Hello my dear students! Welcome to today's mathematics class. I am so happy to see you all here, ready to learn something new and exciting. Today, we are going to explore Chapter 3 of your Mathematics textbook, and the title of this chapter is "Number Play". Now, doesn't that sound interesting? We are going to play with numbers, discover amazing patterns, and learn how numbers can tell us things in ways we never imagined before. Are you ready for this adventure? Let's begin!
Numbers are used in different contexts and in many different ways to organise our lives. We have used numbers to count since we were very young, and we have applied the basic operations of addition, subtraction, multiplication, and division on them to solve problems related to our daily lives. But in this chapter, we will continue this journey by playing with numbers, seeing numbers around us, noticing patterns, and learning to use numbers and operations in new ways.
Think about various situations where you use numbers in your daily life. Can you list five different situations in which numbers are used? You might think about time, when you check the clock in the morning to know when to wake up. You might think about the calendar, to know which day it is or when your birthday is coming. You use numbers when you count your marks in an exam, or when you measure your height and weight at the doctor's clinic. You also use numbers when you count money to buy something from a shop. These are just some examples, but there are countless situations where numbers help us. I want you to think about this and discuss with your classmates what situations you can think of where numbers are used.
Now, let's start with our first exciting concept. We call it "Numbers can tell us things". Imagine some children standing in a line in a park. Each child says a number. What do you think these numbers mean? The children now rearrange themselves, and again each one says a number based on the new arrangement. Can you figure out what these numbers represent?
Here is the hint: Could their heights be playing a role? Let me explain the rule. A child says '1' if there is only one taller child standing next to them. A child says '2' if both the children standing next to them are taller. A child says '0' if neither of the children standing next to them are taller. That is, each person says the number of taller neighbours they have. So, if you are standing in a line with your friends, and you look at the friends immediately to your left and right, you count how many of them are taller than you, and that becomes your number.
Now, let's think about some interesting questions. Can the children rearrange themselves so that the children standing at the ends say '2'? Think about this carefully. The child at the very end of the line has only one neighbour, because there is no one on one side of them. So, even if both their neighbours were taller, they can only say '1' at the most. They can never say '2' because they don't have two neighbours. So the answer is no, it is not possible for the children at the ends to say '2'.
Can we arrange the children in a line so that all would say only 0s? This is possible if all the children in the line are of the same height. If everyone is of equal height, then no one has a taller neighbour, so everyone would say '0'. In real life, this might be hard to find, but mathematically, it is possible.
Can two children standing next to each other say the same number? Yes, this is definitely possible. For example, if you have three children of increasing height standing in a line, the middle child might say '1' because only one neighbour is taller, and one of the end children might also say '1' if they have only one neighbour who is taller.
Now, let's think about a group of 5 children, all of different heights. Can they stand such that four of them say '1' and the last one says '0'? Why or why not? Yes, they can! If the children stand in ascending order of height, from shortest to tallest, then the shortest child at the end has only one neighbour (the second shortest), and that neighbour is taller, so they would say '1'. The second shortest child will have two neighbours - the shortest (shorter) and the middle (taller), so only one neighbour is taller, so they would say '1'. The middle child will have two neighbours - the second shortest (shorter) and the third shortest (taller), so again only one neighbour is taller, so they would say '1'. The third shortest child has two neighbours - the middle (shorter) and the tallest (taller), so only one neighbour is taller, so they would say '1'. The tallest child has two neighbours, both shorter than them, so they say '0'. So actually, we get four children saying '1' and one saying '0'. So yes, it is possible!
For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible? This would mean that all five children say '1'. But think about the tallest child at the end. If they are the tallest, both their neighbours must be shorter, so they would say '0', not '1'. So this sequence is not possible.
Is the sequence 0, 1, 2, 1, 0 possible? Let's think. The first child says '0', which means neither neighbour is taller. So the second child must be taller than the first. The second child says '1', which means exactly one neighbour is taller. Since the first child is shorter, the third child must be taller than the second. The third child says '2', which means both neighbours are taller. So the second and fourth children must both be taller than the third. But we already established that the second child is shorter than the third (because the third is taller than the second). This is a contradiction. So the sequence 0, 1, 2, 1, 0 is not possible. But wait, let me reconsider. What if we arrange them in a different order? Actually, after careful thought, the sequence 0, 1, 2, 1, 0 is indeed possible if we arrange them in descending order of height from left to right. Let me explain: the tallest child at position 1 would have only one neighbour (the second tallest), and that neighbour is shorter, so they would say '0'. The second tallest child would have two neighbours - the tallest (taller) and the third tallest (shorter), so exactly one neighbour is taller, so they say '1'. The third tallest child would have two neighbours - both taller than them, so they say '2'. And so on. So yes, the sequence 0, 1, 2, 1, 0 is possible!
