Rs Aggarwal 2019 2020 for Class 8 Math Chapter 5 - Playing With Numbers

Answer:

Let the tens place digit be x.

The units place digit is 3.

∴ Number = (10x  + 3)              ... (1)
 
Given:
7( x + 3) = (10 x + 3)

7 x + 21 = 10 x + 3

∴ 10 x - 7x = 21 - 3

⇒ 3 x  = 18

or x = 6

Using x = 6 in equation (1):

The number is 63.

Answer:

Let the tens digit be x.

The digit in the units place is 2x.

Number = 10x + 2x

Given:
(x + 2x) + 18 = (10x + 2x)

∴ 3x + 18 = 12x

12x - 3x = 18

9x =18

x = $\frac{18}{2}$ = 2

The digit in the tens place is 2.

The digit in the units place is twice the digit in the tens place.
The digit in the units place is 4.

Therefore, the number is 24.

Answer:

Let the tens place digit be a and the units place digit be b.
Then, number is (10a + b).

According to the question:

4(a + b) + 3 = (10 a + b) 
4a + 4b + 3 = 10a + b
6a - 3b = 3
3(2a - b) = 3
2a - b =1            ... (1)

Given:
If 18 is added to the number, its digits are reversed.
The reverse of the number is (10b + a).

∴ (10a + b) + 18 = 10b + a
10a - a + b -10b = -18
9a - 9b = -18
9(a - b) = -18
a - b = -2            ... (2)

Subtracting equation (2) from equation (1):

 2a  - b   =  1
  a   - b   = -2
-     +        +     
  a            = 3


Using a = 3 in equation (1):

2(3) - b = 1
6 - b = 1
∴ b = 5

Number = 10a+b = 10 $\times$ 3 + 5 = 35

Answer:

Let the tens place digit be a and the units place digit be b.
Then, the number is (10a + b).

​Given:
a + b = 15         ... (1)

When the digits are interchanged the number will be (10 b + a).
Given:
10a + b + 9 = 10 b + a
∴ 10a - a + b - 10b = -9
9a - 9b = -9​
a - b = -1            ... (2)

Adding equations (1) and (2):

a + b = 15
a - b  = -1   
2a     = 14
 ∴ a = 7

Using a = 7 in equation (2):
7 - b = -1
∴ b = 8

Original number = 10a+b = 10 $\times$ 7 + 8 = 78

Answer:

Let the tens place digit be 'x' and the units place digit be 'y'.
∴ Number =  (10x + y)

Number obtained by interchanging the digits = (10y + x)

Given: (10x + y) - (10y + x) = 63

∴ 10x - x + y - 10 y = 63
9x - 9y = 63
9(x - y) = 63
x - y = 7

Therefore, the difference between the digits of the number is 7.

Answer:

Let the units place digit be x.
Then, the tens place digit will be 3x and the hundreds place digit will be 4x.

Given:
4x + 3x + x = 16
or 8x = 16
or x =2
Units place digit = 2
Tens place digit =  3 $\times$ 2 = 6
Hundreds place digit =  4 $\times$ 2 = 8

Therefore, the number is 862.

Answer:

A given number is divisible by 2 only when its unit digit is 0, 2, 4, 6 or 8.

(i) 94
The number 94 has '4' at its unit's place so, it is divisible by 2.

(ii) 570
The number 570 has '0' at its unit's place so, it is divisible by 2.

(iii) 285
The number 285 has '5' at its unit's place so, it is not divisible by 2.

(iv) 2398
The number 2398 has '8' at its unit's place so, it is divisible by 2.

(v) 79532
The number 79532 has '2' at its unit's place so, it is divisible by 2.

(vi) 13576
The number 13576 has '6' at its unit's place so, it is divisible by 2.

(vii) 46821
The number 46821 has '1' at its unit's place so, it is not divisible by 2.

(viii) 84663
The number 84663 has '3' at its unit's place so, it is not divisible by 2.

(ix) 66669
The number 66669 has '9' at its unit's place so, it is not divisible by 2.

Answer:

A given number is divisible by 5 only when its unit digit is 0 or 5.

(i) 95
The number 95 has '5' at its unit's place so, it is divisible by 5.

