RD Sharma solution class 8 chapter 5 Playing with numbers Exercise 5.2

Exercise 5.2

Page-5.20

Question 1:

Given that the number 35α64¯ is divisible by 3, where α is a digit, what are the possible values of α?

Answer 1:

It is given that 35a64¯ is a multiple of 3. (3+5+a+6+4) is a multiple of 3. (a+18) is a multiple of 3. (a+18)=0, 3, 6, 9, 12, 15, 18, 21...But a is a digit of number  35a64 ¯ . So, a can take value 0, 1, 2, 3, 4...9.a+18=18 a=0a+18=21 a=3a+18=24 a=6a+18=27 a=9 a=0, 3, 6, 9

Question 2:

If x is a digit such that the number 18x71¯ is divisible by 3, find possible values of x.

Answer 2:

It is given that 18x71¯ is a multiple of 3.(1+8+x+7+1) is a multiple of 3.(17+x) is a multiple of 3.17+x=0, 3, 6, 9, 12, 15, 18, 21...But x is a digit. So, x can take values 0,1,2,3,4...9.17+x=18 x=117+x=21 x=417+x=24 x=7x=1,4,7

Question 3:

If x is a digit of the number 66784x¯ such that it is divisible by 9, find possible values of x.

Answer 3:

It is given that  66784x¯ is a multiple of 9.Therefore, (6+6+7+8+4+x) is a multiple of 9.And,(31+x) is a multiple of 9. Possible values of (31+x) are 0, 9, 18, 27, 36, 45,...But x is a digit. So, x can only take value 0, 1, 2, 3, 4,...9.31+x=36 x=36-31x=5

Question 4:

Given that the number 67y19¯ is divisible by 9, where y is a digit, what are the possible values of y?

Answer 4:

It is given that 67y19¯ is a multiple of 9. (6+7+y+1+9) is a multiple of 9. (23+y) is a multiple of 9. 23+y=0, 9, 18, 27, 36...But x is a digit. So, x can take values 0,1,2,3,4...9.23+y=27y=4

Question 5:

If 3x2¯ is a multiple of 11, where x is a digit, what is the value of x?

Answer 5:

Sum of the digits at odd places = 3+2= 5Sum of the digit at even place = x Sum of the digit at even place-Sum of the digits at odd places=(x-5) (x-5) must be multiple by 11. Possible values of (x-5) are 0, 11, 22, 33...But x is a digit; therefore x must be  0, 1 ,2, 3...9. x-5=0 x=5

Question 6:

If 98215x2¯ is a number with x as its tens digit such that is is divisible by 4. Find all possible values of x.

Answer 6:

A natural number is divisible by 4 if the number formed by its digits in units and tens places is divisible by 4. 98215x2 will be divisible by 4 if x2 is divisible by 4.  x2 ¯=10x + 2x is a digit; therefore possible values of x are 0, 1, 2, 3...9.x2 ¯=2, 12, 22, 32, 42, 52, 62, 72, 82, 92The numbers that are divisible by 4 are 12, 32, 52, 72, 92.Therefore, the values of x are 1, 3, 5, 7, 9.

Question 7:

If x denotes the digit at hundreds place of the number 67x19¯ such that the number is divisible by 11. Find all possible values of x.

Answer 7:

A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.Sum of digits at odd places-Sum of digits at even places=(6 + x + 9)-(7 + 1)=(15 + x)-8=x + 7 x + 7 =11  x = 4

Question 8:

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

Answer 8:

If a natural number is divided by 5, it has the same remainder when its unit digit is divided by 5.Here, the unit digit of 981547 is 7. When 7 is divided by 5, remainder is 2.Therefore, remainder will be 2 when 981547 is divided by 5.

Question 9:

Find the remainder when 51439786 is divided by 3. Do this without performing actual division.

Answer 9:

Sum of the digits of the number 51439786 = 5 + 1 + 4 + 3 + 9 + 7 + 8 + 6 = 43The remainder of 51439786, when divided by 3, is the same as the remainder when the sum of the digits is divided 3.When 43 is divided by 3, remainder is 1.Therefore, when 51439786 is divided by 3, remainder will be 1.

Question 10:

Find the remainder, without performing actual division, when 798 is divided by 11.

Answer 10:

798 =A multiple of 11 + (Sum of its digits at odd places - Sum of its digits at even places)798 =A multiple of 11 + (7 + 8 - 9)798 =A multiple of 11 + (15 - 9)798 =A multiple of 11 + 6Therefore, the remainder is 6.

Question 11:

Without performing actual division, find the remainder when 928174653 is divided by 11.

Answer 11:

928174653= A multiple of 11 + (Sum of its digits at odd places -Sum of its digits at even places)928174653= A multiple of 11 + {(9 + 8 + 7 + 6 + 3) - (2 + 1 + 4 + 5)}928174653= A multiple of 11 + (33 - 12)928174653= A multiple of 11 + 21928174653= A multiple of 11 + (11 × 1 + 10)928174653= A multiple of 11 + 10Therefore, the remainder is 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 by 32.

Answer 12:

(i) 10
Every number with the structure (4n + 2) is an example of a number that is divisible by 2 but not by 4.

(ii) 15
Every number with the structure (6n + 3) is an example of a number that is divisible by 3 but not by 6.

(iii) 28
Every number with the structure (8n + 4) is an example of a number that is divisible by 4 but not by 8.

(iv) 8
Every number with the  structure (32n + 8), (32n + 16) or (32n + 24) is an example of a number that is divisible by 4 and 8 but not by 32.

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-prime, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4.

Answer 13:

(i) False
Every number with the structures (9n + 3) or (9n + 6) is divisible by 3 but not by 9. Example: 3, 6, 12 etc.
(ii) True
(iii) False
Every number with the structure (8n + 4) is divisible by 4 but not by 8. Example: 4, 12, 20 etc.
(iv) True
(v) False
Example: 24 is divisible by both 3 and 6 but it is not divisible by 18.
(vi) True
(vii) False
Example: 5 divides 10, which is a sum of 3 and 7. However, it neither divides 3 nor 7.
(viii) True
(ix) False
Example: 4 and 9 are co-prime numbers but both are composite numbers too.
(x) True

No comments:

Post a Comment

Contact Form

Name

Email *

Message *