Playing with Numbers

(Factors are numbers that divide the given number exactly without a remainder.)

(a) 24: 1, 2, 3, 4, 6, 8, 12, 24
(Since 1×24, 2×12, 3×8, 4×6)

(b) 15: 1, 3, 5, 15

(c) 21: 1, 3, 7, 21

(d) 27: 1, 3, 9, 27

(e) 12: 1, 2, 3, 4, 6, 12

(f) 20: 1, 2, 4, 5, 10, 20

(g) 18: 1, 2, 3, 6, 9, 18

(h) 23: 1, 23 (Since 23 is a prime number)

(i) 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

(a) Multiples of 5: 5×1, 5×2, 5×3, 5×4, 5×5.
Answer: 5, 10, 15, 20, 25.

(b) Multiples of 8: 8×1, 8×2, 8×3, 8×4, 8×5.
Answer: 8, 16, 24, 32, 40.

(c) Multiples of 9: 9×1, 9×2, 9×3, 9×4, 9×5.
Answer: 9, 18, 27, 36, 45.

Column 1:
(i) 35
(ii) 15
(iii) 16
(iv) 20
(v) 25

Column 2:
(a) Multiple of 8
(b) Multiple of 7
(c) Multiple of 70
(d) Factor of 30
(e) Factor of 50
(f) Factor of 20

Matching:
(i) 35 → (b) Multiple of 7 (7 × 5 = 35)
(ii) 15 → (d) Factor of 30 (15 × 2 = 30)
(iii) 16 → (a) Multiple of 8 (8 × 2 = 16)
(iv) 20 → (f) Factor of 20 (20 × 1 = 20)
(v) 25 → (e) Factor of 50 (25 × 2 = 50)

Multiply 9 by 1, 2, 3, etc., until the product reaches near 100.
9×1 = 9
9×2 = 18
9×3 = 27
9×4 = 36
9×5 = 45
9×6 = 54
9×7 = 63
9×8 = 72
9×9 = 81
9×10 = 90
9×11 = 99

Therefore, the multiples of 9 up to 100 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.

(a) The sum of any two odd numbers is always an Even number.
Example: 3 + 5 = 8 (even). 7 + 11 = 18 (even).

(b) The sum of any two even numbers is always an Even number.
Example: 2 + 4 = 6 (even). 10 + 12 = 22 (even).

(a) The sum of three odd numbers is even.
False. (Odd + Odd = Even. Then Even + Odd = Odd. e.g., 3+5+7=15(odd)).

(b) The sum of two odd numbers and one even number is even.
True. (Odd + Odd = Even. Even + Even = Even. e.g., 3+5+4=12(even)).

(c) The product of three odd numbers is odd.
True. (Odd × Odd = Odd. Odd × Odd = Odd. e.g., 3×5×7=105(odd)).

(d) If an even number is divided by 2, the quotient is always odd.
False. (Example: 4÷2 = 2, which is even. 8÷2=4, which is even).

(e) All prime numbers are odd.
False. (2 is a prime number and it is even).

(f) Prime numbers do not have any factors.
False. (Prime numbers have exactly two factors: 1 and the number itself).

(g) Sum of two prime numbers is always even.
False. (Example: 2 + 3 = 5, which is odd).

(h) 2 is the only even prime number.
True. (All other even numbers are divisible by 2, making them composite).

(i) All even numbers are composite numbers.
False. (2 is an even number but it is prime, not composite).

(j) The product of any two even numbers is always even.
True. (Even × Even = Even).

We check pairs under 100 where reversing digits yields another prime:
- 17 and 71 (both are prime)
- 37 and 73 (both are prime)
- 79 and 97 (both are prime)

So, the pairs are: (17, 71); (37, 73); (79, 97).

Numbers less than 20 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19.

