RD SHARMA SOLUTION CHAPTER – 5 Playing with Numbers | CLASS 8TH MATHEMATICS-EDUGROWN

Exercise 5.1

Question 1.
Without performing actual addition and division, write the quotient when the sum of 69 and 96. is divided by
(i) 11
(ii) 15
Solution:
Two numbers are 69 and 96 whose digits are reversed Here a = 6,= 9
(i) Sum if 69 + 96 is divisible by 11, then quotient = a + 6 = 6 + 9 = 15
(ii) If it is divided by a + b i.e., 6 + 9 = 15, then quotient = 11

Question 2.
Without performing actual computations, find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution:
Two given numbers are 94 and 49. Whose digits are reversed.
(i) If 94 – 49 is divided by 9, then the quotient = a-b = 9-4 = 5
(ii) and when it is divided by a – b i.e. 9-4 = 5, then quotient will be = 9

Question 3.
If sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case.
Solution:
The given number is 985
The other two numbers by arranging its digits
in cyclic order, will be 859, 598 of the form
abc¯¯¯¯¯¯¯,bca¯¯¯¯¯¯¯,cba¯¯¯¯¯¯¯
Therefore,
If 985 + 859 + 598 is divided by 111, then quotient will bea + 6 + c = 9 + 8 + 5 = 22
If this sum is divided by 22, then the quotient = 111
and if it is divided by 37, then quotient = 3 (a + b + c) = 3 (22) = 66

Question 4.
Find the quotient when difference of 985 and 958 is divided by 9.
Solution:
The numbers of three digits are
985 and 958 in which tens and ones digits are reversed, then
abc¯¯¯¯¯¯¯−acb¯¯¯¯¯¯¯ = 9 (b – c)
985 – 958 = 9 (8 – 5) = 9 x 3
i. e., it is divisible by 9, then quotient = b-c =8-5=3

Exercise 5.2

Question 1.
Given that the number  35a64¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3, where a is a digit, what are the possible volues of a ?
Solution:
The number 35a64¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3
∵The sum of its digits will also be divisible by 3
∴ 3 + 5 + a + b + 4 is divisible by 3
⇒ 18 + a is divisible by 3
⇒ a is divisible by 3 (∵ 18 is divisible by 3)
∴ Values of a can be, 0, 3, 6, 9

Question 2.
If x is a digit such that the number 18×71¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3,’ find possible values of x.
Solution:
∵ The number 18×71¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
is divisible by 3
∴ The sum of its digits will also be divisible by 3
⇒ l + 8+ x + 7 + 1 is divisible by 3
⇒ 17 + x is divisible by 3
The sum greater than 17, can be 18, 21, 24, 27…………
∴ x can be 1, 4, 7 which are divisible by 3.

Question 3.
If is a digit of the number 66784x¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ such that it is divisible by 9, find the possible values of x.
Solution:
∵ The number 66784 x is divisible by 9
∴ The sum of its digits will also be divisible by 9
⇒ 6+6+7+8+4+x is divisible by 9
⇒ 31 + x is divisible by 9
Sum greater than 31, are 36, 45, 54………
which are divisible by 9
∴ Values of x can be 5 on 9
∴ x = 5

Question 4.
Given that the number 67y19¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 9, where y is a digit, what are the possible values of y ?
Solution:
∵ The number 67y19¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 9
∴The sum of its digits will also be divisible by 9
⇒ 6 + 7+ y+ 1+ 9 is divisible by 9
⇒ 23 + y is divisible by 9
∴ The numbers greater than 23 are 27, 36, 45,……..
Which are divisible by 9
∴y = A

Question 5.
If 3×2¯¯¯¯¯¯¯¯ is a multiple of 11, where .v is a digit, what is the value of * ?
Solution:
∵ The number 3×2¯¯¯¯¯¯¯¯ is multiple of 11
∴ It is divisible by 11
∴ Difference of the sum of its alternate digits is zero or multiple of 11
∴ Difference of (2 + 3) and * is zero or multiple of 11
⇒ If x – (2 + 3) = 0 ⇒ x-5 = 0
Then x = 5

Question 6.
If 98125×2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is a number with x as its tens digits such that it is divisible by 4. Find all the possible values of x.
Solution:
∵ The number 98125×2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 4
∴ The number formed by tens digit and units digit will also be divisible by 4
∴ x2¯¯¯¯¯ is divisible by 4
∴ Possible number can be 12, 32, 52, 72, 92
∴ Value of x will be 1,3, 5, 7, 9

Question 7.
If x denotes the digit at hundreds place of the number 67×19¯¯¯¯¯¯¯¯¯¯¯¯¯ such that the
number is divisible by 11. Find all possible values of x.
Solution:
∵ The number 67×19¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 11
∴ The difference of the sums its alternate digits will be 0 or divisible by 11
∴ Difference of (9 + x + 6) and (1 + 7) is zero or divisible by 11
⇒ 15+x-8 = 0, or multiple of 11,
7 + x = 0 ⇒ x = -7, which is not possible
∴ 7 + x = 11, 7 + x = 22 etc.
⇒ x=11-7 = 4, x = 22 – 7
⇒ x = 15 which is not a digit
∴ x = 4

