Certainly, here are the rewritten solutions for the math problems:
1. Express each number as a product of its prime factors.
(i) 140
To find the prime factors of 140:
[140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7]
(ii) 156
To find the prime factors of 156:
[156 = 2 \times 2 \times 13 \times 3 = 2^2 \times 13 \times 3]
(iii) 3825
To find the prime factors of 3825:
[3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17]
(iv) 5005
To find the prime factors of 5005:
[5005 = 5 \times 7 \times 11 \times 13 = 5 \times 7 \times 11 \times 13]
(v) 7429
To find the prime factors of 7429:
[7429 = 17 \times 19 \times 23 = 17 \times 19 \times 23]
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
Prime factors of 26: (26 = 2 \times 13)
Prime factors of 91: (91 = 7 \times 13)
HCF(26, 91) = 13
LCM(26, 91) = (2 \times 7 \times 13 = 182)
Verification:
(HCF(26, 91) \times LCM(26, 91) = 13 \times 182 = 2366)
Product of 26 and 91: (26 \times 91 = 2366)
(ii) 510 and 92
Prime factors of 510: (510 = 2 \times 3 \times 5 \times 17)
Prime factors of 92: (92 = 2 \times 2 \times 23)
HCF(510, 92) = 2
LCM(510, 92) = (2 \times 2 \times 3 \times 5 \times 17 \times 23 = 23460)
Verification:
(HCF(510, 92) \times LCM(510, 92) = 2 \times 23460 = 46920)
Product of 510 and 92: (510 \times 92 = 46920)
(iii) 336 and 54
Prime factors of 336: (336 = 2 \times 2 \times 2 \times 2 \times 7 \times 3)
Prime factors of 54: (54 = 2 \times 3 \times 3 \times 3)
HCF(336, 54) = 2 \times 3 = 6
LCM(336, 54) = (2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 = 3024)
Verification:
(HCF(336, 54) \times LCM(336, 54) = 6 \times 3024 = 18144)
Product of 336 and 54: (336 \times 54 = 18144)
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15, and 21
Prime factors of 12: (12 = 2 \times 2 \times 3 = 2^2 \times 3)
Prime factors of 15: (15 = 3 \times 5)
Prime factors of 21: (21 = 3 \times 7)
HCF(12, 15, 21) = 3
LCM(12, 15, 21) = (2^2 \times 3 \times 5 \times 7 = 420)
(ii) 17, 23, and 29
Prime factors of 17: (17 = 17 \times 1)
Prime factors of 23: (23 = 23 \times 1)
Prime factors of 29: (29 = 29 \times 1)
HCF(17, 23, 29) = 1
LCM(17, 23, 29) = (17 \times 23 \times 29 = 11339)
(iii) 8, 9, and 25
Prime factors of 8: (8 = 2 \times 2 \times 2 = 2^3)
Prime factors of 9: (9 = 3 \times 3 = 3^2)
Prime factors of 25: (25 = 5 \times 5 = 5^2)
HCF(8, 9, 25) = 1
LCM(8, 9, 25) = (2^3 \times 3^2 \times 5^2 = 1800)
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Using the relationship (HCF \times LCM = \text{Product of the two given numbers},) we have:
(9 \times LCM(306, 657) = 306 \times 657)
Solving for LCM(306, 657):
(LCM(306, 657) = \frac{306 \times 657}{9} = 22338)
Hence, LCM(306, 657) = 22338.
5. Check whether 6n can end with the digit 0 for any natural number n.
For any number to end with the digit 0, it must be divisible by 10, which means it must have prime factors of 2 and 5. Let’s analyze 6n:
[6n = 2 \times 3 \times n]
The prime factors of 6n are 2 and 3. It does not have the prime factor 5. Therefore, for any natural number n, 6n cannot end with the digit 0.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
To determine if a number is composite, we need to show that it has factors other than 1 and itself.
(i) 7 × 11 × 13 + 13
Factoring out 13, we get:
[7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1)]
Since (
7 \times 11 + 1) is greater than 1 and not equal to 13, it is evident that this expression has factors other than 1 and itself, making it a composite number.
(ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Factoring out 5, we get:
[7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)]
Since (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) is greater than 1 and not equal to 5, this expression also has factors other than 1 and itself, making it a composite number.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?
Since Sonia and Ravi are starting at the same point and going in the same direction, we can find the time it takes for them to meet again at the starting point by finding the least common multiple (LCM) of their times.
LCM(18, 12) = 36
Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.