Number Play Class 8 Extra Questions Maths Chapter 5 - #NCSOLVE 📚

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Class 8 Number Play Extra Questions

Class 8 Maths Chapter 5 Number Play Extra Questions

Class 8 Maths Chapter 5 Extra Questions – Number Play Extra Questions Class 8

Question 1.
Find the value of A and B:

Solution:
Step 1: Here, we have B × 7 = B; this is only possible when either B = 0 or B = 5.
Step 2: If A = 1 and B = 0, then 10 × 17 = 170 ≠ 2110,
and if A = 1 and B = 5, then 15 × 17 = 255 ≠ 2115.
Step 3: If A = 2 and B = 0, then 20 × 27 = 540 ≠ 2110,
and if A = 2 and B = 5, then 25 × 27 = 675 ≠ 2115.
Step 4: If A = 3 and B = 0, then 30 × 37 = 1110 ≠ 2110,
and if A = 3 and B = 5, then 35 × 37 = 1295 ≠ 2115.
Step 5: If A = 4 and B = 0, then 40 × 47 = 1880 ≠ 2110,
and if A = 4 and B = 5, then 45 × 47 = 2115.
Thus, A = 4 and B = 5.

Question 2.
If 31y5 is a multiple of 3, where y is a digit, what could be the value of y?
Solution:
The sum of the digits of a number 31y5 = 3 + 1 + y + 5 = 9 + y
The number is divisible by 3 if the sum of its digits, i.e., (9 + y) is divisible by 3.
This is possible only when y = 0, 3, 6, or 9.
Thus, y could have either values of 0, 3, 6, or 9.

Question 3.
Check the divisibility of the following numbers by 9 as well as 3.
(a) 9061
(b) 5712
Solution:
(a) Given number is 9061.
∴ Sum of digits = 9 + 0 + 6 + 1 = 16, which is not a multiple of 9 or 3.
Hence, 9061 is not divisible by 9 or 3.

(b) Given number is 5712.
∴ Sum of digits = 5 + 7 + 1 + 2 = 15, which is not divisible by 9 but is divisible by 3.
So, 5712 is divisible by 3, but not by 9.

Question 4.
Find the greatest values of a and b so that the odd number ‘8ab4b’ is divisible by both 3 and 5.
Solution:
The number 8ab4b is divisible by 5.
So, b is either 0 or 5.
Since 8ab4b is an odd number,
Therefore, b = 5.
Now, sum of the digits of the number 8ab4b = 8 + a + b + 4 + 5
= 8 + a + 5 + 4 + 5
= a + 22
Since the number 8ab4b is divisible by 3, we have:
a + 22 = 24 or a = 2
a + 22 = 21 or a = 5
a + 22 = 30 or a = 8
Therefore, the required number is 88545.

Question 5.
Write Yes for the number that is divisible by the number on top, and a No which is not divisible.

Solution:

Question 6.
Write the following numbers in generalised form.
(a) ab
(b) 85
(c) 132
(d) 1000
Solution:
(a) ab = 10 × a + 1 × b = 10a + b
(b) 85 = 10 × 8 + 1 × 5 = 10 × 8 + 5
(c) 132 = 100 × 1 + 10 × 3 + 1 × 2
(d) 1000 = 1000 × 1 + 0 × 100 + 0 × 10 + 0 × 1 = 1000 × 1

Question 7.
Write the following in the usual form.
(a) 3 × 100 + 0 × 10 + 6
(b) 5 × 1000 + 3 × 100 + 2 × 10 + 1
Solution:
(a) 3 × 100 + 0 × 10 + 6
= 300 + 0 + 6
= 306

(b) 5 × 1000 + 3 × 100 + 2 × 10 + 1
= 5000 + 300 + 20 + 1
= 5321

Question 8.
Which of the following numbers are divisible by 3?
(a) 106
(b) 726
(c) 915
(d) 1008
Solution:
(a) Sum of the digits of 106 = 1 + 0 + 6 = 7, which is not divisible by 3.
Hence, 106 is not divisible by 3.

(b) Sum of the digits of 726 = 7 + 2 + 6 = 15, which is divisible by 3.
Hence, 726 is divisible by 3.

(c) Sum of the digits of 915 = 9 + 1 + 5 = 15, which is divisible by 3.
Hence, 915 is divisible by 3.