How would you rearrange the five children so that the maximum number of children say '2'? At the most, only two children can say '2', and this happens when the tallest children are in the middle of the line, with shorter children on both sides. So, if you place the tallest child in the middle, with the second and third tallest on either side, and the shortest two at the ends, then the middle child (tallest) would have both neighbours shorter, so they would say '0', not '2'. Actually, to get '2', you need both neighbours to be taller. So you need to place shorter children in the middle and taller children on the sides. The maximum number of children who can say '2' is actually 2, when you arrange them with the shortest in the middle and the two tallest at the ends.
Now, let's move on to our next exciting topic: "Supercells". This is a fascinating concept where we look at numbers in a table and identify special cells.
Observe the numbers written in the table below. Why are some numbers coloured? Let me give you the rule: A cell is coloured if the number in it is larger than its adjacent cells. Let me explain what adjacent cells mean. In a row, the cells immediately to the left and right are adjacent. In a column, the cells immediately above and below are adjacent. For example, in the table shown in your book, the number 626 is coloured because it is larger than 577 (to its left) and 345 (to its right). The number 200 is not coloured because it is smaller than 577. The number 198 is coloured because it has only one adjacent cell with 109 in it, and 198 is larger than 109. So, a supercell is simply a cell that is greater than all of its neighbouring cells.
Now, let's try to colour or mark the supercells in the table given in your book: 6828, 670, 9435, 3780, 3708, 7308, 8000, 5583, 52. Let me think through this with you. We compare each number with its neighbours. 6828 is greater than 670, so it is a supercell. 670 is less than 6828 but we need to check its other neighbour. Wait, in a single row, each cell has at most two neighbours - one on the left and one on the right. So for 670, its neighbours are 6828 and 9435. Since 670 is smaller than both, it is not a supercell. 9435 is greater than 670 and 3780, so it is a supercell. 3780 is less than 9435 but greater than 3708, so it is not a supercell. 3708 is less than 3780 but greater than 7308? No, wait, 3708 is less than 7308, so it is not a supercell. 7308 is greater than 3708 and 8000? No, 7308 is less than 8000, so it is not a supercell. 8000 is greater than 7308 and 5583, so it is a supercell. 5583 is less than 8000 but greater than 52, so it is not a supercell. 52 is less than 5583, so it is not a supercell. So the supercells are 6828, 9435, and 8000.
Now, let's think about some interesting questions about supercells. Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not? Think about this: the cell which has the greatest number among all the numbers will always be a supercell, because it is greater than all its neighbours. No matter where you place the largest number, it will always be greater than its neighbours, so it will always be a supercell. So it is impossible to have a table with no supercells if you are using distinct numbers.
Will the cell having the largest number in a table always be a supercell? Yes, definitely! Because it is larger than every other number, it is certainly larger than its neighbours. So the largest number is always a supercell.
Can the cell having the smallest number in a table be a supercell? No, never! Because the smallest number is smaller than all other numbers, it is certainly smaller than its neighbours. So the smallest number can never be a supercell.
Now, let's explore supercells in a table with multiple rows and columns. Here, the neighbouring cells are those that are immediately to the left, right, top, and bottom. The rule remains the same: a cell becomes a supercell if the number in it is greater than all the numbers in its neighbouring cells. In Table 1 in your book, 8632 is greater than all its neighbours 4580, 8280, 4795, and 1944. So 8632 is a supercell.
Now, let's move on to our next topic: "Patterns of Numbers on the Number Line". We are quite familiar with number lines now. Let's see if we can place some numbers in their appropriate positions on the number line. Here are the numbers: 2180, 2754, 1500, 3600, 9950, 9590, 1050, 3050, 5030, 5300, and 8400. We need to arrange these numbers in order on the number line. The smallest number is 1050, then comes 1500, then 2180, then 2754, then 3050, then 3600, then 5030, then 5300, then 8400, then 9590, and finally 9950. So the order from smallest to largest is: 1050, 1500, 2180, 2754, 3050, 3600, 5030, 5300, 8400, 9590, 9950.
Now, let's identify the numbers marked on some number lines. In the first number line, we have 2010 and 2020 marked. The numbers between them would be 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018, 2019. In the second number line, we have 9996 and 9997 marked. These are consecutive whole numbers, so there are no whole numbers between them. The remaining positions on this number line would be other numbers in the sequence like 9993, 9994, 9995, 9998, 9999, 10000. In the third number line, we have 15077, 15078, and 15083 marked. So the remaining positions between 15078 and 15083 would be 15079, 15080, 15081, 15082, and after 15083 would be 15084, 15085, 15086. In the fourth number line, we have 86705 and 87705 marked. So the remaining positions would be 83705, 84705, 85705, 86705, 87705, 88705, 89705, 90705, 91705, 92705.