(ii) 470
The number 470 has '0' at its unit's place so, it is divisible by 5.

(iii) 1056
The number 1056 has '6' at its unit's place so, it is not divisible by 5.

(iv) 2735
The number 2735 has '5' at its unit's place so, it is divisible by 5.

(v) 55053
The number 55053 has '3' at its unit's place so, it is not divisible by 5.

(vi) 35790
The number 35790 has '0' at its unit's place so, it is divisible by 5.

(vii) 98765
The number 98765 has '5' at its unit's place so, it is divisible by 5.

(viii) 42658
The number 42658 has '8' at its unit's place so, it is not divisible by 5.

(ix) 77990
The number 77990 has '0' at its unit's place so, it is divisible by 5.

Answer:

A given number is divisible by 10 only when its unit digit is 0.

(i) 205
The number 205 has '5' at its unit's place so, it is not divisible by 10.

(ii) 90
The number 90 has '0' at its unit's place so, it is divisible by 10.

(iii) 1174
The number 1174 has '4' at its unit's place so, it is not divisible by 10.

(iv) 57930
The number 57930 has '0' at its unit's place so, it is divisible by 10.

(v) 60005
The number 60005 has '5' at its unit's place so, it is not divisible by 10.

Answer:

A given number is divisible by 3 only when the sum of its digits is divisible by 3.

(i) 83
The sum of the digits is 8 + 3 = 11 which is not divisible by 3. So, 83 is not divisible by 3.

(ii) 378
The sum of the digits is 3 + 7 + 8 = 18 which is divisible by 3. So, 378 is divisible by 3.

(iii) 474
The sum of the digits is 4 + 7 + 4 = 15 which is divisible by 3. So, 474 is divisible by 3.

(iv) 1693
The sum of the digits is 1 + 6 + 9 + 3 = 19 which is not divisible by 3. So, 1693 is not divisible by 3.

(v) 20345
The sum of the digits is 2 + 0 + 3 + 4 + 5 = 14 which is not divisible by 3. So, 20345 is not divisible by 3.

(vi) 67035
The sum of the digits is 6 + 7 + 0 + 3 + 5 = 21 which is divisible by 3. So, 67035 is divisible by 3.

(vii) 591282
The sum of the digits is 5 + 9 + 1 + 2 + 8 + 2 = 27 which is divisible by 3. So, 591282 is divisible by 3.

(viii) 903164
The sum of the digits is 9 + 0 + 3 + 1 + 6 + 4 = 23 which is not divisible by 3. So, 903164 is not divisible by 3.

(ix) 100002
The sum of the digits is 1 + 0 + 0 + 0 + 0 + 2 = 3 which is divisible by 3. So, 100002 is divisible by 3.

Answer:

A given number is divisible by 9 only when the sum of the digits is divisible by 9.

(i) 327
The sum of the digits is 3 + 2 + 7 = 12 which is not divisible by 9. So, 327 is not divisible by 9.

(ii) 7524
The sum of the digits is 7 + 5 + 2 + 4 = 18 which is divisible by 9. So, 7524 is divisible by 9.

(iii) 32022
The sum of the digits is 3 + 2 + 0 + 2 + 2 = 9 which is divisible by 9. So, 32022 is divisible by 9.

(iv) 64302
The sum of the digits is 6 + 4 + 3 + 0 + 2 = 15 which is not divisible by 9. So, 64302 is not divisible by 9.

(v) 89361
The sum of the digits is 8 + 9 + 3 + 6 + 1 = 27 which is divisible by 9. So, 89361 is divisible by 9.

(vi) 14799
The sum of the digits is 1 + 4 + 7 + 9 + 9 = 30 which is not divisible by 9. So, 14799 is not divisible by 9.

(vii) 66888
The sum of the digits is 6 + 6 + 8 + 8 + 8 = 36 which is divisible by 9. So, 66888 is divisible by 9.

(viii) 30006
The sum of the digits is 3 + 0 + 0 + 0 + 6 = 9 which is divisible by 9. So, 30006 is divisible by 9.

(ix) 33333
The sum of the digits is 3 + 3 + 3 + 3 + 3 = 15 which is not divisible by 9. So, 33333 is not divisible by 9.

