RD Sharma Solutions: Playing With Numbers (Exercise 2.1)
Q.1. Define
(i) factor
(ii) multiple
Give four examples of each.
Ans:
(i) Factor: A factor of a number is a whole number which divides that number exactly, leaving no remainder.
For example, 4 exactly divides 32, so 4 is a factor of 32.
Examples of factors are:
2 and 3 are factors of 6 because 2 × 3 = 6.
2 and 4 are factors of 8 because 2 × 4 = 8.
3 and 4 are factors of 12 because 3 × 4 = 12.
3 and 5 are factors of 15 because 3 × 5 = 15.
(ii) Multiple: A multiple of a number is the product obtained when that number is multiplied by any whole number. If a × b = c, then c is a multiple of both a and b.
Examples of multiples are:
6 is a multiple of 2 because 2 × 3 = 6.
8 is a multiple of 4 because 4 × 2 = 8.
12 is a multiple of 6 because 6 × 2 = 12.
21 is a multiple of 7 because 7 × 3 = 21.
Q.2. Write all factors of each of the following numbers:
(i) 60
(ii) 76
(iii) 125
(iv) 729
Ans:
(i) We find factor pairs of 60:
60 = 1 × 60
60 = 2 × 30
60 = 3 × 20
60 = 4 × 15
60 = 5 × 12
60 = 6 × 10
∴ The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
(ii) Factor pairs of 76:
76 = 1 × 76
76 = 2 × 38
76 = 4 × 19
∴ The factors of 76 are 1, 2, 4, 19, 38 and 76.
(iii) Factor pairs of 125:
125 = 1 × 125
125 = 5 × 25
∴ The factors of 125 are 1, 5, 25 and 125.
(iv) Factor pairs of 729:
729 = 1 × 729
729 = 3 × 243
729 = 9 × 81
729 = 27 × 27
∴ The factors of 729 are 1, 3, 9, 27, 81, 243 and 729.
Q.3. Write first five multiples of each of the following numbers:
(i) 25
(ii) 35
(iii) 45
(iv) 40
Ans:
(i) The first five multiples of 25 are obtained by multiplying 25 by 1, 2, 3, 4 and 5:
25 × 1 = 25
25 × 2 = 50
25 × 3 = 75
25 × 4 = 100
25 × 5 = 125
(ii) The first five multiples of 35 are:
35 × 1 = 35
35 × 2 = 70
35 × 3 = 105
35 × 4 = 140
35 × 5 = 175
(iii) The first five multiples of 45 are:
45 × 1 = 45
45 × 2 = 90
45 × 3 = 135
45 × 4 = 180
45 × 5 = 225
(iv) The first five multiples of 40 are:
40 × 1 = 40
40 × 2 = 80
40 × 3 = 120
40 × 4 = 160
40 × 5 = 200
Q.4. Which of the following numbers have 15 as their factor?
(i) 15615
(ii) 123015
Ans:
Method: A number is divisible by 15 if it is divisible by both 3 and 5.
(i) 15615 ends with 5, so it is divisible by 5. Sum of its digits = 1 + 5 + 6 + 1 + 5 = 18, which is divisible by 3. Hence 15 divides 15,615. Indeed, 1,041 × 15 = 15,615.
(ii) 123015 ends with 5, so it is divisible by 5. Sum of its digits = 1 + 2 + 3 + 0 + 1 + 5 = 12, which is divisible by 3. Hence 15 divides 1,23,015. Indeed, 8,201 × 15 = 1,23,015.
Thus, both the given numbers have 15 as a factor.
Disclaimer: The answer given in the book is incorrect.
Q.5. Which of the following numbers are divisible by 21?
(i) 21063
(ii) 20163
Ans: Since 21 = 3 × 7, a number is divisible by 21 if it is divisible by both 3 and 7.
(i) Sum of digits of 21,063 = 2 + 1 + 0 + 6 + 3 = 12, which is divisible by 3. So 21,063 is divisible by 3.
For divisibility by 7 we use the test: subtract twice the unit digit from the number formed by the remaining digits. If the result is divisible by 7, the original number is divisible by 7.