Now, let's move on to "Playing with Digits". We start writing numbers from 1, 2, 3, and so on. There are nine 1-digit numbers, from 1 to 9. How many numbers have two digits? They are from 10 to 99, so there are 90 two-digit numbers. How many numbers have three digits? They are from 100 to 999, so there are 900 three-digit numbers. How many numbers have four digits? They are from 1000 to 9999, so there are 9000 four-digit numbers. How many numbers have five digits? They are from 10000 to 99999, so there are 90000 five-digit numbers. So the table becomes: 1-digit numbers: 9, 2-digit numbers: 90, 3-digit numbers: 900, 4-digit numbers: 9000, 5-digit numbers: 90000. Do you see a pattern? Each time we increase the number of digits, we multiply the count by 10. This makes sense because for each additional digit, we have 10 choices (0 through 9), except for the first digit which cannot be 0.
Now, let's talk about "Digit sums". Komal observes that when she adds up the digits of certain numbers, the sum is the same. For example, adding the digits of the number 68 gives 6 + 8 = 14. Adding the digits of 176 gives 1 + 7 + 6 = 14. Adding the digits of 545 gives 5 + 4 + 5 = 14. So all these numbers have a digit sum of 14. The digit sum is simply the sum of all the digits in a number.
Let's explore digit sum 14. Write other numbers whose digits add up to 14. There are many! For example, 59 gives 5 + 9 = 14. 68 gives 6 + 8 = 14. 77 gives 7 + 7 = 14. 86 gives 8 + 6 = 14. 95 gives 9 + 5 = 14. We can also have three-digit numbers like 149 (1 + 4 + 9 = 14), 158 (1 + 5 + 8 = 14), 167, 176, 185, 194, 239, 248, and so on. There are countless numbers whose digits add up to 14.
What is the smallest number whose digit sum is 14? To get the smallest number, we want to use the smallest digits possible in the leftmost positions. The smallest number would be 59, because 5 + 9 = 14, and any number with a digit greater than 5 in the tens place would be larger. What about 68? 68 is larger than 59. What about 77? Larger still. So 59 is the smallest.
What is the largest 5-digit number whose digit sum is 14? To get the largest 5-digit number, we want the leftmost digit to be as large as possible. But we also need the digit sum to be 14. So we can put 9 in the first digit, then another 9 would give us 18, which is too much. So we put 9 in the first digit, then we need the remaining four digits to sum to 5. The largest arrangement would be 95000, because 9 + 5 + 0 + 0 + 0 = 14. But wait, is 95000 the largest? What about 86000? 8 + 6 + 0 + 0 + 0 = 14, and 86000 is smaller than 95000. What about 80060? 8 + 0 + 0 + 6 + 0 = 14, but 80060 is smaller than 86000. So 95000 is indeed the largest 5-digit number with digit sum 14.
How big a number can you form having the digit sum of 14? Can you make an even bigger number? Well, we can keep adding zeros to the right, like 950000, 9500000, and so on. These are all 6-digit, 7-digit numbers and beyond, and they all have digit sum 14. So there is no limit to how big the number can be!
Now, let's calculate the digit sums of 3-digit numbers whose digits are consecutive. For example, 345 gives 3 + 4 + 5 = 12. What about 234? 2 + 3 + 4 = 9. What about 456? 4 + 5 + 6 = 15. Do you see a pattern? The digit sums are 6, 9, 12, 15, 18, 21, 24. These are all multiples of 3! This pattern will continue because when you have three consecutive digits, say n, n+1, and n+2, their sum is 3n + 3, which is always divisible by 3. So yes, the digit sums of 3-digit numbers with consecutive digits are always multiples of 3.
Now, let's be "Digit Detectives". After writing numbers from 1 to 100, Dinesh wondered how many times he would have written the digit '7'! Among the numbers 1 to 100, how many times will the digit '7' occur? Let's think about this. In the ones place, '7' appears in 7, 17, 27, 37, 47, 57, 67, 77, 87, 97 - that's 10 times. In the tens place, '7' appears in 70, 71, 72, 73, 74, 75, 76, 77, 78, 79 - that's 10 times. But wait, we counted 77 twice - once for the ones place and once for the tens place. So the total is 10 + 10 - 1 = 19 times. But what about 77? In 77, we have two sevens. So when we count the ones place, we count one seven from 77, and when we count the tens place, we count another seven from 77. So we need to add that extra one. So actually, it's 10 + 10 = 20 times. Let me list them: 7, 17, 27, 37, 47, 57, 67, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 87, 97. That's 19 numbers, but 77 has two sevens, so total sevens = 19 + 1 = 20. Yes, the answer is 20 times.
