NCERT Solutions for Class 8 Math Chapter 16 - Playing with Numbers

Answer:

25 = 20 + 5 = (2 × 10) + 5 × 1

Answer:

73 = 70 + 3 = (7 × 10) + 3 × 1

Answer:

129 = 100 + 20 + 9
= (1 × 100) + (2 × 10) + 9 × 1

Answer:

302 = 300 + 0 + 2
= (3 × 100) + (10 × 0) + 2 × 1

Answer:

10 × 5 + 6 = 50 + 6 = 56

Answer:

100 × 7 + 10 × 1 + 8 = 700 + 10 + 8 = 718

Answer:

100 × a + 10 × c + b = 100a + 10c + b = acb

Answer:

27
Here, 27 + 72 = 99
$\therefore$ 99 $÷$ 11 = 9
Also, 2 + 7 = 9
i.e. the quotient is equal to the sum of the digits of the number he chose.

Answer:

39
Here, 39 + 93 = 132
$\therefore$ 132 $÷$ 11 = 12
Also, 3 + 9 = 12
i.e., the quotient is equal to the sum of the digits of the number he chose.

Answer:

64
Here, 64 + 46 = 110
$\therefore$ 110 $÷$ 11 = 10
Also, 6 + 4 = 10
i.e., the quotient is equal to the sum of the digits of the number he chose.

Answer:

17
Here, 17 + 71 = 88
$\therefore$ 88 $÷$ 11 = 8
Also, 1 + 7 = 8
i.e., the quotient is equal to the sum of the digits of the number he chose.

Answer:

17
On reversing the digits, we have : 71
$\therefore$ 71 - 17 = 54
Now, 54 $÷$ 9 = 6.
Also, 7 - 1 = 6.
i.e., the quotient is the difference between the
digits of the number he chose.

Answer:

21
On reversing the digits, we have; 12
$\therefore$ 21 - 12 = 9
Now, 9 $÷$ 9 = 1
Also, 2 - 1 = 1
i.e., the quotient is the difference between the digits of the number he chose.

Answer:

96
On reversing the digits, we have : 69
$\therefore$ 96 - 69 = 27
Now, 27 $÷$ 9 = 3
Also, 9 - 6 = 3
i.e., the quotient is the difference between the digits of the number he chose.

Answer:

37
On reversing the digits, we have : 73.
$\therefore$ 73 - 37 = 36
Now, 36 $÷$ 9 = 4
Also, 7 - 3 = 4
i.e., the quotient is the difference between the digits of the number he chose.

Answer:

132
On reversing the digits, we have : 231
$\therefore$ Difference = 231 - 132 = 99
Now, 99 $÷$ 99 = 1
Also, 2 - 1 = 1
i.e., the quotient is equal to the difference between the hundreds digit and the ones digit.

Answer:

469
On reversing the digits, we have; 964.
$\therefore$ Difference = 964 - 469 = 495
Now, 495 $÷$ 99 = 5
Also, 9 - 4 = 5
i.e., the quotient is equal to the difference between the hundreds digits and the ones digit.

Answer:

737
On reversing the digits, we have; 737
$\therefore$ Difference = 737 - 737 = 0
Now, 0 $÷$ 99 = 0
Also, 7 - 7 = 0
i.e., the quotient is equal to the difference between the hundreds digits and the ones digit.

Answer:

901
On reversing the digits, we have; 109.
$\therefore$ Difference = 901 - 109 = 792
Now, 792 $÷$ 99 = 8
Also, 9 - 1 = 8
i.e., the quotient is equal to the difference between the hundreds digits and the ones digit.

Answer:

417
Here; 417
Now take; 741
[i.e., the ones digit shifted to the “left end” of the number.]
Then; 174
[i.e., the hundreds digit shifted to the “right end” of the number.]
On adding all these three numbers, we have;

On dividing this number by 37, we have;
i.e., 1332 $÷$ 37 = 36 (No remainder).

