Class 6 Exam  >  Class 6 Notes  >  Mathematics (Maths) Class 6  >  RD Sharma Solutions: Playing With Numbers (Exercise 2.5)

Playing With Numbers (Exercise 2.5) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download

Q.1. Test the divisibility of the following numbers by 2:

(i) 6520

(ii) 984325

(iii) 367314

Ans: Rule: A natural number is divisible by 2 if its unit digit is 0, 2, 4, 6, or 8.

(i) Here, the unit's digit = 0

Thus, the given number is divisible by 2.

(ii) Here, the unit's digit = 5

Thus, the given number is not divisible by 2.

(iii) Here, the unit's digit = 4

Thus, the given number is divisible by 2.


Q.2. Test the divisibility of the following numbers by 3:

(i) 70335

(ii) 607439

(iii) 9082746

Ans: Rule: A number is divisible by 3 if the sum of its digits is divisible by 3.

(i) Here, the sum of the digits in the given number = 7 + 0 + 3 + 3 + 5 = 18 which is divisible by 3.

Thus, 70,335 is divisible by 3.

(ii) Here, the sum of the digits in the given number = 6 + 0 + 7 + 4 + 3 + 9 = 29 which is not divisible by 3.

Thus, 6,07,439 is not divisible by 3.

(iii) Here, the sum of the digits in the given number = 9 + 0 + 8 + 2 + 7 + 4 + 6 = 36 which is divisible by 3.

Thus, 90,82,746 is divisible by 3.


Q.3. Test the divisibility of the following numbers by 6:

(i) 7020

(ii) 56423

(iii) 732510

Ans: Rule: A number is divisible by 6 if it is divisible by 2 as well as 3.

(i)  Here, the unit's digit = 0

Thus, the given number is divisible by 2.

Also, the sum of the digits = 7 + 0 + 2 + 0 = 9 which is divisible by 3. So, the given number is divisible by 3.

Hence, 7,020 is divisible by 6.

(ii)  Here, the unit's digit = 3

Thus, the given number is not divisible by 2.

Also, the sum of the digits = 5 + 6 + 4 + 2 + 3 = 20 which is not divisible by 3. So, the given number is not divisible by 3.

Since 3,56,423 is neither divisible by 2 nor by 3, it is not divisible by 6.

(iii)  Here, the unit's digit = 0

Thus, the given number is divisible by 2.

Also, the sum of the digits = 7 + 3 + 2 + 5 + 1 + 0 = 18 which is divisible by 3. So, the given number is divisible by 3.

Hence, 7,32,510 is divisible by 6.


Q.4. Test the divisibility of the following numbers by 4:

(i) 786532

(ii) 1020531

(iii) 9801523

Ans: Rule: A natural number is divisible by 4 if the number formed by its last two digits is divisible by 4.

(i) Here, the number formed by the last two digits is 32 which is divisible by 4.

Thus, 7,86,532 is divisible by 4.

(ii) Here, the number formed by the last two digits is 31 which is not divisible by 4.

Thus, 10,20,531 is not divisible by 4.

(iii) Here, the number formed by the last two digits is 23 which is not divisible by 4.

Thus, 98,01,523 is not divisible by 4.


Q.5. Test the divisibility of the following numbers 8:

(i) 8364

(ii) 7314

(iii) 36712

Ans: Rule: A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

(i) The given number = 8364

The number formed by its last three digit is 364 which is not divisible by 8.

Therefore, 8,364 is not divisible by 8.

(ii) The given number = 7314

The number formed by its last three digit is 314 which is not divisible by 8.

Therefore, 7,314 is not divisible by 8.

(iii) The given number = 36712

Since the number formed by its last three digit = 712 which is divisible by 8.

Therefore, 36,712 is divisible by 8.


Q.6. Test the divisibility of the following numbers by 9:

(i) 187245

(ii) 3478

(iii) 547218

Ans: Rule: A number is divisible by 9 if the sum of its digits is divisible by 9.

(i) The given number = 187245

The sum of the digits in the given number = 1 + 8 + 7 + 2 + 4 + 5 = 27 which is divisible by 9.

Therefore, 1,87,245 is divisible by 9.

(ii) The given number = 3478

The sum of the digits in the given number = 3 + 4 + 7 + 8 = 22 which is not divisible by 9.

Therefore, 3,478 is not divisible by 9.

(iii) The given number = 547218

The sum of the digits in the given number = 5 + 4 + 7 + 2 + 1 + 8 = 27 which is divisible by 9.

Therefore, 5,47,218 is divisible by 9.


Q.7. Test the divisibility of the following numbers by 11:

(i) 5335

(ii) 70169803

(iii) 10000001

Ans: 

(i) The given number is 5,335.

The sum of the digit at the odd places = 5 + 3 = 8

The sum of the digits at the even places = 3 + 5 = 8

Their difference = 8 − 8 = 0

∴ 5,335 is divisible by 11.

(ii) The given number is 7,01,69,803.