Among the numbers 1 to 1000, how many times will the digit '7' occur? This is a bit more complicated, but we can think about it systematically. In the ones place, '7' appears in every ten numbers, so from 1 to 1000, there are 100 tens, so 100 times. In the tens place, '7' appears in every hundred numbers, ten times each hundred, so 10 × 10 = 100 times. In the hundreds place, '7' appears in the numbers 700 to 799, which is 100 numbers. So total = 100 + 100 + 100 = 300 times. But wait, we need to be careful about numbers like 777, which has three sevens. Let me think again. For the ones place: from 0 to 999, there are 1000 numbers, and in each block of 10 numbers, the ones digit is 7 once. So 1000 ÷ 10 = 100 times. For the tens place: in each block of 100 numbers, the tens digit is 7 ten times. So 1000 ÷ 100 = 10 blocks, and 10 × 10 = 100 times. For the hundreds place: in each block of 1000 numbers, the hundreds digit is 7 for 100 numbers (700-799). So 1000 ÷ 1000 = 1 block, and 1 × 100 = 100 times. So total = 100 + 100 + 100 = 300 times. But we haven't counted the number 1000 itself, which doesn't have a 7. So the answer is 300 times.
Now, let's explore "Pretty Palindromic Patterns". What pattern do you see in these numbers: 66, 848, 575, 797, 1111? These numbers read the same from left to right and from right to left. Try and see. Such numbers are called "palindromes" or "palindromic numbers". For example, 66 reads the same forwards and backwards. 848 reads the same forwards and backwards. 575 reads the same forwards and backwards. Even 1111 reads the same forwards and backwards because all the digits are the same.
Now, let's find all palindromes using the digits 1, 2, and 3. The numbers 121, 313, 222 are some examples of palindromes using these digits. Write all possible 3-digit palindromes using these digits. For a 3-digit palindrome, the first and last digits must be the same. So we can choose the first digit as 1, 2, or 3. The middle digit can be 1, 2, or 3. So the possibilities are: 111, 121, 131, 222, 212, 232, 313, 323, 333. That's 9 palindromes in total.
Now, let's explore "Reverse-and-add palindromes". Look at these additions. Try to figure out what is happening. The steps to follow are: Start with a 2-digit number. Add this number to its reverse. Stop if you get a palindrome, or else repeat the steps of reversing the digits and adding. For example, let's start with 12. Reverse it to get 21. Add them: 12 + 21 = 33, which is a palindrome! So we stopped after one step. Let's try another number, say 47. Reverse it to get 74. Add them: 47 + 74 = 121, which is a palindrome! So we stopped after one step again. Let's try 18. Reverse it to get 81. Add them: 18 + 81 = 99, which is a palindrome! It seems like many numbers become palindromes in just one step. But there are some numbers for which you have to repeat this a large number of times. Are there numbers for which you do not reach a palindrome at all? This is actually an interesting question in mathematics. For 2-digit numbers, it is believed that you will always eventually reach a palindrome. But for 3-digit numbers, it is suspected that starting with 196 never yields a palindrome, but this has not been proven yet! This is one of the unsolved mysteries of mathematics.
Now, let's solve a puzzle. Write the number in words: "I am a 5-digit palindrome. I am an odd number. My 't' digit is double of my 'u' digit. My 'h' digit is double of my 't' digit. Who am I?" Let's think about this. We have a 5-digit palindrome, so it looks like ABCBA, where A is the first digit, B is the second, C is the middle, B is the fourth, and A is the last. The number is odd, so the last digit (which is A) must be odd. The 't' digit is double of the 'u' digit. Let's assume the 'u' digit is the units digit, which is A. So the 't' digit, which is the tens digit (B), is double of A. So B = 2A. Since A is a digit from 0 to 9, and B must also be a digit from 0 to 9, and B = 2A, the possible values are: if A = 1, then B = 2; if A = 2, then B = 4; if A = 3, then B = 6; if A = 4, then B = 8; if A = 5, then B = 10, which is not a digit. So the possible pairs are (A, B) = (1, 2), (2, 4), (3, 6), (4, 8). Now, the 'h' digit is double of the 't' digit. The 'h' digit is the hundreds digit (which is C), and the 't' digit is the tens digit (which is B). So C = 2B. If B = 2, then C = 4; if B = 4, then C = 8; if B = 6, then C = 12, which is not a digit; if B = 8, then C = 16, which is not a digit. So the only possibilities are (A, B, C) = (1, 2, 4) or (2, 4, 8). But the number must be odd, so A must be odd. So A = 1, B = 2, C = 4. So the number is 12421. Let's check: it is a 5-digit palindrome, it is odd, the tens digit (2) is double of the units digit (1), and the hundreds digit (4) is double of the tens digit (2). So the answer is 12421, which in words is "twelve thousand four hundred twenty-one".
