Playing with Numbers Class 8 RS Aggarwal Exe-4B Goyal Brothers ICSE Maths Solutions

Playing with Numbers Class 8 RS Aggarwal Exe-4B Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-4.  We provide step by step Solutions of cisce prescribe textbook / publications to develop skill and confidence. In this articles you would learn how to check divisibility test. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.

Playing with Numbers Class 8 RS Aggarwal Exe-4B Goyal Brothers ICSE Maths Solutions

Playing with Numbers Class 8 RS Aggarwal Exe-4B Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-4

Board ICSE
Publications Goyal Brothers Prakshan
Subject Maths
Class 8th
writer RS Aggarwal
Book Name Foundation
Ch-4 Playing with Numbers
Exe-4B divisibility test of number
Edition 2024-2025

divisibility test of number

Playing with Numbers Class 8 RS Aggarwal Exe-4B Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-4

Page- 62,63

Exercise- 4B

divisibility test of number

Que-1: Test the divisibility of each of the following number by 2 :

(i) 96  (ii) 830  (iii) 367  (iv) 922 (v) 881  (vi) 934            (vii) 1089   (viii) 13058

Solution- All numbers whose last digit is even is divisible by 2.
(i) 96 ; last digit 6 ; even, divisible
(ii) 830 it is even hence divisible by 2
(iii) 367 its last digit is 7 ;not even; hence not divisible by 2
(iv) 922 ; Last digit 2; even ; divisible
(v) 881 ; last digit 1; odd, not divisible
(vi) 934 ; last digit 4, even, divisible
(vii) 1089 ; last digit 9; odd, not divisible
(viii) 13058 ; last digit 8; even, divisible

Que-2: Test the divisibility of each of the following number by 3 :

(i) 258  (ii) 774    (iii) 584    (iv) 1033 (v) 2571  (vi) 6654     (vii) 10385    (viii) 35742

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

(i) 258
The sum of the digits is 2+5+8 = 15 is divisible by 3. So, 258 is divisible by 3.

(ii) 774
The sum of the digits is 7+7+4 = 18 is divisible by 3. So, 774 is divisible by 3.

(iii) 584
The sum of the digits is 5+8+4 = 17 is not divisible by 3. So, 584 is not divisible by 3.

(iv) 1033
The sum of the digits is 1+0+3+3 = 7 is not divisible by 3. So, 1033 is not divisible by 3.

(v) 2571
The sum of the digits is 2+5+7+1 = 15 is divisible by 3. So, 2571 is divisible by 3.

(vi) 6654
The sum of the digits is 6+6+5+4 = 21 is divisible by 3. So, 6654 is divisible by 3.

(vii) 10385
The sum of the digits is 1+0+3+8+5 = 17 is not divisible by 3. So, 10385 is not divisible by 3.

(viii) 35742
The sum of the digits is 3+5+7+4+2 = 21 is divisible by 3. So, 35742 is divisible by 3.

Que-3: Test the divisibility of each of the following number by 4 :

(i) 562     (ii) 8176    (iii) 56318 (iv) 35344   (v) 909032       (vi) 77752

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

(i) 562
The last two digits of 562 are ’62’ which is not divisible by 4. Hence, 562 is not divisible by 4.

(ii) 8176
The last two digits of 8176 are ’76’ which is divisible by 4. Hence, 8176 is divisible by 4.

(iii) 56318
The last two digits of 56318 are ’18’ which is not divisible by 4. Hence, 56318 is not divisible by 4.

(iv) 35344
The last two digits of 35344 are ’44’ which is divisible by 4. Hence, 35344 is divisible by 4.

(v) 909032
The last two digits of 909032 are ’32’ which is divisible by 4. Hence, 909032 is divisible by 4.

(vi) 77752
The last two digits of 77752 are ’52’ which is divisible by 4. Hence, 77752 is divisible by 4.

Que-4: Test the divisibility of each of the following number by 5 :

(i) 1095   (ii) 2746   (iii) 8185 (iv) 4040   (v) 98762              (vi) 77770

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

(i) 1095
The number 1095 has ‘5’ at its unit’s place so, it is divisible by 5.

(ii) 2746
The number 2746 has ‘6’ at its unit’s place so, it is not divisible by 5.

(iii) 8185
The number 8185 has ‘5’ at its unit’s place so, it is divisible by 5.

(iv) 4040
The number 4040 has ‘0’ at its unit’s place so, it is divisible by 5.