Here, 2106 - 2×3 = 2106 - 6 = 2100, and 2100 = 7 × 300, so it is divisible by 7. Therefore 21,063 is divisible by 7 and hence by 21.
(ii) Sum of digits of 20,163 = 2 + 0 + 1 + 6 + 3 = 12, which is divisible by 3. So 20,163 is divisible by 3.
Check for 7: 2016 - 2×3 = 2016 - 6 = 2010. Since 2010 ÷ 7 = 287 remainder 1, 2010 is not divisible by 7. Therefore 20,163 is not divisible by 7 and so it is not divisible by 21.
Q.6. Without actual division show that 11 is a factor of each of the following numbers:
(i) 1111
(ii) 11011
(iii) 110011
(iv) 1100011
Ans: Use the divisibility rule for 11: If the difference between the sum of digits in odd places and the sum of digits in even places (counting from the unit's place as position 1) is 0 or a multiple of 11, the number is divisible by 11.
(i) 1,111
Sum of digits at odd places = 1 + 1 = 2 (positions 1 and 3)
Sum of digits at even places = 1 + 1 = 2 (positions 2 and 4)
Difference = 2 - 2 = 0 ⇒ divisible by 11. Hence 1,111 is divisible by 11.
(ii) 11,011
Sum of digits at odd places = 1 + 0 + 1 = 2
Sum of digits at even places = 1 + 1 = 2
Difference = 2 - 2 = 0 ⇒ divisible by 11. Hence 11,011 is divisible by 11.
(iii) 1,10,011
Sum of digits at odd places = 1 + 0 + 1 = 2
Sum of digits at even places = 1 + 0 + 1 = 2
Difference = 2 - 2 = 0 ⇒ divisible by 11. Hence 1,10,011 is divisible by 11.
(iv) 11,00,011
Sum of digits at odd places = 1 + 0 + 0 + 1 = 2
Sum of digits at even places = 1 + 0 + 1 = 2
Difference = 2 - 2 = 0 ⇒ divisible by 11. Hence 11,00,011 is divisible by 11.
Q.7. Without actual division show that each of the following numbers is divisible by 5:
(i) 55
(ii) 555
(iii) 5555
(iv) 50005
Ans: A number is divisible by 5 if its unit's digit is either 0 or 5.
(i) 55 ends with 5, so it is divisible by 5.
(ii) 555 ends with 5, so it is divisible by 5.
(iii) 5,555 ends with 5, so it is divisible by 5.
(iv) 50,005 ends with 5, so it is divisible by 5.
Q.8. Is there any natural number having no factor at all?
Ans: No. Every natural number has at least one factor - itself. In fact, every natural number n has at least the factors 1 and n (except 1, which has only 1 as its factor). Thus there is no natural number with no factor at all.
Q.9. Find numbers between 1 and 100 having exactly three factors.
Ans: The numbers between 1 and 100 that have exactly three factors are 4, 9, 25 and 49.
Reason: A number has exactly three positive factors when it is the square of a prime number. The prime numbers whose squares lie between 1 and 100 are 2, 3, 5 and 7. Their squares are 4, 9, 25 and 49 respectively.
The factors of 4 are 1, 2 and 4.
The factors of 9 are 1, 3 and 9.
The factors of 25 are 1, 5 and 25.
The factors of 49 are 1, 7 and 49.
Q.10. Sort out even and odd numbers:
(i) 42
(ii) 89
(iii) 144
(iv) 321
Ans:
A number is called even if it is exactly divisible by 2. Therefore, 42 and 144 are even numbers because 42 ÷ 2 = 21 and 144 ÷ 2 = 72.
A number is called odd if it is not exactly divisible by 2. Therefore, 89 and 321 are odd numbers because they leave a remainder 1 when divided by 2 (89 ÷ 2 = 44 remainder 1, 321 ÷ 2 = 160 remainder 1).
FAQs on RD Sharma Solutions: Playing With Numbers (Exercise 2.1)
| 1. What is the concept of playing with numbers in mathematics? |  |
| 2. How can playing with numbers help in improving mathematical skills? | |
| 3. What are the divisibility rules in playing with numbers? | |
| 4. How can prime and composite numbers be identified while playing with numbers? | |
| 5. How can playing with numbers be applied in real-life situations? | |
