Page 1 PLAYING WITH NUMBERS 249 16.1 Introduction Y ou have studied various types of numbers such as natural numbers, whole numbers, integers and rational numbers. Y ou have also studied a number of interesting properties about them. In Class VI, we explored finding factors and multiples and the relationships among them. In this chapter, we will explore numbers in more detail. These ideas help in justifying tests of divisibility . 16.2 Numbers in General Form Let us take the number 52 and write it as 52 = 50 + 2 = 10 × 5 + 2 Similarly, the number 37 can be written as 37 = 10 × 3 + 7 In general, any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b What about ba? ba = 10 × b + a = 10b + a Let us now take number 351. This is a three digit number. It can also be written as 351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1 Similarly 497 = 100 × 4 + 10 × 9 + 1 × 7 In general, a 3digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c In the same way, cab = 100c + 10a + b bca = 100b + 10c + a and so on. Playing with Numbers CHAPTER 16 Here ab does not mean a × b! 201920 Page 2 PLAYING WITH NUMBERS 249 16.1 Introduction Y ou have studied various types of numbers such as natural numbers, whole numbers, integers and rational numbers. Y ou have also studied a number of interesting properties about them. In Class VI, we explored finding factors and multiples and the relationships among them. In this chapter, we will explore numbers in more detail. These ideas help in justifying tests of divisibility . 16.2 Numbers in General Form Let us take the number 52 and write it as 52 = 50 + 2 = 10 × 5 + 2 Similarly, the number 37 can be written as 37 = 10 × 3 + 7 In general, any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b What about ba? ba = 10 × b + a = 10b + a Let us now take number 351. This is a three digit number. It can also be written as 351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1 Similarly 497 = 100 × 4 + 10 × 9 + 1 × 7 In general, a 3digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c In the same way, cab = 100c + 10a + b bca = 100b + 10c + a and so on. Playing with Numbers CHAPTER 16 Here ab does not mean a × b! 201920 250 MATHEMATICS TRY THESE 1. Write the following numbers in generalised form. (i) 25 (ii) 73 (iii) 129 (iv) 302 2. Write the following in the usual form. (i) 10 × 5 + 6 (ii) 100 × 7 + 10 × 1 + 8 (iii) 100 × a + 10 × c + b 16.3 Games with Numbers (i) Reversing the digits – two digit number Minakshi asks Sundaram to think of a 2digit number, and then to do whatever she asks him to do, to that number. Their conversation is shown in the following figure. Study the figure carefully before reading on. It so happens that Sundaram chose the number 49. So, he got the reversed number 94; then he added these two numbers and got 49 + 94 = 143. Finally he divided this number by 11 and got 143 ÷ 11 = 13, with no remainder. This is just what Minakshi had predicted. 201920 Page 3 PLAYING WITH NUMBERS 249 16.1 Introduction Y ou have studied various types of numbers such as natural numbers, whole numbers, integers and rational numbers. Y ou have also studied a number of interesting properties about them. In Class VI, we explored finding factors and multiples and the relationships among them. In this chapter, we will explore numbers in more detail. These ideas help in justifying tests of divisibility . 16.2 Numbers in General Form Let us take the number 52 and write it as 52 = 50 + 2 = 10 × 5 + 2 Similarly, the number 37 can be written as 37 = 10 × 3 + 7 In general, any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b What about ba? ba = 10 × b + a = 10b + a Let us now take number 351. This is a three digit number. It can also be written as 351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1 Similarly 497 = 100 × 4 + 10 × 9 + 1 × 7 In general, a 3digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c In the same way, cab = 100c + 10a + b bca = 100b + 10c + a and so on. Playing with Numbers CHAPTER 16 Here ab does not mean a × b! 201920 250 MATHEMATICS TRY THESE 1. Write the following numbers in generalised form. (i) 25 (ii) 73 (iii) 129 (iv) 302 2. Write the following in the usual form. (i) 10 × 5 + 6 (ii) 100 × 7 + 10 × 1 + 8 (iii) 100 × a + 10 × c + b 16.3 Games with Numbers (i) Reversing the digits – two digit number Minakshi asks Sundaram to think of a 2digit number, and then to do whatever she asks him to do, to that number. Their conversation is shown in the following figure. Study the figure carefully before reading on. It so happens that Sundaram chose the number 49. So, he got the reversed number 94; then he added these two numbers and got 49 + 94 = 143. Finally he divided this number by 11 and got 143 ÷ 11 = 13, with no remainder. This is just what Minakshi had predicted. 201920 PLAYING WITH NUMBERS 251 TRY THESE TRY THESE Check what the result would have been if Sundaram had chosen the numbers shown below . 