(v) 98762
The number 98762 has ‘2’ at its unit’s place so, it is not divisible by 5.

(vi) 77770
The number 77770 has ‘0’ at its unit’s place so, it is divisible by 5.

Que-5: Test the divisibility of each of the following number by 8 :

(i) 7304   (ii) 8612 (iii) 44440   (iv) 265472

Solution- We know that if number formed by last three digits of a given number is divisible by 8, then entire given number is divisible by 8.

(i) 7304
The last three digits of 7304 are ‘304’ which is divisible by 8. Hence, 7304 is divisible by 8.

(ii) 8612
The last three digits of 8612 are ‘612’ which is not divisible by 8. Hence, 8612 is divisible not by 8.

(iii) 44440
The last three digits of 44440 are ‘440’ which is divisible by 8. Hence, 44440 is divisible by 8.

(iv) 265472
The last three digits of 265472 are ‘472’ which is divisible by 8. Hence, 265472 is divisible by 8.

Que-6: Test the divisibility of each of the following number by 10 :

(i) 4005    (ii) 30303 (iii) 64610    (iv) 85730

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

(i) 4005
The number 4005 has ‘5’ at its unit’s place so, it is not divisible by 10.

(ii) 30303
The number 30303 has ‘3’ at its unit’s place so, it is not divisible by 10.

(iii) 64610
The number 64610 has ‘0’ at its unit’s place so, it is divisible by 10.

(iv) 85730
The number 85730 has ‘0’ at its unit’s place so, it is divisible by 10.

Que-7: Test the divisibility of each of the following number by 11 :

(i) 1452  (ii) 9416 (iii) 818532   (iv) 84524

Solution- 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) 1452
For the given number,
sum of the digits at odd places = 1 + 5 = 6
sum of digits at even places = 4 + 2 = 6
Difference of the above sums = 6 − 6 = 0
Since the difference is 0 and a number divisible by 11 so, the 1452 is divisible by 11.

(ii) 9416
For the given number,
sum of the digits at odd places = 9 + 1 = 10
sum of digits at even places = 4 + 6 = 10
Difference of the above sums = 10 − 10 = 0
Since the difference is 0 and a number divisible by 11 so, the 9416 is divisible by 11.

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

(iv) 84524
For the given number,
sum of the digits at odd places = 8 + 5 + 4 = 17
sum of digits at even places = 4 + 2 = 6
Difference of the above sums = 17 − 6 = 11
Since the difference is 11 and a number divisible by 11 so, the 84524 is divisible by 11.

Que-8: Find all possible values of x for which the number 534 x 2 is divisible by 3. Also, find each such number.

Solution- The number 534×2 is divisible by 3 if the sum of its digits is divisible by 3.
So, the sum of its digits is 5+3+4+x+2 = 14+x.
If x = 1 then,
14 + x = 14 + 1 = 15 is divisible by 3
If x = 4 then,
14 + x = 14 + 4 = 18 is divisible by 3.
If x = 7 then,
14 + x = 14 + 7 = 21 is divisible by 3.
So, x = 1,4 or 7
and the numbers is : 53412, 53442, 53472

Que-9: Find the least number which should replace * in the number 89*2 to make it exactly divisible by : 

(i) 3   (ii) 9

Solution- (i)  A no. is divisible by 3 if the sum of its all digits is divisible by 3
89*2
Replacing *
8+9+1+2 = 20 →no
8+9+2+2 = 21→ yes
So , 2 should replace * in the number 89*2 to make it divisible by 3.

(ii)  The divisibility rule of 9 states that the sum of the digits must be divisible by nine.
8+9+2 = 19, which is not divisible by nine.
19 + 8 = 27, which is divisible by nine.
Therefore, 8 is the smallest number to be added.

Que-10: Find the value of z for which the number 417 z 8 is divisible by 9. Also, the number.

Solution-  Given number is 417Z8
⇒ the sum of digits of given number = 20 + z; to be a multiple of 9, z should be 7
⇒ So, the number is 41778
∴ the value of Z = 7.

Que-11: Give five examples of numbers, each one of which is divisible by 3 but not divisible by 9.

Solution- 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.

Que-12: Give five examples of numbers, each one of which is divisible by 4 but not divisible by 8.

Solution- 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 12, 20, 28, 36, 44.

— : End of Playing with Numbers Class 8 RS Aggarwal Exe-4B Goyal Brothers Prakashan ICSE Foundation Maths Solutions :–

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