Now, let's explore "The Magic Number of Kaprekar". D.R. Kaprekar was a mathematics teacher in a government school in Devlali, Maharashtra. He liked playing with numbers very much and found many beautiful patterns in numbers that were previously unknown. In 1949, he discovered a fascinating and magical phenomenon when playing with 4-digit numbers.
Follow these steps and experience the magic for yourselves! Pick any 4-digit number having at least two different digits, say 6382. First, make the largest number from these digits. Arrange the digits in descending order to get 8632. Call it A. Then, make the smallest number from these digits. Arrange the digits in ascending order to get 2368. Call it B. Now, subtract B from A: C = 8632 - 2368 = 6264. What happens if we continue doing this? Now, take the number 6264. Make the largest number: 6642. Make the smallest number: 2466. Subtract: 6642 - 2466 = 4176. Continue: take 4176, largest is 7641, smallest is 1467, subtract: 7641 - 1467 = 6174. Now, if we continue with 6174, we get: largest is 7641, smallest is 1467, subtract: 7641 - 1467 = 6174 again! So we are stuck at 6174. This is amazing! No matter what 4-digit number you start with (as long as it has at least two different digits), you will always eventually reach the magic number 6174! The number 6174 is now called the "Kaprekar constant". This is truly magical, isn't it?
You can try this with different 4-digit numbers and see what happens. You will always reach the magic number 6174! This is one of the most beautiful discoveries in mathematics, and it was made by an Indian mathematician!
Carry out these same steps with a few 3-digit numbers. What number will start repeating? Let's try with 321. Largest: 321, smallest: 123, difference: 321 - 123 = 198. Next: largest: 981, smallest: 189, difference: 981 - 189 = 792. Next: largest: 972, smallest: 279, difference: 972 - 279 = 693. Next: largest: 963, smallest: 369, difference: 963 - 369 = 594. Next: largest: 954, smallest: 459, difference: 954 - 459 = 495. Next: largest: 954, smallest: 459, difference: 954 - 459 = 495 again! So we get stuck at 495. So for 3-digit numbers, the magic constant is 495.
Now, let's explore "Clock and Calendar Numbers". On the usual 12-hour clock, there are timings with different patterns. For example, 4:44, 2:22, 3:33, 10:10, 11:11, 12:12, 09:09 are times where all the digits shown are the same. Try and find out all possible times on a 12-hour clock of each of these types. For times like 4:44, the hour and minute digits are all the same. So possible times are 1:11, 2:22, 3:33, 4:44, 5:55, 9:09, 10:10, 11:11, and 12:12. For times like 10:01 and 12:21, the hour and minute digits are reversed. So possible times are 01:10, 02:20, 03:30, 04:40, 05:50, 10:01, and 12:21.
Now, let's talk about dates. Manish has his birthday on 20/12/2012 where the digits '2', '0', '1', and '2' repeat in that order. Find some other dates of this form from the past. We need dates where the day, month, and year have repeating digits in order. For example, 20/04/2004 would be 20, 04, 2004, and the digits are 2, 0, 0, 4, 2, 0, 0, 4. That's not quite right. What about 21/12/2012? That's 21, 12, 2012, and the digits are 2, 1, 1, 2, 2, 0, 1, 2. That's not it either. Let me think of 30/03/3003? That would be in the future. What about 10/02/1002? That would be in the past. Actually, let's look for dates like 20/02/2002: 20, 02, 2002 gives digits 2, 0, 0, 2, 2, 0, 0, 2. That's 2, 0, 0, 2, 2, 0, 0, 2. That's not quite right. What about 21/02/2100? No. Let me think differently. We want the digits to repeat in order. For example, 20/12/2012 gives 2, 0, 1, 2, 2, 0, 1, 2. So the pattern is 2012 repeated. So we need dates where the day-month-year gives a repeating pattern. For example, 01/01/1111 would be 0, 1, 1, 1, 1, 1, 1, 1. That's 01111111. Not quite. What about 11/11/1111? That's 1, 1, 1, 1, 1, 1, 1, 1. That's all ones! So 11/11/1111 is a date where all digits are 1. But that's in the past. What about 12/21/2012? That's 1, 2, 2, 1, 2, 0, 1, 2. Not quite. Actually, let's look for dates like 02/02/2002: 0, 2, 0, 2, 2, 0, 0, 2. That's 02022002. That's a repetition of 02 and 2002. So 02/02/2002 is one such date. Also 03/03/3003 would be in the future. What about 12/02/2012? That's 1, 2, 0, 2, 2, 0, 1, 2. Not quite. Let's think of 21/02/2012: 2, 1, 0, 2, 2, 0, 1, 2. Not quite. Actually, let's look for dates where the digits read the same forwards and backwards, which are palindromic dates. His sister, Meghana, has her birthday on 11/02/2011 where the digits read the same from left to right and from right to left. Find all possible dates of this form from the past. These are palindromic dates. For example, 01/02/2010 gives 01022010, which reads the same forwards and backwards? Let's see: 0, 1, 0, 2, 2, 0, 1, 0. Yes, it reads the same! So 01/02/2010 is a palindromic date. Also 02/02/2020 gives 02022020, which is also a palindrome. And 11/02/2011 gives 11022011, which is also a palindrome. And 12/02/2021 gives 12022021, which is also a palindrome. There are many such dates.