Prime Numbers: 2, 3, 5, 7, 11, 13, 17, 19.
(Numbers having only two factors)

Composite Numbers: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18.
(Numbers having more than two factors)

Note: 1 is neither prime nor composite.

Prime numbers between 1 and 10 are: 2, 3, 5, 7.

Therefore, the greatest prime number among these is 7.

(Goldbach's conjecture says every even number > 2 can be written as the sum of two primes)

(a) 44 = 3 + 41  (or let's take 13 + 31; both are valid).

(b) 36 = 5 + 31  (or 13 + 23; 17 + 19).

(c) 24 = 5 + 19  (or 7 + 17; 11 + 13).

(d) 18 = 5 + 13  (or 7 + 11).

Three pairs of twin primes are:
(3, 5)
(5, 7)
(11, 13)

(a) 23 uses factors: 1, 23 → Prime.

(b) 51 uses factors: 1, 3, 17, 51 → Composite (since 17×3=51).

(c) 37 uses factors: 1, 37 → Prime.

(d) 26 uses factors: 1, 2, 13, 26 → Composite.

Therefore, (a) 23 and (c) 37 are prime numbers.

We need a gap of 7 between two primes under 100.
The primes under 100 are: 2, 3, 5, 7 ... 79, 83, 89, 97.
Between 89 and 97, there are 7 consecutive composite numbers.
They are: 90, 91, 92, 93, 94, 95, 96.

(a) 21 = 3 + 5 + 13  (or 3 + 7 + 11)

(b) 31 = 3 + 5 + 23  (or 5 + 7 + 19)

(c) 53 = 13 + 17 + 23

(d) 61 = 7 + 13 + 41  (or 11 + 13 + 37)

Primes less than 20: 2, 3, 5, 7, 11, 13, 17, 19.
We need sums to be 5, 10, 15, 20, 25, 30, 35.

1. 2 + 3 = 5 (Divisible by 5)
2. 3 + 17 = 20 (Divisible by 5)
3. 7 + 13 = 20 (Divisible by 5)
4. 3 + 7 = 10 (Divisible by 5)
5. 11 + 19 = 30 (Divisible by 5)

The pairs are: (2, 3), (3, 7), (3, 17), (7, 13), (11, 19).

(a) A number which has only two factors is called a prime number.

(b) A number which has more than two factors is called a composite number.

(c) 1 is neither prime nor composite.

(d) The smallest prime number is 2.

(e) The smallest composite number is 4.

(f) The smallest even number is 2.

Rule for 4: Last 2 digits must be divisible by 4.
Rule for 8: Last 3 digits must be divisible by 8.

(a) 572:
Last two digits (72): 72 ÷ 4 = 18. So, divisible by 4.
Last three digits (572): 572 ÷ 8 = 71.5. Not divisible by 8.
Divisible by 4, not by 8.

(b) 726352:
Last two digits (52): 52 ÷ 4 = 13. Divisible by 4.
Last three digits (352): 352 ÷ 8 = 44. Divisible by 8.
Divisible by 4 and by 8.

(c) 5500:
Ends in 00. Divisible by 4.
Last 3 digits (500). Not divisible by 8 (500÷8=62.5).
Divisible by 4, not by 8.

(d) 6000:
Ends in 00. Divisible by 4.
Ends in 000. Divisible by 8.
Divisible by 4 and by 8.

(e) 12159:
Last digit is odd (9). It is not divisible by 2, therefore cannot be divisible by 4 or 8.
Not divisible by 4 or 8.

(f) 14560:
Last two digits (60). Divisible by 4.
Last three (560). Divisible by 8 (560/8=70).
Divisible by 4 and 8.

(g) 21084:
Last two (84). Divisible by 4.
Last three (084). Not divisible by 8.
Divisible by 4, not by 8.

(h) 31795072:
Last two (72). Divisible by 4.
Last three (072). Divisible by 8.
Divisible by 4 and 8.

