NCERT Textbook- Playing with Numbers Class 8 Notes | EduRev

Mathematics (Maths) Class 8

Created by: Indu Gupta

Class 8 : NCERT Textbook- Playing with Numbers Class 8 Notes | EduRev

 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 3-digit 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!
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 3-digit 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!
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 2-digit 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.
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 3-digit 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!
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 2-digit 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.
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 2-digit
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 2-digit 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 2-digit 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).
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 3-digit 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!
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 2-digit 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.
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 2-digit
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 2-digit 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 2-digit 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).
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 3-digit 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 3-digit 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 3-digit 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 3-digit numbers.
(iii) Forming three-digit numbers with given three-digits.
Now it is Minakshi’s turn once more.
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 3-digit 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!
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 2-digit 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.
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 2-digit
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 2-digit 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 2-digit 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).
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 3-digit 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 3-digit 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 3-digit 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 3-digit numbers.
(iii) Forming three-digit numbers with given three-digits.
Now it is Minakshi’s turn once more.
PLAYING WITH NUMBERS  253
TRY THESE
Minakshi: Think of any 3-digit number.
Sundaram: Alright, I have done so.
Minakshi: Now use this number to form two more 3-digit 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 3-digit 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 3-digit 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?
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