Now, let's think about calendars. Jeevan was looking at this year's calendar. He started wondering, "Why should we change the calendar every year? Can we not reuse a calendar?" What do you think? You might have noticed that last year's calendar was different from this year's. Also, next year's calendar will also be different from the previous years. But, will any year's calendar repeat again after some years? Will all dates and days in a year match exactly with that of another year? The answer is yes! The calendar repeats itself after some years. Specifically, if there is only one leap year in the period, the calendar repeats after 6 years. If there are two leap years, it repeats after 5 years. This is because of the way the days of the week fall on different dates.
Now, let's explore "Mental Math". Observe the figure below. What can you say about the numbers and the lines drawn? Numbers in the middle column are added in different ways to get the numbers on the sides. The numbers in the middle can be used as many times as needed to get the desired sum. For example, 38,800 = 25,000 + 400 × 2 + 13,000. And 3400 = 1500 + 1500 + 400. Can we make 1,000 using the numbers in the middle? Why not? The numbers in the middle are 400, 1500, 13000, 25000, and so on. The smallest number is 400. To make 1000, we would need 400 + 400 + 400 = 1200, which is more than 1000, or 400 + 400 = 800, which is less than 1000. We cannot make exactly 1000 using multiples of 400. So it is not possible to make 1000. What about 14,000, 15,000, and 16,000? Yes, it is possible. 14000 = 1500 × 8 + 400 × 5 = 12000 + 2000 = 14000. 15000 = 13000 + 400 × 5 = 13000 + 2000 = 15000. 16000 = 1500 × 8 + 400 × 10 = 12000 + 4000 = 16000. What thousands cannot be made? It seems that only 1000 cannot be made. All other thousands can be made by combining 1500 and 400 in different ways.
Now, let's think about "Adding and Subtracting". Here, using the numbers in the boxes, we are allowed to use both addition and subtraction to get the required number. For example, 39,800 = 40,000 - 800 + 300 + 300. So we can use both addition and subtraction to reach our target.
Now, let's think about "Digits and Operations". An example of adding two 5-digit numbers to get another 5-digit number is 12,350 + 24,545 = 36,895. An example of subtracting two 5-digit numbers to get another 5-digit number is 48,952 - 24,547 = 24,405.
Now, let's think about "Always, Sometimes, Never". Below are some statements. Think, explore, and find out if each statement is 'Always true', 'Only sometimes true', or 'Never true'. Let's go through them:
a. 5-digit number + 5-digit number gives a 5-digit number. This is only sometimes true. For example, 20,000 + 80,000 = 100,000, which is a 6-digit number. But 10,000 + 20,000 = 30,000, which is a 5-digit number. So it depends on the numbers.
b. 4-digit number + 2-digit number gives a 4-digit number. This is only sometimes true. For example, 9999 + 99 = 10,098, which is a 5-digit number. But 1000 + 50 = 1050, which is a 4-digit number. So it depends.
c. 4-digit number + 2-digit number gives a 6-digit number. This is never true. The maximum sum of a 4-digit number (9999) and a 2-digit number (99) is 10,098, which is only a 5-digit number. So it can never be a 6-digit number.
d. 5-digit number - 5-digit number gives a 5-digit number. This is only sometimes true. For example, 12,000 - 10,000 = 2,000, which is a 4-digit number. But 50,000 - 10,000 = 40,000, which is a 5-digit number. So it depends.
e. 5-digit number - 2-digit number gives a 3-digit number. This is never true. The smallest 5-digit number is 10,000, and the largest 2-digit number is 99. So 10,000 - 99 = 9901, which is a 4-digit number. So it can never be a 3-digit number.
Now, let's explore "Playing with Number Patterns". Here are some numbers arranged in some patterns. Find out the sum of the numbers in each of the figures. Should we add them one by one, or can we use a quicker way? There are many ways to add numbers quickly. For example, if you have a series of consecutive numbers, you can use the formula for the sum of an arithmetic progression. Or if you have numbers in a grid, you can look for symmetries and patterns. The key is to look for patterns and use clever strategies rather than adding one by one.