(i) 1700:
Ends in 00. Divisible by 4.
Last three (700). Not divisible by 8.
Divisible by 4, not by 8.

(j) 2150:
Last two (50). Not divisible by 4.
Not divisible by 4, so not by 8.

Rule for 6: A number must be divisible by 2 (ends in 0,2,4,6,8) AND divisible by 3 (sum of digits is a multiple of 3).

(a) 297144:
Ends in 4 → Divisible by 2.
Sum of digits = 2+9+7+1+4+4 = 27 → Divisible by 3.
Since divisible by both, it is divisible by 6.

(b) 1258:
Ends in 8 → Divisible by 2.
Sum of digits = 1+2+5+8 = 16 → Not divisible by 3.
Not divisible by 6.

(c) 4335:
Ends in 5 → Not divisible by 2.
Not divisible by 6.

(d) 61233:
Ends in 3 → Not divisible by 2.
Not divisible by 6.

(e) 901352:
Ends in 2 → Divisible by 2.
Sum of digits = 9+0+1+3+5+2 = 20 → Not divisible by 3.
Not divisible by 6.

(f) 438750:
Ends in 0 → Divisible by 2.
Sum of digits = 4+3+8+7+5+0 = 27 → Divisible by 3.
Divisible by 6.

(g) 1790184:
Ends in 4 → Divisible by 2.
Sum of digits = 1+7+9+0+1+8+4 = 30 → Divisible by 3.
Divisible by 6.

(h) 12583:
Ends in 3 → Not divisible by 2.
Not divisible by 6.

(i) 639210:
Ends in 0 → Divisible by 2.
Sum of digits = 6+3+9+2+1+0 = 21 → Divisible by 3.
Divisible by 6.

(j) 17852:
Ends in 2 → Divisible by 2.
Sum of digits = 1+7+8+5+2 = 23 → Not divisible by 3.
Not divisible by 6.

Rule for 11: Find the difference between the sum of the digits at odd places and sum of digits at even places (from the right). If the difference is either 0 or a multiple of 11, then the number is divisible by 11.

(a) 5445:
Sum of odds (from right): 5 + 4 = 9
Sum of evens: 4 + 5 = 9
Difference: 9 - 9 = 0. Divisible by 11.

(b) 10824:
Sum of odds: 4 + 8 + 1 = 13
Sum of evens: 2 + 0 = 2
Difference: 13 - 2 = 11. Divisible by 11.

(c) 7138965:
Sum of odds: 5 + 9 + 3 + 7 = 24
Sum of evens: 6 + 8 + 1 = 15
Difference: 24 - 15 = 9. Not divisible by 11.

(d) 70169308:
Sum of odds: 8 + 3 + 6 + 0 = 17
Sum of evens: 0 + 9 + 1 + 7 = 17
Difference: 17 - 17 = 0. Divisible by 11.

(e) 10000001:
Sum of odds: 1 + 0 + 0 + 0 = 1
Sum of evens: 0 + 0 + 0 + 1 = 1
Difference: 1 - 1 = 0. Divisible by 11.

(f) 901153:
Sum of odds: 3 + 1 + 0 = 4
Sum of evens: 5 + 1 + 9 = 15
Difference: 15 - 4 = 11. Divisible by 11.

Rule for 3: Sum of digits must be divisible by 3.

(a) _ 6724:
Sum of given digits = 6 + 7 + 2 + 4 = 19.
For sum to be multiple of 3, next multiples are 21, 24, 27.
For 21: Added digit = 21 - 19 = 2 (Smallest digit)
For 27: Added digit = 27 - 19 = 8 (Greatest digit)
Answer: Smallest = 2, Greatest = 8.

(b) 4765 _ 2:
Sum of given digits = 4 + 7 + 6 + 5 + 2 = 24.
24 is already divisible by 3. Next multiples are 24, 27, 30, 33.
For 24: Added digit = 0 (Smallest digit)
For 33: Added digit = 9 (Greatest digit)
Answer: Smallest = 0, Greatest = 9.