(d) Sum of the digits of 1008 = 1 + 0 + 0 + 8 = 9, which is divisible by 3.
Hence, 1008 is divisible by 3.

Question 9.
Prove that the sum of the given numbers and the numbers obtained by reversing their digits is divisible by 11.
(a) 89
(b) ab
(c) 69
(d) 54
Solution:
(a) Given number = 89
Number obtained by reversing the order of digits = 98
Sum = 89 + 98 = 187 ÷ 11 = 17
Hence proved.

(b) Given number = ab = 10a + b
Number obtained by reversing the digits = 10b + a
Sum = (10a + b) + (10b + a)
= 10a + b + 10b + a
= 11a + 11b
= 11(a + b) ÷ 11
= a + b
Hence proved.

(c) Given number = 69
Number obtained by reversing the digits = 96
Sum = 69 + 96 = 165 ÷ 11 = 15
Hence proved.

(d) Given number = 54
Number obtained by reversing the digits = 45
Sum = 54 + 45 = 99 ÷ 11 = 9
Hence proved.

Question 10.
Prove that the difference of the given numbers and the numbers obtained by reversing their digits is divisible by 9.
(a) 59
(b) xy
(c) xyz
(d) 203
Solution:
(a) Given number = 59
Number obtained by reversing the digits = 95
Difference = 95 – 59 = 36 ÷ 9 = 4
Hence proved.

(b) Given number = xy = 10x + y
Number obtained by reversing the digits = 10y + x
Difference = (10x + y) – (10y + x)
= 10x + y – 10y – x
= 9x – 9y
= 9(x -y) ÷ 9
= x – y
Hence proved.

(c) Given number = xyz = 100x + 10y + z
Number obtained by reversing the digits = 100z + 10y + x
Difference = (100x + 10y + z) – (100z + 10y + x)
= 100x + 10y + z – 100z – 10y – x
= 99x – 99z
= 99(x – z)
= 99(x – z) ÷ 9
= 11(x – z)
Hence proved.

(d) Given number = 203
Number obtained by reversing the digits = 302
Difference = 302 – 203 = 99 ÷ 9 = 11
Hence proved.

Question 11.
If a, b, and c are three digits of a three-digit number, prove that abc + cab + bca is a multiple of 37.
Solution:
We have abc + cab + bca
abc = 100a +10b + c
cab = 100c + 10a + b
bca = 100b + 10c + a
Adding abc + cab + bca = 111a + 111b + 111c
= 111(a + b + c)
= 37 × 3(a + b + c), which is a multiple of 37.
Hence proved.

Question 12.
Observe the following patterns:
1 × 9 – 1 = 8
21 × 9 – 1 = 188
321 × 9 – 1 = 2888
4321 × 9 1 = 38888
Find the value of 87654321 × 9 – 1
Solution:
From the pattern, we observe that there are as many eights in the result as the first digit from the right, which is to be multiplied by 9 and reduced by 1.
87654321 × 9 – 1 = 788888888

Question 13.
Find the numbers if the sum of three consecutive numbers is 24.
Solution:
Let the numbers be x, x + 1, and x + 2.
⇒ x + (x + 1) + (x + 2) = 24
⇒ 3x + 3 = 24
⇒ 3x = 24 – 3 = 21
⇒ 3x = 21
⇒ x = 7
∴ x + 1 = 7 + 1 = 8
∴ x + 2 = 7 + 2 = 9
Thus, the numbers are 7, 8, and 9.

Question 14.
Find the remainder when 981547 is divided by 5. Do this without doing actual division.
Solution:

We know that when a natural number is divided by 5, the remainder is the same as the remainder obtained on dividing its unit digit, here, 7, by 5, and here this remainder is 2.

Question 15.
If y is a digit of the number 66784y such that it is divisible by 9, find possible values of y.
Solution:
Because the number 66784y is divisible by 9, therefore by rule of divisibility,
sum of its digits = 6 + 6 + 7 + 8 + 4 + y = y + 31 is also divisible by 9.
∴ y + 31 = 0, 9, 18, 27, 36, 45, …..(i)
But as we can observe that y is a digit in the given number 66784y, therefore y takes only values 0, 1, 2, 3, 4,…9.
Adding 31 to each of these values, y + 31 can take values 31, 32, 33, 34, 35, 36, 37, 38, 39, 40,….. (ii)
Both (i) and (ii) give values of y + 31.
But the only common value in (i) and (ii) is 36.
y + 31 = 36
⇒ y = 36 – 31 = 5
∴ The only possible value of y is 5.