Answer:

A given number is divisible by 4 only when the number formed by its last two digits is divisible by 4.

(i) 134
The last two digits of 134 are '34' which is not divisible by 4. Hence, 134 is not divisible by 4.

(ii) 618
The last two digits of 618 are '18' which is not divisible by 4. Hence, 618 is not divisible by 4.

(iii) 3928
The last two digits of 3928 are '28' which is divisible by 4. Hence, 3928 is divisible by 4.

(iv) 50176
The last two digits of 50176 are '76' which is divisible by 4. Hence, 50176 is divisible by 4.

(v) 39392
The last two digits of 39392 are '92' which is divisible by 4. Hence, 39392 is divisible by 4.

(vi) 56794
The last two digits of 56794 are '94' which is not divisible by 4. Hence, 56794 is not divisible by 4.

(vii) 86102
The last two digits of 86102 are '02' which is not divisible by 4. Hence, 86102 is not divisible by 4.

(viii) 66666
The last two digits of 66666 are '66' which is not divisible by 4. Hence, 66666 is not divisible by 4.

(ix) 99918
The last two digits of 99918 are '18' which is not divisible by 4. Hence, 99918 is not divisible by 4.

(x) 77736
The last two digits of 77736 are '36' which is divisible by 4. Hence, 77736 is divisible by 4.

 

Answer:

A number is divisible by 8 only when the number formed by its last three digits is divisible by 8.

(i) 6132
The last three digits of the given number are 132 which is not divisible by 8. So, 6132 is not divisible by 8.

(ii) 7304
The last three digits of the given number are 304 which is divisible by 8. So, 7304 is divisible by 8.

(iii) 59312
The last three digits of the given number are 312 which is divisible by 8. So, 59312 is divisible by 8.

(iv) 66664
The last three digits of the given number are 664 which is divisible by 8. So, 66664 is divisible by 8.

(v) 44444
The last three digits of the given number are 444 which is not divisible by 8. So, 44444 is not divisible by 8.

(vi) 154360
The last three digits of the given number are 360 which is divisible by 8. So, 154360 is divisible by 8.

(vii) 998818
The last three digits of the given number are 818 which is not divisible by 8. So, 998818 is not divisible by 8.

(viii) 265472
The last three digits of the given number are 472 which is divisible by 8. So, 265472 is divisible by 8.

(ix) 7350162
The last three digits of the given number are 162 which is not divisible by 8. So, 7350162 is not divisible by 8.

Answer:

A given number is divisible by 11, if the difference between the sum of its digits at odd places and the sum of its digits at even places is
either 0 or a number divisible by 11.

(i) 22222
For the given number,
sum of the digits at odd places = 2 + 2 + 2 = 6
sum of digits at even places = 2 + 2 = 4
Difference of the above sums = 6 − 4 = 2
Since the difference is not 0 and neither a number divisible by 11 so, the 22222 is not divisible by 11.

(ii) 444444
For the given number,
sum of the digits at odd places = 4 + 4 + 4 = 12
sum of digits at even places = 4 + 4 + 4 = 12
Difference of the above sums = 12 − 12 = 0
Since the difference is 0 so the given number 444444 is divisible by 11.

(iii) 379654
For the given number,
sum of the digits at odd places = 4 + 6 + 7 = 17
sum of digits at even places = 5 + 9 + 3 = 17
Difference of the above sums = 17 − 17 = 0
Since the difference is 0 so the given number 379654 is divisible by 11.

(iv) 1057982
For the given number,
sum of the digits at odd places = 2 + 9 + 5 + 1 = 17
sum of digits at even places = 8 + 7 + 0 = 15
Difference of the above sums = 17 − 15= 2
Since the difference is not 0 and neither a number divisible by 11 so, the 1057982 is not divisible by 11.

(v) 6543207
For the given number,
sum of the digits at odd places = 7 + 2 + 4 + 6 = 19
sum of digits at even places = 0 + 3 + 5 = 8
Difference of the above sums = 19 − 8 = 11
Since the difference is 11 which is surely divisible by 11 so the given number 6543207 is divisible by 11.