(a) 92 _ 389:
Let digit be 'x'. Number: 9, 2, x, 3, 8, 9.
Sum of odd places (from right): 9 + 3 + 2 = 14.
Sum of even places: 8 + x + 9 = 17 + x.
Difference: (17 + x) - 14 = 3 + x.
For divisibility by 11, (3 + x) must be 0, 11, etc.
If 3 + x = 11, then x = 8.
Answer: 8.

(b) 8 _ 9484:
Let digit be 'y'. Number: 8, y, 9, 4, 8, 4.
Sum of odd places (from right): 4 + 4 + y = 8 + y.
Sum of even places: 8 + 9 + 8 = 25.
Difference: 25 - (8 + y) = 17 - y.
For divisibility by 11, (17 - y) must be 11.
17 - y = 11 → y = 6.
Answer: 6.

(a) Factors of 20 = 1,2,4,5,10,20. Factors of 28 = 1,2,4,7,14,28. Common = 1, 2, 4

(b) Factors of 15 = 1,3,5,15. Factors of 25 = 1,5,25. Common = 1, 5

(c) Factors of 35 = 1,5,7,35. Factors of 50 = 1,2,5,10,25,50. Common = 1, 5

(d) Factors of 56 = 1,2,4,7,8,14,28,56. Factors of 120 = 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120. Common = 1, 2, 4, 8

(a) LCM = 24. First three = 24, 48, 72

(b) LCM = 36. First three = 36, 72, 108

(a) Co-prime   (b) Co-prime   (c) Not co-prime (both divisible by 5)

(d) Not co-prime (17 divides 68)   (e) Co-prime (consecutive)   (f) Co-prime

5 and 12 are co-prime numbers, so any number divisible by both is divisible by 5 × 12 = 60.

Greatest 4-digit number = 9999.
9999 ÷ 3 = 3333 → ÷ 3 = 1111 → ÷ 11 = 101 (prime).
9999 = 3 × 3 × 11 × 101.

(a) 1729 = 7 × 13 × 19 → 7, 13, 19

(b) 4 = 2 × 2 → 2

(c) 27 = 3 × 3 × 3 → 3

(d) 100 = 2 × 2 × 5 × 5 → 2, 5

(a) 18 = 2×3×3; 48 = 2×2×2×2×3. HCF = 6

(b) 30 = 2×3×5; 42 = 2×3×7. HCF = 6

(c) 18 = 2×3×3; 60 = 2×2×3×5. HCF = 6

(d) 27 = 3×3×3; 63 = 3×3×7. HCF = 9

(e) 36 = 2×2×3×3; 84 = 2×2×3×7. HCF = 12

(a) 1   (b) 2   (c) 1

75 = 3×5×5; 69 = 3×23. HCF = 3 kg.

63 = 3×3×7; 70 = 2×5×7; 77 = 7×11.
LCM = 2×3×3×5×7×11 = 6930 cm (69 m 30 cm).

HCF(825, 675, 450): common prime factors = 3×5×5 = 75 cm.

LCM(6, 8, 12) = 24. 100 ÷ 24 = 4 rem 4. Next multiple = 24 × 5 = 120.

LCM(8, 10, 12) = 120. 999 ÷ 120 = 8 rem 39. Answer = 999 - 39 = 960.

(a) 36   (b) 60   (c) 30   (d) 60

Yes — each pair is co-prime (HCF = 1), so LCM = product of the numbers.

144 = 2×2×2×2×3×3
180 = 2×2×3×3×5
192 = 2×2×2×2×2×2×3
Common prime factors = 2×2×3 = 12.

12 = 2×2×3; 15 = 3×5; 20 = 2×2×5; 30 = 2×3×5.
LCM = 2×2×3×5 = 60.

Divide 1645 by 7: 1645 ÷ 7 = 235. Since the division gives no remainder, yes, 7 is a factor of 1645.