Answer:

632
Here; 632
Now take; 263
[i.e., the ones digit shifted to the “left end” of the number.]
Then; 326
[i.e., the hundreds digit shifted to the “right end” of the number.]
On adding all these three numbers, we have;
On dividing this number by 37, we have;

i.e., 1221 $÷$ 37 = 33 (No remainder).

Answer:

117
Here; 117
Now take; 711
[i.e., the ones digit shifted to the “left end” of the number.]
Then; 171
[i.e., the hundreds digit shifted to the “right end” of the number.]
On adding all these three numbers, we have;

On dividing this number by 37, we have; i.e., 999 $÷$ 37 = 27 (No remainder).

Answer:

937
Here; 937
Now take; 793
[i.e., the ones digit shifted to the “left end” of the number.]
Then; 379
[i.e., the hundreds digit shifted to the “right end” of the number.]
On adding all these three numbers, we have;

On dividing this number by 2109 by 37, we have;
i.e., 2109 $÷$ 37 = 57 (No remainder)

Answer:

Let the two digit number be ab and the number obtained by reversing its digits is ba.
Let their sum a 3-digit number be dad.
$\therefore$ ab + ba = dad
$\Rightarrow$ (10a + b) + (10b + a) = dad
$\Rightarrow$ 11(a + b) = dad
The sum a + b cannot exceed 18, because the greatest 2-digit number can be 99 so that 99 + 99 = 198.
Yes, these number 198 is clearly a multiple of 11.
All the 3-digit numbers which are the multiples of 11 are 110, 121, 132, 143, 154, 165, 176, 187 and 198.
Clearly, dad = 121
$\Rightarrow$ a = 2, d = 1.

Answer:

Here, we have to find the values of letters A and B.
Since, the ones digits of (A + 5) is 2.
Which is possible only when A is 7 i.e., 7 + 5
= 12
Also, Ten’s digit and carry digit = 3 + 2 + 1 = B
($\therefore$ carry is 1)
$\Rightarrow$ 5 + 1 = B
$\Rightarrow$ B = 6
$\therefore$ A = 7, B = 6.

Answer:

Here, we have to find the values of letters A, B and C.
Since the ones digit of (A + 8) is 3.
Which is possible, only when, A is 5 i.e., (5 + 8)
= 13.
Also, Ten’s digit and hundreds digit
= 4 + 9 = CB
= 13 + 1 = CB [$\therefore$ carry is 1]
$\Rightarrow$ 14 = CB
$\Rightarrow$ C = 1, B = 4
$\therefore$ A = 5, B = 4, C = 1.

Answer:

Here, we have to find the value of letter A. Since, the ones digit of A × A = A.
$\therefore$ A may have the values either A = 1, 5, 6.
But the tens digit is 9, which is possible only for value A = 6.

Answer:

Here, B + 7 = A and A + 3 = 6
which means, we have
A = 2, B = 5

Answer:

Here, we have to find the values of letters A, B and C.
The unit and ten’s digit of product AB × 3 = CAB
Which is possible only when; 50 × 3 = 150
$\therefore$ A = 5, B = 0 and C = 1.

Answer:

Here, AB × 5 = CAB
It means, the units digit and hundreds digit
should be same as that of the two digit no.
which is to be multiplied by 5, which is
possible only if; A = 2, B = 5 and C = 1
Alternate Solution
AB × 5 = CAB
Which is possible if
A = 5, B = 0 and C = 2

So, AB = 50
and CAB = 250

Answer:

Here, we have two missing entries A and B.
As it is given that; AB × 6 = BBB,
which is possible only when, A = 7 and B = 4
i.e., 74 × 6 = 444

Answer:

Here, it is given that; 1 + B = 0,
which is only possible, when B = 9 i.e., 1 + 9 = 10
and A + 1 = B $\Rightarrow$ A + 1 = 9
which means A must be 7
Because, 7 + 1 + 1 (carry) = 9
i.e., A = 7 and B = 9

Answer:

