### Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.1)

**Tags:**RD Sharma Class 6 Math Solution Chapter 2 Playing with Numbers Exercise 2.1, Solution of RD Sharma for class 6,RD Sharma solutions Grade VI, Solution of RD Sharma exercises for sixth standard, R.D. Sharma chapter Playing With Numbers Exercise 2.1 (Factors and Multiples)Playing With Numbers

Exercise 2.1 (Factors and Multiples)

Question 1 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

Factor – A factor of a number is an exact divisor of that number.

__Example:__Each of the number 1, 3, 5 and 15 is a factor of 15. However, none of the numbers 2, 4, 6, 7, 8, 9, 10, 11, 12, 13 and 14 is a factor of 12.Multiple – A multiple of a number is a number obtained by multiplying it by a natural number.

__Example:__If we multiply 4 by 1, 2, 3, 4, 5, 6, ….., we get4x1=4, 4x2=8, 4x3=12, 4x4=16, 4x5=20, 4x6=24 …….

Thus, 4, 8, 12, 16, 20, 24 ……… are multiples of 4.

Question 2 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 60

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 are the numbers which exactly divides the given number.

Therefore, Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

ii. 76

1, 2, 4, 19, 38 and 76 are the numbers which exactly divides the given number.

Therefore, Factors of 76 are 1, 2, 4, 19, 38 and 76.

iii. 125

1, 5, 25 and 125 are the numbers which exactly divides the given number.

Therefore, Factors of 125 are 1, 5, 25 and 125.

iv. 729

1, 3, 9, 27, 81, 243 and 729 are the numbers which exactly divides the given number.

Therefore, Factors of 729 are 1, 3, 9, 27, 81, 243 and 729.

Question 3 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 25

In order to find first five multiples of 25, we multiply the given number by 1, 2, 3, 4 and 5 respectively.

We have,

25x1 = 25

25x2 = 50

25x3 = 75

25x4 = 100

25x5 = 125

Hence, first five multiples of 25 are 25, 50, 75, 100 and 125 respectively.

ii. 35

In order to find first five multiples of 35, we multiply the given number by 1, 2, 3, 4 and 5 respectively.

We have,

35x1 = 35

35x2 = 70

35x3 = 105

35x4 = 140

35x5 = 175

Hence, first five multiples of 35 are 35, 70, 105, 140 and 175 respectively.

iii. 45

In order to find first five multiples of 45, we multiply the given number by 1, 2, 3, 4 and 5 respectively.

We have,

45x1 = 45

45x2 = 90

45x3 = 135

45x4 = 180

45x5 = 225

Hence, first five multiples of 45 are 45, 90, 135, 180 and 225 respectively.

iv. 40

In order to find first five multiples of 40, we multiply the given number by 1, 2, 3, 4 and 5 respectively.

We have,

40x1 = 40

40x2 = 80

40x3 = 120

40x4 = 160

40x5 = 200

Hence, first five multiples of 40 are 40, 80, 120, 160 and 200 respectively.

Question 4 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 15625

15625 ÷ 15 = 1041.67

As the given number is not exactly divisible by 15, therefore 15 is not the factor of 15625.

ii. 123015

123015 ÷ 15 = 8201

As the given number is exactly divisible by 15, therefore 15 is the factor of 123015.

Question 5 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 21063

21063 ÷ 21 = 1003

As the given number is exactly divisible by 21, therefore 21 is the factor of 21063.

ii. 20163

20163 ÷ 21 = 960.14

As the given number is not exactly divisible by 21, therefore 21 is not the factor of 20163.

Question 6 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

__Divisibility Rule of 11__To check the divisibility of a number by 11, the rule is, find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number. If the difference is either 0 or divisible by 11, then the number is divisible by 11.

i. 1111

Sum of digits at odd places from the right = 1+1 = 2

Sum of digits at even places from the right = 1+1 = 2

Difference = 2 – 2 = 0

The difference is 0. Therefore, 1111 is divisible by 11

ii. 11011

Sum of digits at odd places from the right = 1+0+1 = 2

Sum of digits at even places from the right = 1+1 = 2

Difference = 2 – 2 = 0

The difference is 0. Therefore, 11011 is divisible by 11

iii. 110011

Sum of digits at odd places from the right = 1+0+1 = 2

Sum of digits at even places from the right = 1+0+1 = 2

Difference = 2 – 2 = 0

The difference is 0. Therefore, 110011 is divisible by 11

iv. 1100011

Sum of digits at odd places from the right = 1+0+0+1 = 2

Sum of digits at even places from the right = 1+0+1 = 2

Difference = 2 – 2 = 0

The difference is 0. Therefore, 1100011 is divisible by 11

Question 7 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

__Divisibility Rule of 5__To check the divisibility of a number by 5, the rule is, a number which has either 0 or 5 in its ones place is divisible by 5.

i. 55

The given number has 5 in its ones place. Therefore ‘55’ is divisible by 5.

ii. 555

The given number has 5 in its ones place. Therefore, ‘555’ is divisible by 5.

iii. 5555

The given number has 5 in its ones place. Therefore, ‘5555’ is divisible by 5.

iv. 50005

The given number has 5 in its ones place. Therefore, ‘50005’ is divisible by 5.

Question 8 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

No, there no natural number having no factor at all.

Question 9 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

Numbers between 1 and 100 having exactly three factors are 4, 9, 25 and 49.

Question 10 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

Even Numbers – 42 and 144

Odd Numbers – 89 and 321