7 × 11 × 13 + 13
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
Since the number equals 13 × 78, it has factors other than 1 and itself. Therefore it is a composite number.

9 = 3 × 3 and 25 = 5 × 5. These two numbers are co-prime (HCF = 1).
If a number is divisible by two co-prime numbers, it must be divisible by their product.
Product = 9 × 25 = 225. Hence proved.

First find LCM of 6, 15, and 18.
6 = 2×3; 15 = 3×5; 18 = 2×3×3.
LCM = 2×3×3×5 = 90.

The required number = LCM + remainder = 90 + 5 = 95.

We need HCF of 513 and 783.
Using Euclid's method: 783 = 513 × 1 + 270.
513 = 270 × 1 + 243.
270 = 243 × 1 + 27.
243 = 27 × 9 + 0. Remainder = 0.
HCF = 27.

LCM of 6, 12, 18.
6 = 2×3; 12 = 2×2×3; 18 = 2×3×3.
LCM = 2×2×3×3 = 36 minutes.
Time = 2:00 PM + 36 minutes = 2:36 PM.

Divide 1056 by 23: 1056 ÷ 23 = 45 remainder 21.
The remainder is 21, so we need to add (23 - 21) = 2 to make it divisible by 23.

Rule for 4: Last 2 digits must be divisible by 4. Last 2 digits of 4598 = 98. 98 ÷ 4 = 24 remainder 2. Not divisible by 4.
Since not divisible by 4, it cannot be divisible by 8 either. Not divisible by 8.

90 ÷ 2 = 45 → 45 ÷ 3 = 15 → 15 ÷ 3 = 5 → 5 is prime.
90 = 2 × 3 × 3 × 5.

48 = 2×2×2×2×3; 72 = 2×2×2×3×3.
HCF = 2×2×2×3 = 24.
LCM = 2×2×2×2×3×3 = 144.

Verification: HCF × LCM = 24 × 144 = 3456.
Product of numbers = 48 × 72 = 3456. Verified!

Sum of digits of 3654 = 3 + 6 + 5 + 4 = 18. Since 18 is divisible by 9, 3654 is divisible by 9.

One possibility: The numbers share HCF 10, so they could be 10a and 10b where a and b are co-prime.
LCM = 10ab = 150 → ab = 15.
Co-prime factor pairs for 15: (1, 15) or (3, 5).
Numbers = 30 and 50 (using 10×3 and 10×5) or 10 and 150 (using 10×1 and 10×15).

LCM of 64, 80 and 96.
64 = 2×2×2×2×2×2; 80 = 2×2×2×2×5; 96 = 2×2×2×2×2×3.
LCM = 2×2×2×2×2×2×3×5 = 960 cm (9 m 60 cm).

= 11 × (2 × 3 × 5 × 7 + 1) = 11 × (210 + 1) = 11 × 211.
Since the number has a factor (11) other than 1 and itself, it is composite.

(a) 5918:
Sum of odd place digits (from right): 8 + 9 = 17.
Sum of even place digits: 1 + 5 = 6.
Difference = 17 - 6 = 11. Since 11 is a multiple of 11, 5918 is divisible by 11.

(b) 70169:
Sum of odd places: 9 + 1 + 7 = 17.
Sum of even places: 6 + 0 = 6.
Difference = 17 - 6 = 11. Divisible by 11.

The numbers are 3k and 4k, where k = HCF = 8.
Numbers = 3 × 8 = 24 and 4 × 8 = 32.
LCM = (product of numbers) ÷ HCF = (24 × 32) ÷ 8 = 768 ÷ 8 = 96.

If two numbers are co-prime, their HCF = 1.
Therefore, HCF = 1.
LCM = Product of the numbers (since HCF = 1) = 2028.

Prime numbers between 1 and 50:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

Total count = 15 prime numbers.