Here, the missing entries are A and B.
As B + 1 = 8 $\therefore$ the possible value of B is 7
and A + B = 1, also B = 7
$\therefore$ A must be 4, which gives sum as 11.
$\therefore$ 2 + A = B $\Rightarrow$ 2 + 4 + (1 carry) = 7
i.e., A = 4 and B = 7

Answer:

Here, the missing entries are A and B.
It is given that; 2 + A = 0 $\Rightarrow$ 2 + 8 = 10
Also, 1 + 6 = A
But A = 8; 1 + 6 + (1 carry) = 8 = A
$\therefore$ And A + B = 9 $\Rightarrow$ 8 + B = 9 $\Rightarrow$ B = 1
i.e., A = 8, B = 1

Answer:

The ones digit, when divided by 5, must leave a remainder of 3. So, the ones digit must be either 3 or 8.

Answer:

The ones digit, when divided by 5, must leave a remainder of 1. So, the ones digit must be either 1 or 6.

Answer:

The ones digit, when divided by 5, must leave a remainder of 4. So, the ones digit must be either 4 or 9.

Answer:

N is odd, so its ones digit is odd.
Therefore, the ones digit must be 1, 3, 5, 7 or 9.

Answer:

N is an even number, so its units digit is also an even number. Therefore, the ones digit must be 2, 4, 6, 8 or 0.

Answer:

1. 108.
The sum of the digits of 108 is 1 + 0 + 8 = 9.
As this number is divisible by 9.
$\therefore$ The number 108 is divisible by 9.
2. 616.
The sum of the digits of 616 is 6 + 1 + 6 = 13.
As this number is not divisible by 9.
$\therefore$ The number 616 is not divisible by 9.
3. 294.
The sum of the digits of 294 is 2 + 9 + 4 = 15.
As this number is not divisible by 9.
$\therefore$ The number 294 is not divisible by 9.
4. 432.
The sum of the digits of 432 is 4 + 3 + 2 = 9.
As this number is divisible by 9.
$\therefore$ The number 432 is divisible by 9.
5. 927.
The sum of the digits of 927 is 9 + 2 + 7 = 18
As this number is divisible by 9.
$\therefore$ The number 927 is divisible by 9.

Answer:

Yes, we can say that if a number is divisible by any number m, then it will also be divisible by each of the factors of m.

Answer:

(i) It is necessary that (a + c - b) should be divisible by 11.
(ii) Yes, we can say that if the number abcd is divisible by 11, then [(b + d) - (a + c)] must be divisible by 11.
(iii) Yes, from (i) and (ii) above we can say that a number will be divisible by 11 if the difference between the sum of digits of its odd places and that of digits at the even places is divisible by 11.

Answer:

Answer:

Since 21y5 is a multiple of 9
$\therefore$ Its sum of digits 2 + 1 + y + 5 = 8 + y is a multiple of 9.
As y is a digit.
So, 8 + y = 9 $\Rightarrow$ y = 1
Therefore, y = 1

Answer:

Since, 31z 5 is a multiple of 9.
$\therefore$ Its sum of digits 3 + 1 + z + 5 = 9 + z is a multiple of 9.
As z is a digit.
So, 9 + z is either 0 or 9.
$\therefore$ 9 + 0 = 9 and 9 + 9 = 18 = 1 + 8 = 9
$\therefore$ z = 0 or z = 9.

Answer:

Since 24x is a multiple of 3.
$\therefore$ Its sum of digits = 2 + 4 + x = 6 + x is a multiple of 3.
So, 6 + x is one of these numbers; 0, 3, 6, 9, 12,
15, 18 ............,
But, since x is a digit, it can only be that ;
6 + x = 6 or 9 or 12 or 15.
Therefore, x = 0 or 3 or 6 or 9.
Thus, x can have any four different values.

Answer:

Since, 31z 5 is a multiple of 3.
$\therefore$ Its sum of digits = 3 + 1 + z + 5 = 9 + z.
Since, the sum (9 + z) is a multiple of 3 and z is a digit.
$\therefore$ z can have the values either 0 or 3 or 6 or 9.