1. 27 2. 39 3. 64 4. 17 Now, let us see if we can explain Minakshi’s “trick”. Suppose Sundaram chooses the number ab, which is a short form for the 2digit number 10a + b. On reversing the digits, he gets the number ba = 10b + a. When he adds the two numbers he gets: (10a + b) + (10b + a) = 11a + 11b = 11 (a + b). So, the sum is always a multiple of 11, just as Minakshi had claimed. Observe here that if we divide the sum by 11, the quotient is a + b, which is exactly the sum of the digits of chosen number ab. Y ou may check the same by taking any other two digit number. The game between Minakshi and Sundaram continues! Minakshi: Think of another 2digit number, but don’t tell me what it is. Sundaram: Alright. Minakshi: Now reverse the digits of the number, and subtract the smaller number from the larger one. Sundaram: I have done the subtraction. What next? Minakshi: Now divide your answer by 9. I claim that there will be no remainder! Sundaram: Y es, you are right. There is indeed no remainder! But this time I think I know how you are so sure of this! In fact, Sundaram had thought of 29. So his calculations were: first he got the number 92; then he got 92 – 29 = 63; and finally he did (63 ÷ 9) and got 7 as quotient, with no remainder. Check what the result would have been if Sundaram had chosen the numbers shown below . 1. 17 2. 21 3. 96 4. 37 Let us see how Sundaram explains Minakshi’ s second “trick”. (Now he feels confident of doing so!) Suppose he chooses the 2digit number ab = 10a + b. After reversing the digits, he gets the number ba = 10b + a. Now Minakshi tells him to do a subtraction, the smaller number from the larger one. • If the tens digit is larger than the ones digit (that is, a > b), he does: (10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b). 201920 Page 4 PLAYING WITH NUMBERS 249 16.1 Introduction Y ou have studied various types of numbers such as natural numbers, whole numbers, integers and rational numbers. Y ou have also studied a number of interesting properties about them. In Class VI, we explored finding factors and multiples and the relationships among them. In this chapter, we will explore numbers in more detail. These ideas help in justifying tests of divisibility . 16.2 Numbers in General Form Let us take the number 52 and write it as 52 = 50 + 2 = 10 × 5 + 2 Similarly, the number 37 can be written as 37 = 10 × 3 + 7 In general, any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b What about ba? ba = 10 × b + a = 10b + a Let us now take number 351. This is a three digit number. It can also be written as 351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1 Similarly 497 = 100 × 4 + 10 × 9 + 1 × 7 In general, a 3digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c In the same way, cab = 100c + 10a + b bca = 100b + 10c + a and so on. Playing with Numbers CHAPTER 16 Here ab does not mean a × b! 201920 250 MATHEMATICS TRY THESE 1. Write the following numbers in generalised form. (i) 25 (ii) 73 (iii) 129 (iv) 302 2. Write the following in the usual form. (i) 10 × 5 + 6 (ii) 100 × 7 + 10 × 1 + 8 (iii) 100 × a + 10 × c + b 16.3 Games with Numbers (i) Reversing the digits – two digit number Minakshi asks Sundaram to think of a 2digit number, and then to do whatever she asks him to do, to that number. Their conversation is shown in the following figure. Study the figure carefully before reading on. It so happens that Sundaram chose the number 49. So, he got the reversed number 94; then he added these two numbers and got 49 + 94 = 143. Finally he divided this number by 11 and got 143 ÷ 11 = 13, with no remainder. This is just what Minakshi had predicted. 201920 PLAYING WITH NUMBERS 251 TRY THESE TRY THESE Check what the result would have been if Sundaram had chosen the numbers shown below . 1. 27 2. 39 3. 64 4. 17 Now, let us see if we can explain Minakshi’s “trick”. Suppose Sundaram chooses the number ab, which is a short form for the 2digit number 10a + b. On reversing the digits, he gets the number ba = 10b + a. When he adds the two numbers he gets: (10a + b) + (10b + a) = 11a + 11b = 11 (a + b). So, the sum is always a multiple of 11, just as Minakshi had claimed. Observe here that if we divide the sum by 11, the quotient is a + b, which is exactly the sum of the digits of chosen number ab. Y ou may check the same by taking any other two digit number. The game between Minakshi and Sundaram continues! Minakshi: Think of another 2digit number, but don’t tell me what it is. Sundaram: Alright. Minakshi: Now reverse the digits of the number, and subtract the smaller number from the larger one. Sundaram: I have done the subtraction. What next? Minakshi: Now divide your answer by 9. I claim that there will be no remainder! Sundaram: Y es, you are right. There is indeed no remainder! But this time I think I know how you are so sure of this! In fact, Sundaram had thought of 29. So his calculations were: first he got the number 92; then he got 92 – 29 = 63; and finally he did (63 ÷ 9) and got 7 as quotient, with no remainder. Check what the result would have been if Sundaram had chosen the numbers shown below . 