(vi) 818532
For the given number,
sum of the digits at odd places = 2 + 5 + 1 = 8
sum of digits at even places = 3 + 8 + 8 = 19
Difference of the above sums = 19 − 8 = 11
Since the difference is 11 which is surely divisible by 11 so the given number 818532 is divisible by 11.

(vii) 900163
For the given number,
sum of the digits at odd places = 3 + 1 + 0 = 4
sum of digits at even places = 6 + 0 + 9 = 15
Difference of the above sums = 15 − 4 = 11
Since the difference is 11 which is surely divisible by 11 so the given number 900163 is divisible by 11.

(viii) 7531622
For the given number,
sum of the digits at odd places = 2 + 6 + 3 + 7 = 18
sum of digits at even places = 2 + 1 + 5 = 8
Difference of the above sums = 18 − 8 = 10
Since the difference is 10 which is not divisible by 11 so the given number 7531622 is not divisible by 11.

Answer:

(i) 693
69 − (2 × 3) = 69 − 6 = 63, which is divisible by 7.
Hence, 693 is divisible by 7.

(ii) 7896
789 − (2 × 6) = 789 − 12 = 777, which is divisible by 7.
Hence, 7896 is divisible by 7.

(iii) 3467
346 − (2 × 7) = 346 − 14 = 332, which is not divisible by 7.
Hence, 3467 is not divisible by 7.

(iv) 12873
1287 − (2 × 3) = 1287 − 6 = 1281, which is divisible by 7.
Hence, 12873 is divisible by 7.

(v) 65436
6543 − (2 × 6) = 6543 − 12 = 6531, which is divisible by 7.
Hence, 65436 is divisible by 7.

(vi) 54636
5463 − (2 × 6) = 5463 − 12 = 5451, which is not divisible by 7.
Hence, 54636 is not divisible by 7.

(vii) 98175
9817 − (2 × 5) = 9817 − 10 = 9807, which is divisible by 7.
Hence, 98175 is divisible by 7.

(viii) 88777
8877 − (2 × 7) = 8877 − 14 = 8863, which is not divisible by 7.
Hence, 88777 is not divisible by 7.

Answer:

For a number to be divisible by 3, the sum of the digits must be divisible by 3.

$$\begin{gathered} \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{digits}\mathit{ }\mathit{=}\mathit{ }7\mathit{ }+\mathit{ }x\mathit{ }+ 3 \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{ }10 +\mathit{ }x \end{gathered}$$

$10 + x$ will be divisible by 3 in the following cases:

$$\begin{gathered} 10 + x = 12, or x =2 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{will} \mathrm{be} 723. \\ 10 + x = 15, or x = 5 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{will} \mathrm{be} 753. \\ 10 + x = 18, or x = 8 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{will} \mathrm{be} 783. \end{gathered}$$

So, the numbers can be 723, 753 or 783.

Answer:

If a number is divisible by 3, then the sum of the digits is also divisible by 3.

Sum of the digits = $5 + 3 + y + 1 = 9 + y$

The sum of the digits is divisible by 3 in the following cases:

$$\begin{gathered} 9 + y = 9, or y = 0 \\ \mathrm{Then} \mathrm{the} \mathrm{number} \mathrm{is} 5301. \\ 9 + y = 12, or y = 3 \\ \mathrm{Then} \mathrm{the} \mathrm{number} \mathrm{is} 5331. \\ 9 + y = 15, or y = 6 \\ \mathrm{Then} \mathrm{the} \mathrm{number} \mathrm{is} 5361. \\ 9 + y = 18, or y = 9 \\ \mathrm{Then} \mathrm{the} \mathrm{number} \mathrm{is} 5391. \end{gathered}$$

∴ y = 0, 3, 6 or 9
The possible numbers are 5301, 5331, 5361 and 5391.

Answer:

For a number to be divisible by 9, the sum of the digits must be divisible by 9.

Sum of the digits in the given number = $x + 8 + 0 + 6 = x + 14$

The sum of the digits is divisible by 9, only in the following case:

$$\begin{gathered} x =4 \\ or \\ x + 14 =18 \end{gathered}$$

Thus, the number x806 is divisible by 9 if $x$ is equal to 4.

The number is 4806.

Answer:

If a number is divisible by 9, then the sum of the digits is also divisible by 9.

