the same both ways. The date 11/02/2011 is a palindrome too: 11022011 read backwards is the same.

D. R. Kaprekar was a mathematics teacher at a government school in Devlali, Maharashtra, who loved playing with numbers. In 1949 he found an amazing pattern with 4-digit numbers. Try it.

• Pick a 4-digit number whose digits are not all the same, say 6382.
• Arrange its digits to make the largest number, A = 8632.
• Arrange them to make the smallest number, B = 2368.
• Subtract: C = A − B = 6264.
• Repeat the steps with C.

Once we reach 6174, it reproduces itself: 7641 − 1467 = 6174 again. Whatever 4-digit number you start with (with at least two different digits), you always reach 6174, and in at most 7 rounds. This number is now called the Kaprekar constant.

A longer example. Start with 5683: 8653 − 3568 = 5085; 8550 − 0558 = 7992; 9972 − 2799 = 7173; 7731 − 1377 = 6354; 6543 − 3456 = 3087; 8730 − 0378 = 8352; 8532 − 2358 = 6174. That is 7 rounds, the most any number needs.

Keep the zeros. If a difference has only 3 digits, such as 999, treat it as 0999. The largest arrangement is then 9990 and the smallest 0999. Forgetting the zero breaks the magic.

Three digits. Do the same with 3-digit numbers: 352 gives 532 − 235 = 297, then 972 − 279 = 693, then 963 − 369 = 594, then 954 − 459 = 495, and 495 repeats forever. So 495 is the magic number for three digits.

Why digits must differ. For 3333, the largest and smallest arrangements are equal and the difference is 0, so the process stops at once.

Your turn. Try your year of birth. Kaprekar’s discovery shows that simple steps that anyone can follow can hide a deep pattern. Following a fixed set of steps exactly like this is called an algorithm.

Clock patterns. On a 12-hour clock, some times look special: 4:44 has one repeated digit, 10:10 repeats a pair, and 12:21 is a palindrome. If it is 10:01 now, the next palindromic time is 11:11, which is 70 minutes later, and the one after that is 12:21, another 70 minutes.

Calendar patterns. Manish was born on 20/12/2012, where the digits 2, 0, 1, 2 appear in that order twice. His sister Meghana was born on 11/02/2011, and 11022011 is a palindrome. Can a calendar be reused? A year’s calendar repeats when a later year starts on the same day of the week and is also a leap year or not a leap year, like it. The calendar for 2026 can be used again in 2037.

Mental maths with friendly numbers. To add or subtract big numbers in your head, break them into easy parts. 38,800 = 25,000 + 13,000 + 400 + 400. 39,800 = 40,000 − 800 + 300 + 300. Rounding to a nearby friendly number and then adjusting is a strong mental trick: 4,998 + 3,215 = 5,000 + 3,215 − 2 = 8,213.

Largest and smallest from digits. Pratibha uses the digits 4, 7, 3 and 2 to make the largest number 7432 and the smallest 2347. Their difference is 5085 and their sum is 9779. Using different digits changes these: digits 9, 8, 1, 0 give 9810 − 1089 = 8721.

Always, sometimes or never? Test each statement with examples and a reason.

Smart adding. When the same number repeats, multiply instead of adding: 32 written 32 times is 32 × 32 = 1024. Grouping numbers that make 10 or 100 also speeds things up.

Here is a very simple rule. Start with any whole number. If it is even, halve it. If it is odd, multiply by 3 and add 1. Repeat with the answer.

Every one of these reaches 1. (After 1 the rule gives 4, 2, 1 again, so we stop at 1.) Notice how the numbers jump up and down like hailstones before falling to 1: starting from 7, the sequence climbs to 52 before coming down.

The conjecture. In 1937 the German mathematician Lothar Collatz guessed that this always happens: whatever whole number you start with, you will reach 1. A guess that has not been proved is called a conjecture. Computers have checked an enormous number of starting values and found no exception, yet nobody has proved that there is none. It is one of the most famous unsolved problems in mathematics.

Powers of 2 are easy. If you start with a power of 2, such as 64, every number is even, and halving takes you 64, 32, 16, 8, 4, 2, 1. So the conjecture is true for all powers of 2. Many sequences end by falling onto a power of 2: 21 jumps to 64 and then slides down.

Try 100. 100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. It reaches 1 after 25 steps.

What we learn. A problem can be easy to state, so that a Class 6 student understands it fully, and yet be unsolved by the best mathematicians in the world. Checking many cases is evidence, not proof. Maybe one day someone, perhaps you, will settle it.

Often we do not need an exact number; a good estimate is enough. Paromita’s class has three sections with 32, 29 and 35 children. She thinks ‘about 30 each, so about 100 in Class 6’. Her school has Classes 6 to 10, each with 3 sections of similar size, so she estimates about 5 × 100 = 500 students. The exact total does not matter for her purpose.

How to estimate.

• Round the numbers to easy values.
• Break a big question into small steps you can guess.
• Use what you know: your step is about half a metre; a day has 24 hours; a year has about 200 school days.
• Check whether the answer is sensible.

Checking a claim. Sheetal in Class 6 says she has spent about 13,000 hours in school. Let us estimate. A school day has about 6 hours; a year has about 200 school days, which is about 1,200 hours a year. By Class 6 she has done about 7 years (including pre-primary), so roughly 7 × 1,200 = 8,400 hours. 13,000 hours is too many; even 13,000 hours spread over 7 years would need over 1,800 school hours a year.

Shopping. Roshan wants milk and three fruits for a custard for 5 people and guesses ₹100. Milk alone for 5 people may cost about ₹30 to ₹40, and three kinds of fruit may cost ₹100 or more, so ₹100 is probably too little.

Distances. Gandhinagar in Gujarat and Kohima in Nagaland are on opposite sides of India. On a map, the gap is roughly 2,200 km in a straight line, and more by road. Walking at about 5 km an hour for 8 hours a day covers 40 km a day, so the walk would take about 55 days, nearly two months.

Different estimates are fine. Two good estimators may give different answers. What matters is a sensible method that you can explain. Make your own estimation questions: how many leaves on a tree, litres in a bucket, words in your textbook.

Numbers can help you win games. Try the famous game of 21 with a friend.

Rules. The first player says 1, 2 or 3. Players then take turns, each adding 1, 2 or 3 to the last number said. Whoever says 21 wins.

Work backwards. If you say 17, your friend must say 18, 19 or 20, and then you can reach 21. So 17 is a winning number. In the same way, if you say 13, your friend can only reach 14, 15 or 16, and you can then say 17. Going back in steps of 4, the winning numbers are 21, 17, 13, 9, 5, 1. Why steps of 4? Whatever your friend adds (1, 2 or 3), you add the amount that makes 4 (3, 2 or 1).

Who wins? The first player can say 1, a winning number, and then keep to 5, 9, 13, 17, 21. So the first player can always win by playing correctly.

The 99 game. Here players add a number from 1 to 10, so the step is 1 + 10 = 11. Counting back from 99 in 11s gives 88, 77, …, 22, 11. The first player must start with a number from 1 to 10 and cannot say 11. The second player then says 11 and wins by always making the total jump by 11.

The general secret. If each turn may add 1 up to some number k, the key step is k + 1. Find the remainder when the target is divided by k + 1. If the remainder is not 0, the first player says that remainder and wins. If it is 0, the second player wins. For 21 and step 4: remainder 1. For 22: remainder 2. For 99 and step 11: remainder 0.

Make your own. Change the target or how much may be added, then find the winning numbers. Thinking of a plan that always works is part of computational thinking: designing a set of steps that solves every case.