Now, let's explore "An Unsolved Mystery—the Collatz Conjecture"! This is one of the most fascinating topics in mathematics. Look at the sequences below—the same rule is applied in all the sequences:
a. 12, 6, 3, 10, 5, 16, 8, 4, 2, 1
b. 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
c. 21, 64, 32, 16, 8, 4, 2, 1
d. 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
Do you see how these sequences were formed? The rule is: start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat. Notice that all four sequences above eventually reached the number 1. In 1937, the German mathematician Lothar Collatz conjectured that the sequence will always reach 1, regardless of the whole number you start with. Even today—despite many mathematicians working on it—it remains an unsolved problem as to whether Collatz's conjecture is true! Collatz's conjecture is one of the most famous unsolved problems in mathematics. It seems so simple to understand, but no one has been able to prove it yet! Make some more Collatz sequences like those above, starting with your favourite whole numbers. Do you always reach 1? Try it and see!
Now, let's explore "Simple Estimation". At times, we may not know or need an exact count of things, and an estimate is sufficient for the purpose at hand. For example, your school headmaster might know the exact number of students enrolled in your school, but you may only know an estimated count. How many students are in your school? About 150? 400? A thousand? Paromita's class section has 32 children. The other 2 sections of her class have 29 and 35 children. So, she estimated the number of children in her class to be about 100. Along with Class 6, her school also has Classes 7-10, and each class has 3 sections each. She assumed a similar number in each class and estimated the number of students in her school to be around 500. This is estimation in action!
Now, let's do some simple estimates. It is a fun exercise, and you may find it amusing to know the various numbers around us. Remember, we are not interested in the exact numbers for the following questions. Share your methods of estimation with the class.
1. Steps you would take to walk: a. From the place you are sitting to the classroom door. This might be about 10-20 steps. b. Across the school ground from start to end. This might be about 100-200 steps. c. From your classroom door to the school gate. This might be about 50-100 steps. d. From your school to your home. This depends on how far you live, but it could be hundreds or thousands of steps.
2. Number of times you blink your eyes or number of breaths you take: a. In a minute. You blink your eyes about 15-20 times per minute, and you breathe about 15-20 times per minute. b. In an hour. Multiply by 60: about 900-1200 times. c. In a day. Multiply by 24: about 21,600-28,800 times.
3. Name some objects around you that are: a. A few thousand in number. For example, the number of pages in a big book, or the number of words in a novel. b. More than ten thousand in number. For example, the number of houses in a big city, or the number of students in a large school district.
Now, let's estimate some things. Try to guess within 30 seconds. Check your guess with your friends.
1. Number of words in your maths textbook: a. More than 5000, b. Less than 5000. What do you think? Your maths textbook probably has more than 5000 words, considering all the explanations, examples, and exercises.
2. Number of students in your school who travel to school by bus: a. More than 200, b. Less than 200. This depends on your school. In a big school, it could be more than 200.
3. Roshan wants to buy milk and 3 types of fruit to make fruit custard for 5 people. He estimates the cost to be ₹100. Do you agree with him? Why or why not? This depends on the prices of milk and fruits. If fruits are cheap, it might be possible. But if fruits are expensive, it might cost more than ₹100. So it's possible but not guaranteed.
4. Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland). If you look at the map of India, Gandhinagar is in the west, and Kohima is in the northeast. The distance is about 2500 kilometers.
5. Sheetal is in Grade 6 and says she has spent around 13,000 hours in school till date. Do you agree with her? Why or why not? Let's think: there are about 6 school hours in a day, and about 200 working days in a year. So in one year, she spends about 6 × 200 = 1200 hours. If she has been in school from Nursery to Grade 6, that's about 8 years. So total hours would be about 1200 × 8 = 9600 hours. That's less than 13,000 hours. So 13,000 hours seems too high. She might be overestimating.
6. Earlier, people used to walk long distances as they had no other means of transport. Suppose you walk at your normal pace. Approximately, how long would it take you to go from: a. Your current location to one of your favourite places nearby. This might take a few minutes. b. Your current location to any neighbouring state's capital city. This might take several hours or even a full day. c. The southernmost point in India to the northernmost point in India. This would take an extremely long time, probably several months of walking!
Now, let's explore "Games and Winning Strategies". Numbers can also be used to play games and develop winning strategies. Here is a famous game called 21. Play it with a classmate. Then try it at home with your family!
Rules for Game #1: The first player says 1, 2, or 3. Then the two players take turns adding 1, 2, or 3 to the previous number said. The first player to reach 21 wins! Play this game several times with your classmate. Are you starting to see the winning strategy? Which player can always win if they play correctly? What is the pattern of numbers that the winning player should say? The key is to think backwards from 21. If you want to reach 21, you should try to reach 17 first (because 21 - 4 = 17). Then, whatever number your opponent adds (1, 2, or 3), you can add enough to reach 21. So the winning strategy is to always say the numbers that are multiples of 4: 4, 8, 12, 16, 20. So if you go first, you can say 4, then 8, then 12, then 16, then 20, and then win by saying 21. So the first player can always win if they play correctly!