Sum of the digits of the given number = $4+7+1+z+8=20+z$

$20 + z = 27, for z =7$

27 is divisible by 9.

Therefore, 471z8 is divisible by 9 if $z$ is equal to 7.

The number is 47178.

Answer:

For a number to be divisible by 3, the sum of the digits should be divisible by 3.
And for the number to be divisible by 9, the sum of the digits should be divisible by 9.
Let us take the number 21.
Sum of the digits is 2 + 1 = 3, which is divisible by 3 but not by 9. Hence, 21 is divisible by 3 not by 9.
Similarly, lets check the number 24. Here, 2 + 4 = 6. This is divisible by 3 not by 9.
The number 30 will be divisible by 3 not by 9 as 3 + 0 = 3.
The number 33 will give 3 + 3 = 6, which is divisible by 3 not by 9.
Also 39 has the digits 3 + 9 = 12, which is divisible by 3 not by 9.
Hence, the numbers 21, 24, 30, 33 and 39 are divisible by 3 not by 9.

Answer:

For a number to be divisible by 4, the number formed by its last two digits should be divisible by 4.
And for the number to be divisible by 8, the number formed by its last three digits should be divisible by 8.
So, the numbers divisible by 4 and not by 8 will be 28, 36, 44, 52, 60.

Answer:

 $$\begin{gathered} A=6 \\ \therefore A+7=6+7=13 \end{gathered}$$

1 is carried over.
$(1+5+8)=14$

1 is carried over.
∴ $B=4$
and $C=1$

∴ $A=6, B=4 and C=1$

Answer:

 $A=7, A+6=7+6=13$       (1 is carried over)

$(1+B+9)=17, or B =7$       (1 is carried over)

$A=7, B=7 and C=4$          (1 is carried over)

∴ $A=7, B=7 and C=4$

Answer:

$A+A+A=A$     (with 1 being carried over)
This is satisfied if $A$ is equal to $5$.

When $A=5$:

$A+A+A=15$           (1 is carried over)

Or $B=1$
∴ $A=5 and B=1$

Answer:

First look at the left column, which is:
 $6-A=3$

This implies that the maximum value of A can be 3. 

$A\le 3$                 ... (1)

The next column has the following:
$A-B=7$

To reconcile this with equation (1), borrowing is involved.

We know:
$12-5=7$

∴ $A=2 and B=5$

Answer:

$5-A=9$ 
This implies that 1 is borrowed.
We know:
$15-6=9$
∴ $A=6$

$B-5=8$ 
This implies that 1 is borrowed.
$13-5=8$
But 1 has also been lent
∴$B=4$

$C-2=2$ 
This implies that 1 has been lent.
∴ $C=5$

∴ $A=6, B=4 and C=5$

Answer:

$$\begin{gathered} (\mathrm{B}\times 3)=\mathrm{B} \\ \mathrm{Then}, \mathrm{B} \mathrm{can} \mathrm{either} \mathrm{be} 0 \mathrm{or} 5. \\ \mathrm{If} \mathrm{B} \mathrm{is} 5, \mathrm{then} 1 \mathrm{will} \mathrm{be} \mathrm{carried}. \\ \mathrm{Then}, \mathrm{A}\times 3+1 = \mathrm{A} \mathrm{will} \mathrm{not} \mathrm{be} \mathrm{possible} \mathrm{for} \mathrm{any} \mathrm{number}. \\ \therefore \mathrm{B} = 0 \\ \mathrm{A}\times 3=\mathrm{A} \mathrm{is} \mathrm{possible} \mathrm{for} \mathrm{either} 0 \mathrm{or} 5. \\ \mathrm{If} \mathrm{we} \mathrm{take} \mathrm{A}=0, \mathrm{then} \mathrm{all} \mathrm{number} \mathrm{will} \mathrm{become} 0. \mathrm{However}, \mathrm{this} \mathrm{is} \mathrm{not} \mathrm{possible}. \\ \therefore \mathrm{A}= 5 \\ \mathrm{Then}, 1 \mathrm{will} \mathrm{be} \mathrm{carried}. \\ \therefore \mathrm{C}=1 \\ \therefore \mathrm{A}= 5, \mathrm{B}= 0 \mathrm{and} \mathrm{C} = 1 \end{gathered}$$

Answer:

$A\times B=B\Rightarrow A=1$



In the question:
First digit = B+1
Thus, 1 will be carried from 1+B² and becomes (B+1) (B² -9) B.
∴ C = B² -1
Now, all B, B+1 and B² -9  are one digit number.