1. 17 2. 21 3. 96 4. 37 Let us see how Sundaram explains Minakshi’ s second “trick”. (Now he feels confident of doing so!) Suppose he chooses the 2digit number ab = 10a + b. After reversing the digits, he gets the number ba = 10b + a. Now Minakshi tells him to do a subtraction, the smaller number from the larger one. • If the tens digit is larger than the ones digit (that is, a > b), he does: (10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b). 201920 252 MATHEMATICS TRY THESE • If the ones digit is larger than the tens digit (that is, b > a), he does: (10b + a) – (10a + b) = 9(b – a). • And, of course, if a = b, he gets 0. In each case, the resulting number is divisible by 9. So, the remainder is 0. Observe here that if we divide the resulting number (obtained by subtraction), the quotient is a – b or b – a according as a > b or a < b. Y ou may check the same by taking any other two digit numbers. (ii) Reversing the digits – three digit number. Now it is Sundaram’s turn to play some tricks! Sundaram: Think of a 3digit number, but don’t tell me what it is. Minakshi: Alright. Sundaram: Now make a new number by putting the digits in reverse order, and subtract the smaller number from the larger one. Minakshi: Alright, I have done the subtraction. What next? Sundaram: Divide your answer by 99. I am sure that there will be no remainder! In fact, Minakshi chose the 3digit number 349. So she got: • Reversed number: 943; • Difference: 943 – 349 = 594; • Division: 594 ÷ 99 = 6, with no remainder. Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end. 1. 132 2. 469 3. 737 4. 901 Let us see how this trick works. Let the 3digit number chosen by Minakshi be abc = 100a + 10b + c. After reversing the order of the digits, she gets the number cba = 100c + 10b + a. On subtraction: • If a > c, then the difference between the numbers is (100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c). • If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a). • And, of course, if a = c, the difference is 0. In each case, the resulting number is divisible by 99. So the remainder is 0. Observe that quotient is a – c or c – a. Y ou may check the same by taking other 3digit numbers. (iii) Forming threedigit numbers with given threedigits. Now it is Minakshi’s turn once more. 201920 Page 5 PLAYING WITH NUMBERS 249 16.1 Introduction Y ou have studied various types of numbers such as natural numbers, whole numbers, integers and rational numbers. Y ou have also studied a number of interesting properties about them. In Class VI, we explored finding factors and multiples and the relationships among them. In this chapter, we will explore numbers in more detail. These ideas help in justifying tests of divisibility . 16.2 Numbers in General Form Let us take the number 52 and write it as 52 = 50 + 2 = 10 × 5 + 2 Similarly, the number 37 can be written as 37 = 10 × 3 + 7 In general, any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b What about ba? ba = 10 × b + a = 10b + a Let us now take number 351. This is a three digit number. It can also be written as 351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1 Similarly 497 = 100 × 4 + 10 × 9 + 1 × 7 In general, a 3digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c In the same way, cab = 100c + 10a + b bca = 100b + 10c + a and so on. Playing with Numbers CHAPTER 16 Here ab does not mean a × b! 201920 250 MATHEMATICS TRY THESE 1. Write the following numbers in generalised form. (i) 25 (ii) 73 (iii) 129 (iv) 302 2. Write the following in the usual form. (i) 10 × 5 + 6 (ii) 100 × 7 + 10 × 1 + 8 (iii) 100 × a + 10 × c + b 16.3 Games with Numbers (i) Reversing the digits – two digit number Minakshi asks Sundaram to think of a 2digit number, and then to do whatever she asks him to do, to that number. Their conversation is shown in the following figure. Study the figure carefully before reading on. It so happens that Sundaram chose the number 49. So, he got the reversed number 94; then he added these two numbers and got 49 + 94 = 143. Finally he divided this number by 11 and got 143 ÷ 11 = 13, with no remainder. This is just what Minakshi had predicted. 201920 PLAYING WITH NUMBERS 251 TRY THESE TRY THESE Check what the result would have been if Sundaram had chosen the numbers shown below . 1. 27 2. 39 3. 64 4. 