Question 16.
Find the remainder when 51439786 is divided by 3. Do this without performing actual division.
Solution:
Given: The number 51439786 is divided by 3.

∴ By the divisibility rule, the remainder will be the same as the remainder obtained on dividing the sum of digits = 5 + 1 + 4 + 3 + 9 + 7 + 8 + 6 = 43 by 3, and this remainder is 1.

Question 17.
Two-digit numbers with equal digits, i.e., 11, 22, 33, …….., 88, 99 are divisible by 11.
Solution:
(a) Let us consider a three-digit number 286.
Sum of digits in even places – Sum of digits in odd places
= 8 – (2 + 6)
= 0
∴ 286 is divisible by 11.

(b) Again, let us consider the number 61809.
Sum of digits in odd places = 6 + 8 + 9 = 23.
Sum of digits in even places = 1 + 0 = 1
Larger sum – Smaller sum = 23 – 1 = 22, which is divisible by 11.
∴ 61809 is divisible by 11.

Question 18.
If x denotes the digit at the hundreds place of the number 67×19, such that the number is divisible by 11. Find all possible values of x.
Solution:
The given number is 67×19.
The digits at odd places are 6, x, 9, and the digits at even places are 7 and 1.
Sum of digits at odd places = 6 + x + 9 = 15 + x
Sum of digits at even places = 7 + 1 = 8
Clearly, 15 + x is greater than 8.
∴ Larger sum – Smaller sum = Sum of digits at odd places – Sum of digits at even places
= 15 + x – 8
= x + 7
Because the given number 67×19 is divisible by 11, therefore by Divisibility by 11.
sum of digits at odd places – Sum of digits at even places = x + 7 is divisible by 11
∴ x + 7 = 0, 11, 22,……..(i)
But as given that x is a digit at the hundred place,
Therefore, x can take only the values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
∴ x + 7 can take values 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,…..(ii)
Both (i) and (ii) give values of x + 7.
But the only common value in (i) and (ii) is 11.
∴ x + 7 = 11
⇒ x = 11 – 7 = 4
∴ The only possible value of x is 4.

Practice Questions

Question 1.
If 4xy4 is divisible by 36, list all possible pairs of values for x and y.
Solution:
(2, 8)

Question 2.
Find three consecutive numbers such that the first is divisible by 7, the second is divisible by 5, and the third is divisible by 8.
Solution:
14, 15, 16

Question 3.
Riya says, “My number is divisible by 18. If reversed, it is still divisible by 18.”
(a) Is her claim true for multiples of 18?
(b) Are other digits possible for this to hold?
Solution:
(a) Sometimes true
(b) Yes, but rare.

Question 4.
If 59a76b is divisible by 24, list all possible pairs of values for a and b.
Solution:
(1, 8)

Question 5.
Find three consecutive numbers such that the first is divisible by 3, the second by 4, and the third by 5.
Solution:
3, 4, 5

Question 6.
Write five multiples of 39 between 70,000 and 70,500.
Solution:
70039, 70078, 70117, 70156, 70195

Question 7.
The middle number in a sequence of 7 consecutive odd numbers is 71. Write the sequence.
Solution:
65, 67, 69, 71, 73, 75, 77

Question 8.
Neha claims, “There exist multiples of 14 that when doubled are still multiples of 14, but not all multiples of 14 when tripled are multiples of 14.” Check her claim.
Solution:
Incorrect

Question 9.
Decide if these statements are always true, sometimes true, or never true:
(a) The sum of two multiples of 6 is a multiple of 12.
(b) The product of three consecutive integers is divisible by 6.
(c) If abcdef is a multiple of 9, then fedcba is also a multiple of 9.
Solution:
(a) Sometimes True
(b) Always True
(c) Always True

Question 10.
Pick any 4 numbers. When their sum is divided by 13, what are the possible remainders?
Solution:
0 to 12

Question 11.
Choose any 5 numbers. When their sum is divided by 17, what is the range of possible remainders?
Solution:
0 to 16

Question 12.
If 7mn5 is divisible by 15, list all possible pairs (m, n).
Solution:
(0, 0), (5, 5)

The post Number Play Class 8 Extra Questions Maths Chapter 5 appeared first on Learn CBSE.

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