Question 8.
Find the remainder when 981547 is divided by 5. Do this without doing actual division.
Solution:
A number is divisible by 5 if its units digit is 0 or 5
But in number 981547, units digit is 7
∴ Dividing the number by 5,
Then remainder will be 7 – 5 = 2

Question 9.
Find the remainder when 51439786 is divided by 3. Do this without performing actual division.
Solution:
In the number 51439786, sum of digits is 5 + 1+ 4 + 3 + 9 + 7 + 8 + 6 = 43 and the given number is divided by 3.
∴ The sum of digits must by divisible by 3
∴ Dividing 43 by 3, the remainder will be = 1
Hence remainder = 1

Question 10.
Find the remainder, without performing actual division when 798 is divided by 11.
Solution:
Let n = 798 = a multiple of 11 + [7 + 8 – 9] 798 = a multiple of 11 + 6
∴ Remainder = 6

Question 11.
Without performing actual division, find the remainder when 928174653 is divided by 11.
Solution:
Let n = 928174653
= A multiple of 11+(9 + 8 + 7 + 6 + 3)-(2 + 1+4 + 5)
= A multiple of 11 + 33 – 12
= A multiple of 11 + 21
= A multiple of 11 + 11 + 10
= A multiple of 11 + 10
∴ Remainder =10

Question 12.
Given an example of a number which is divisible by :
(i) 2 but not by 4.
(ii) 3 but not by 6.
(iii) 4 but not by 8.
(iv) both 4 and 8 but not 32.
Solution:
(i) 2 but not by 4
A number is divisible by 2 if units do given is even but it is divisible by 4 if the number formed by tens digit and ones digit is divisible by 4.
∴ The number can be 222, 342 etc.
(ii) 3 but not by 6
A number is divisible by 3 if the sum of its digits is divisible by 3
But a number is divisible by 6, if it is divided by 2 and 3 both
∴ The numbers can be 333, 201 etc.
(iii) 4 but not by 8
A number is divisible by 4 if the number formed by the tens digit and ones digit is divisible by 4 but a number is divisible by 8, if the number formed by hundreds digit, tens digit and ones digit is divisible by 8.
∴ The number can be 244, 1356 etc.
(iv) Both 4 and 8 but not by 32
A number in which the number formed by the hundreds, tens and one’s digit, is divisible by 8 is divisible by 8. It will also divisible by 4 also.
But a number when is divisible by, 4 and 8 both is not necessarily divisible by 32 e.g., 328, 5400 etc.

Question 13.
Which of the following statements are true ?
(i) If a number is divisible by 3, it must be divisible by 9.
(ii) If a number is divisible by 9, it must be divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
(iv) If a number is divisible by 8, it must be divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
(ix) If two numbers are co-priirie, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4.
Solution:
(i) False, it is not necessarily that it must divide by 9.
(ii) Trae.
(iii) False, it is not necessarily that it must divide by 8.
(iv) True.
(v) False, it must be divisible by 9 and 2 both.
(vi) True.
(vii) False, it is not necessarily.
(viii)True.
(ix) False. It is not necessarily.
(x) True.

Exercise 5.3

Solve each of the following cryptarithms.
Question 1.

Solution:

Values of A and B be from 0 to 9 In ten’s digit 3 + A = 9
∴ A = 6 or less.
∴ 7 + B = A = 6 or less
∴ 7 + 9 or 8 = 16 or 15
∴ But it is two digit number
B = 8
Then A = 5

Question 2.

Solution:
Values of A and B can be between 0 and 9
In tens digit, A + 3 = 9
∴ A = 9 – 3 = 6 or less than 6
In ones unit B + 7 = A = 6or less
∴ 7 + 9 or 8 = 16 or 15
But it is two digit number
∴ B = 8 and
∴ A = 5

Question 3.

Solution:

Value of A and B can be between 0 and 9 In units place.
1+B = 0 ⇒1+B = 10
∴ B = 10 – 1 = 9
and in tens place
1 + A + 1 = B ⇒ A + 2 = 9
⇒ A = 9 – 2 = 7

Question 4.

Solution:
Values of A and.B can be between 0 and 9
In units place, B+1 = 8 ⇒ B = 8-1=7
In tens place A + B= 1 or A + B = 11
⇒ A + 7 = 11 ⇒ A =11-7 = 4

Question 5.

Solution:

Values of A and B can be between 0 and 9
In tens place, 2 + A = 0 or 2 + A=10
A = 10-2 = 8
In units place, A + B = 9
⇒ 8 + B = 9 ⇒ B = 9- 8 = 1

Question 6.

Solution:

Values of A and B can be between 0 and 9
In hundreds place,

Question 7.
Show that cryptarithm 4 x AB¯¯¯¯¯¯¯¯=CAB¯¯¯¯¯¯¯¯¯¯¯ does not have any solution.
Solution:

It means that 4 x B is a numebr whose units digit is B
Clearly, there is no such digit
Hence the given cryptarithm has no solution.