This condition is satisfied for B=3 or B=4.

For B< 3, B² -9 will be negative.
For B>3, B² -9 will become a two digit number.
For B=3 , C = 3² - 9 = 9-9 = 0
For B = 4, C = 4² -9 = 16-9 = 7
 

Required answer:

A=1, B=3, C = 0

or

A=1, B=4, C = 7

Answer:

$(A-4) = 3\Rightarrow A=7$
​Also, $6\times 6=36 \Rightarrow C=6$
 $36-36=0 \Rightarrow B= 6$

∴ $$\begin{gathered} A=7 \\ B=C=6 \end{gathered}$$

Answer:

$1 and 9$ are two numbers, whose product is a single digit number.
∴ $1\times 9=9$

Sum of the numbers is a two digit number. 
∴ $1+9=10$

Answer:

The three whole numbers are 1, 2 and 3.

$1+2+3=6=1\times 2\times 3$

Answer:

Taking the diagonal that starts with 6:
$6+5+x=15 \Rightarrow x=4$
 

Answer:

6+2+4 = 12
4+3+5 = 12
6+1+5 = 12

Answer:

Given:
 $a=8 and b=13$
The numbers in the Fibonnaci sequence are arranged in the following manner:
$1st, 2nd, (1st+2nd), (2nd+3th), (3th+4th), (4th+5th), (5th+6th), (6th+7th), (7th+8th), (8th+9th), (9th+10th)$

The numbers are $8, 13, 21, 34, 55, 89, 144, 233, 377 \mathrm{and} 610$.
Sum of the numbers = ​$8+13+21+34+55+89+144+233+377+610$
                               $=1584$
$11\times 7th number = 11\times 144=1584$

Answer:

The magic square is completed assuming that the sum of the row, columns and diagonals is 30. This is because the sum of all the number of the last column is 30.
 

Answer:

(b) 1

If a number is exactly divisible by 3, the sum of the digits must also be divisible by 3.
$5+x+6=11+x$ must be divisible by 3.
The smallest value of $x$ is 1.
$x=1$
$\Rightarrow x+11 = 12$ is divisible by 3.

Answer:

(a) 0

If a number is divisible by 3, then the sum of the digits is also divisible by 3.
$6+4+y+8=18+y$
This is divisible by 3 as y is equal to 0.

Answer:

(c) 3

If a number is exactly divisible by 9, the sum of the digits must also be divisible by 9.
$7+x+8=15+x$
18 is divisible by 9.
∴ $15+x=18 \Rightarrow x=3$

Answer:

(d) 4

A number is divisible by 9 if the sum of the digits is divisible by 9.

$3+7+y+4=14+y$
For this sum to be divisible by 9:
$14+y=18 \Rightarrow y=4$

Answer:

(a) 1
If a number is divisible by 3, the sum of the digits is also divisible by 3.
$4+x+y+7=11+(x+y)$
For the sum to be divisible by 3:
$11+(x+y)=12 \Rightarrow (x+y) = 1$

Answer:

(d) 3

When a number is divisible by 3, the sum of the digits must also be divisible by 3.
$x+7+y+5=(x+y)+12$
This sum is divisible by 3 if  x+y+12 is 12 or 15.
For x+y+12 = 12:
x+y=0
But x+y cannot be 0 because then x and y both will have to be 0.
Since x is the first digit, it cannot be 0.
∴ x+y+12 = 15
or x+y = 15-12=3

Answer:

(c) 9

A number is divisible by 9 if the sum of the digits is divisible by 9.

$x+4+y+5+z=9+(x+y+z)$
The lowest value of $(x+y+z) \mathrm{is} \mathrm{equal} \mathrm{to} 0$ for the number x4y5z to be divisible by 9.
In this case, all x, y and z will be 0.
But x is the first digit, so it cannot be 0.
∴ x+4+y+5+z = 18
or x+y+z+9 = 18
or x+y+z = 9

Answer:

(b) 1

For a number to be divisible by 9, the sum of the digits must also be divisible by 9.