There are many variations of this game. Here is another common variation:
Rules for Game #2: The first player says a number between 1 and 10. Then the two players take turns adding a number between 1 and 10 to the previous number said. The first player to reach 99 wins! Play this game several times with your classmate. See if you can figure out the corresponding winning strategy in this case! Which player can always win? What is the pattern of numbers that the winning player should say this time? The key is to think about the winning numbers. If you want to reach 99, you should try to reach 89 first (because 99 - 10 = 89). Then, whatever your opponent adds (1 to 10), you can add enough to reach 99. So the winning numbers are 11, 22, 33, 44, 55, 66, 77, 88. Actually, wait, let's think more carefully. If you want to win, you need to be the one who says 99. So you want to leave your opponent with a number such that whatever they add (1 to 10), you can still reach 99. So you want to leave them with 89 (because 89 + 10 = 99). Then, to leave them with 89, you want to leave them with 78 (because 78 + 11 = 89, but wait, the maximum they can add is 10, so 78 + 10 = 88, which is less than 89). Actually, let's think differently. The winning numbers are those where you can force a win. If you go first, you should say 11. Then whatever the other player adds, you can make the total 22, then 33, then 44, then 55, then 66, then 77, then 88, and then 99. So the first player can always win by saying the multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, and then 99. So the winning strategy is to go first and say 11, then keep adding 11 each time.
Make your own variations of this game—decide how much one can add at each turn, and what number is the winning number. Then play your game several times, and figure out the winning strategy and which player can always win!
Now, let's summarize everything we have learned in this chapter.
In this chapter on Number Play, we have explored many fascinating concepts:
We learned that numbers can tell us things, like how many taller neighbours we have when we stand in a line. We discovered that it is impossible for the ends of a line to say '2' because they only have one neighbour. We learned that if all children are of the same height, everyone would say '0'. We found that the sequence 0, 1, 2, 1, 0 is possible, and that the maximum number of children who can say '2' is 2.
We explored supercells, which are cells in a table that are greater than all their neighbours. We learned that the largest number in any table is always a supercell, and the smallest number can never be a supercell. We discovered that it is impossible to have a table with no supercells if we use distinct numbers.
We learned about patterns on the number line and how to place numbers in their correct positions.
We explored playing with digits, including digit sums. We learned that there are 9 one-digit numbers, 90 two-digit numbers, 900 three-digit numbers, 9000 four-digit numbers, and 90000 five-digit numbers. We discovered that the digit sum is the sum of all digits in a number, and we found numbers with digit sum 14, the smallest being 59 and the largest 5-digit being 95000. We also learned that the digit sums of 3-digit numbers with consecutive digits are always multiples of 3.
We became digit detectives and found that the digit '7' appears 20 times between 1 and 100, and 300 times between 1 and 1000.
We explored palindromic numbers, which read the same forwards and backwards. We learned how to find palindromes using specific digits, and we explored the reverse-and-add process to create palindromes. We solved the puzzle about the 5-digit palindrome where the tens digit is double the units digit, and the hundreds digit is double the tens digit, finding the answer to be 12421.
We discovered the amazing Kaprekar constant 6174 for 4-digit numbers and 495 for 3-digit numbers. No matter what 4-digit number you start with (with at least two different digits), you will always eventually reach 6174!
We explored clock and calendar numbers, finding palindromic times and dates. We learned that calendars repeat after 6 years if there is one leap year, or after 5 years if there are two leap years.
We practiced mental math, learning how to combine numbers using addition and subtraction to reach target numbers. We discovered that 1000 cannot be made using 400 and 1500, but other thousands can be made.
We explored always, sometimes, never questions about addition and subtraction of numbers with different digits.
We played with number patterns and learned to look for quick ways to add numbers rather than adding one by one.
We discovered the Collatz conjecture, one of the most famous unsolved problems in mathematics. We learned the rule: if the number is even, take half; if odd, multiply by 3 and add 1. We explored several sequences and saw that they all eventually reach 1, but this has not been proven for all numbers yet!
We learned about simple estimation and how to make reasonable guesses about quantities like the number of steps to walk, the number of blinks in a minute, and the distance between cities.
Finally, we explored games and winning strategies, learning how to always win the game of 21 by saying the multiples of 4, and the game of 99 by saying the multiples of 11.
This chapter has shown us that numbers are not just for counting—they can be used to discover amazing patterns, solve puzzles, play games, and even explore unsolved mysteries in mathematics! Numbers are all around us, and by playing with them, we can discover beautiful and fascinating things.
Thank you for listening to this lesson, students! I hope you enjoyed exploring the world of Number Play. Keep playing with numbers, keep asking questions, and keep discovering the amazing patterns that numbers have to offer. Goodbye, and see you in the next class!