17 Now, let us see if we can explain Minakshi’s “trick”. Suppose Sundaram chooses the number ab, which is a short form for the 2digit number 10a + b. On reversing the digits, he gets the number ba = 10b + a. When he adds the two numbers he gets: (10a + b) + (10b + a) = 11a + 11b = 11 (a + b). So, the sum is always a multiple of 11, just as Minakshi had claimed. Observe here that if we divide the sum by 11, the quotient is a + b, which is exactly the sum of the digits of chosen number ab. Y ou may check the same by taking any other two digit number. The game between Minakshi and Sundaram continues! Minakshi: Think of another 2digit number, but don’t tell me what it is. Sundaram: Alright. Minakshi: Now reverse the digits of the number, and subtract the smaller number from the larger one. Sundaram: I have done the subtraction. What next? Minakshi: Now divide your answer by 9. I claim that there will be no remainder! Sundaram: Y es, you are right. There is indeed no remainder! But this time I think I know how you are so sure of this! In fact, Sundaram had thought of 29. So his calculations were: first he got the number 92; then he got 92 – 29 = 63; and finally he did (63 ÷ 9) and got 7 as quotient, with no remainder. Check what the result would have been if Sundaram had chosen the numbers shown below . 1. 17 2. 21 3. 96 4. 37 Let us see how Sundaram explains Minakshi’ s second “trick”. (Now he feels confident of doing so!) Suppose he chooses the 2digit number ab = 10a + b. After reversing the digits, he gets the number ba = 10b + a. Now Minakshi tells him to do a subtraction, the smaller number from the larger one. • If the tens digit is larger than the ones digit (that is, a > b), he does: (10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b). 201920 252 MATHEMATICS TRY THESE • If the ones digit is larger than the tens digit (that is, b > a), he does: (10b + a) – (10a + b) = 9(b – a). • And, of course, if a = b, he gets 0. In each case, the resulting number is divisible by 9. So, the remainder is 0. Observe here that if we divide the resulting number (obtained by subtraction), the quotient is a – b or b – a according as a > b or a < b. Y ou may check the same by taking any other two digit numbers. (ii) Reversing the digits – three digit number. Now it is Sundaram’s turn to play some tricks! Sundaram: Think of a 3digit number, but don’t tell me what it is. Minakshi: Alright. Sundaram: Now make a new number by putting the digits in reverse order, and subtract the smaller number from the larger one. Minakshi: Alright, I have done the subtraction. What next? Sundaram: Divide your answer by 99. I am sure that there will be no remainder! In fact, Minakshi chose the 3digit number 349. So she got: • Reversed number: 943; • Difference: 943 – 349 = 594; • Division: 594 ÷ 99 = 6, with no remainder. Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end. 1. 132 2. 469 3. 737 4. 901 Let us see how this trick works. Let the 3digit number chosen by Minakshi be abc = 100a + 10b + c. After reversing the order of the digits, she gets the number cba = 100c + 10b + a. On subtraction: • If a > c, then the difference between the numbers is (100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c). • If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a). • And, of course, if a = c, the difference is 0. In each case, the resulting number is divisible by 99. So the remainder is 0. Observe that quotient is a – c or c – a. Y ou may check the same by taking other 3digit numbers. (iii) Forming threedigit numbers with given threedigits. Now it is Minakshi’s turn once more. 201920 PLAYING WITH NUMBERS 253 TRY THESE Minakshi: Think of any 3digit number. Sundaram: Alright, I have done so. Minakshi: Now use this number to form two more 3digit numbers, like this: if the number you chose is abc, then • ‘the first number is cab (i.e., with the ones digit shifted to the “left end” of the number); • the other number is bca (i.e., with the hundreds digit shifted to the “right end” of the number). Now add them up. Divide the resulting number by 37. I claim that there will be no remainder. Sundaram: Y es. Y ou are right! In fact, Sundaram had thought of the 3digit number 237. After doing what Minakshi had asked, he got the numbers 723 and 372. So he did: 2 3 7 + 7 2 3 + 3 7 2 1 3 3 2 Then he divided the resulting number 1332 by 37: 1332 ÷ 37 = 36, with no remainder. Check what the result would have been if Sundaram had chosen the numbers shown below. 1. 417 2. 632 3. 117 4. 937 Will this trick always work? Let us see. abc = 100a + 10b + c cab = 100c + 10a + b bca = 100b + 10c + a abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37 16.4 Letters for Digits Here we have puzzles in which letters take the place of digits in an arithmetic ‘sum’, and the problem is to find out which letter represents which digit; so it is like cracking a code. Here we stick to problems of addition and multiplication. Form all possible 3digit numbers using all the digits 2, 3 and 7 and find their sum. Check whether the sum is divisible by 37! Is it true for the sum of all the numbers formed by the digits a, b and c of the number abc? 201920Read More
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