The sum of the digit at the odd places = 7 + 1 + 9 + 0 = 17

The sum of the digits at the even places = 0 + 6 + 8 + 3 = 17

Their difference = 17 − 17 = 0

∴ 7,01,69,803 is divisible by 11.

(iii) The given number is 1,00,00,001.

The sum of the digit at the odd places = 1 + 0 + 0 + 0 = 1

The sum of the digits at the even places = 0 + 0 + 0 + 1 = 1

Their difference = 1 − 1 = 0

∴ 1,00,00,001 is divisible by 11.


Q.8. In each of the following numbers, replace * by the smallest number to make it divisible by 3:

(i) 75 * 5

(ii) 35 * 64

(iii) 18 * 71

Ans: We can replace the * by the smallest number to make the given numbers divisible by 3 as follows:

(i) 75*5 = 7515

As 7 + 5 + 1 + 5 = 18, it is divisible by 3.

(ii) 35*64 = 35064

As 3 + 5 +6 + 4 = 18, it is divisible by 3.

(iii) 18*71 = 18171

As 1 + 8 + 1 + 7 + 1 = 18, it is divisible by 3.


Q.9.In each of the following numbers, replace * by the smallest number to make it divisible by 9:

(i) 67 * 19

(ii) 66784 *

(iii) 538 * 8

Ans: 

(i) Sum of the given digits = 6 + 7 + 1 + 9 = 23

The multiple of 9 which is greater than 23 is 27.

Therefore, the smallest required number = 27 − 23 = 4

(ii) Sum of the given digits = 6 + 6 + 7 + 8 + 4 = 31

The multiple of 9 which is greater than 31 is 36.

Therefore, the smallest required number = 36 − 31 = 5

(iii) Sum of the given digits = 5 + 3 + 8 + 8 = 24

The multiple of 9 which is greater than 24 is 27.

Therefore, the smallest required number = 27 − 24 = 3


Q.10. In each of the following numbers. replace * by the smallest number to make it divisible by 11:

(i) 86 * 72

(ii) 467 * 91

(iii) 9 * 8071

Ans: Rule: A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11.

(i) 86 × 72

Sum of the digits at the odd places = 8 + missing number + 2 = missing number + 10

Sum of the digits at the even places = 6 + 7 = 13

Difference = [missing number + 10 ] − 13 = Missing number − 3

According to the rule, missing number − 3 = 0        [∵ the missing number is a single digit]

Thus, missing number = 3

Hence, the smallest required number is 3.

(ii) 467 × 91

Sum of the digits at the odd places = 4 + 7 + 9 = 20

Sum of the digits at the even places = 6 + missing number + 1 = missing number + 7

Difference = 20 − [missing number + 7] = 13 − missing number

According to rule, 13 − missing number = 11        [∵ the missing number is a single digit]

Thus, missing number = 2

Hence, the smallest required number is 2.

(iii) 9 × 8071

Sum of the digits at the odd places = 9 + 8 + 7 = 24

Sum of the digits at the even places =  missing number + 0 + 1 = missing number + 1

Difference = 24 − [missing number + 1] = 23 − missing number

According to rule, 23 − missing number = 22              [∵ 22 is a multiple of 11 and the missing number is a single digit]

Thus, missing number = 1

Hence, the smallest required number is 1.


Q.11. 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.

 Ans: 

(i) A number which is divisible by 2 but not by 4 is 6.

(ii) A number which is divisible by 3 but not by 6 is 9.

(iii) A number which is divisible by 4 but not by 8 is 28.

(iv) A number which is divisible by 4 and 8 but not by 32 is 48.


Q.12. 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.

Ans: 

(i) False. 12 is divisible by 3 but not by 9.

(ii) True.

(iii) False. 20 is divisible by 4 but not by 8.

(iv) True.

(v) False. 12 is divisible by both 3 and 6 but it is not divisible by 18.

(vi) True.

(vii) False. 10 divides the sum of 18 and 2 (i.e., 20) but 10 divides neither 18 nor 2.

(viii) True.

(ix) False. 4 and 9 are co-primes and both are composite numbers.

(x) True.

The document Playing With Numbers (Exercise 2.5) RD Sharma Solutions | Mathematics (Maths) Class 6 is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6
94 videos|347 docs|54 tests

Top Courses for Class 6

94 videos|347 docs|54 tests
Download as PDF
Explore Courses for Class 6 exam

Top Courses for Class 6

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Viva Questions

,

Sample Paper

,

Exam

,

shortcuts and tricks

,

pdf

,

past year papers

,

mock tests for examination

,

MCQs

,

Semester Notes

,

video lectures

,

Playing With Numbers (Exercise 2.5) RD Sharma Solutions | Mathematics (Maths) Class 6

,

ppt

,

Free

,

Objective type Questions

,

Playing With Numbers (Exercise 2.5) RD Sharma Solutions | Mathematics (Maths) Class 6

,

Important questions

,

Playing With Numbers (Exercise 2.5) RD Sharma Solutions | Mathematics (Maths) Class 6

,

Previous Year Questions with Solutions

,

practice quizzes

,

Extra Questions

,

study material

,

Summary

;