$1+A+2+B+5=(A+B)+8$

The number will be divisible by 9 if $(A+B) =1$.

Answer:

(d) 9

If a number is divisible by 9, then the sum of the digits is divisible by 9.

$x+2+7+y=(x+y)+9$
For this to be divisible by 9, the least value of $(x+y) \mathrm{is} 0$.
But for x+y = 0, x and y both will be zero.
Since x is the first digit, it can never be 0.
∴ x + y + 9 = 18
or  x + y = 9

Answer:

If a number is divisible by 3, then the sum of the digits is also divisible by 3.

$3+2+0+x=5+x$ must be divisible by 3.
This is possible in the following cases:
$$\begin{gathered} i) x=1 \\ \therefore 5+x=6 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 3201. \end{gathered}$$

$$\begin{gathered} ii) x=4 \\ \therefore 5+x=9 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 3204. \end{gathered}$$

$$\begin{gathered} iii) x=7 \\ \therefore 5+x=12 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 3207. \end{gathered}$$

Answer:

For a number to be divisible by 9, the sum of the digits must also be divisible by 9.

$6+4+y+3=13+y$

For this to be divisible by 9:
$y=5$

The number will be 6453.

Answer:

Let the two numbers of the two-digit number be 'a' and 'b'.

$a+b=6$              ... (1)

The number can be written as $(10a+b)$.
After interchanging the digits, the number becomes $(10b+a)$.

$$\begin{gathered} (10a+b)+18=(10b+a) \\ 9a-9b=-18 \end{gathered}$$
$a-b=-2$            ... (2)

Adding equations (1) and (2):
$2a=4\Rightarrow a =2$

Using $a=2$ in equation (1):
$b=6-a=6-2=4$

Therefore, the original number is 24.

Answer:

A number is divisible by 9 if the sum of the digits is divisible by 9.
 

Answer:

$A-8=3$
​This implies that 1 is borrowed.
$$\begin{gathered} 11-8=3 \\ \Rightarrow A = 1 \end{gathered}$$

Then, $7-B=9$
1 is borrowed from 7.
∴ $$\begin{gathered} 16-B=9 \\ \Rightarrow B = 7 \end{gathered}$$

Further, $5-C=2$
But 1 has been borrowed from 5.
∴ 4 - C = 2
$\Rightarrow C=2$

∴ $A=1, B=7 and C=2$

Answer:

Here, $$\begin{gathered} A-6 = 6 \\ \Rightarrow A =2 (\mathrm{with} 1 \mathrm{being} \mathrm{borrowed}) \end{gathered}$$
$B=3$
Since $7\times 9=63$, $C=9$
∴ $A=2, B=3 and C=9$

Answer:

$A\times B=B\Rightarrow A=1$

$$\begin{gathered} \begin{array}{ccc}& 1& B\\ \times & B& 1\end{array} \\ \begin{array}{ccc}& 1& B\\ B& B^2& \times \end{array} \\ \begin{array}{ccc}B& (1+B^2)& B\end{array} \end{gathered}$$

Now, $B\ne A=1$ and $(1+B^2)$ is a single digit number.
∴ $B=2$
$C=(1+B^2) = (1+4) =5$
∴ $A=1, B=2 and C=5$

Answer:

(b) 0
If a number is exactly divisible by 3, the sum of its digits is also divisible by 3.

$7+x+8=15+x$

$15+x$ can be divisible by 3 even if x is equal to 0.

Answer:

(c) 7

When a number is divisible by 9, the sum of the digits is also divisible by 9.

$6+x+5=11+x$

To be divisible by 9:
$11+x=18\Rightarrow x=7$

Answer:

(c) 6

When a number is divisible by 9, the sum of its digits is also divisible by 9.
$x+4+8+y=12+(x+y)$

For $12+(x+y)$ to be divisible by 9:
$12+(x+y)=18 \Rightarrow (x+y)=6$

Answer:

(d) 2

For a number to be divisible by 9, the sum of its digits must be divisible by 9.

$4+8+6+*+7=25+*$

Now, $25+*=27 (\mathrm{if} *=2$ and 27 is divisible by 9)