📝 Solution:

1 hour = 60 minutes

40 = 2 × 2 × 2 × 5 = 23 × 5

60 = 2 × 2 × 3 × 5 = 22 × 3 × 5

∴ LCM (40, 60) = 23 × 3 × 5 = 120

120 minutes = 2 hours

Hence, the two bells will ring together again at

9:00 + 2:00 = 11:00 a.m.

📝 Solution:

17 × 11 × 2 + 17 × 11 × 5

= 17 × 11 × (2 + 5)

= 17 × 11 × 7

Since, 17 × 11 × 2 + 17 × 11 × 5 can be expressed as a product of primes, therefore, it is a composite number.

📝 Solution:

No, because HCF always divide LCM but here 24 does not divide 7290.

📝 Solution:

Let, us suppose that 14n ends with the digit 0 for some n ∈ N

∴ 14n is divisible by 5

But, prime factors of 14 are 2 and 7.

∴ Prime factor of (14)n are (2 × 7)n

⇒ It is clear that in prime factorisation of 14″ there is no place for 5.

∴ By Fundamental theorem of Arithmetic

Every composite no. can be expressed as a product of primes and this factorisation is unique, a part from the order in which the prime factor occur.

∴ Our Supposition is wrong.

Hence, there exists no natural number n for which 14n ends with the digit zero.

📝 Solution:

Prime factors of 10010 = 2 × 5 × 7 × 11 × 13

Prime factors of 140 = 2 × 2 × 5 × 7

📝 Solution:

Let p = 2m + 1, q = 2n + 1, where m and n are natural numbers.

Now, (p² – q²) = (p – q)(p + q) ,

But, p- q = (2m + 1) – (2n + 1) = 2(m – n)

p + q = (2m + 1) + (2n + 1) = 2 (m + n + 1)

Clearly both factors of (p² – q²) are even i.e., none of the divisors (p – q) and (p + q) is equal to 1.

⇒ (p² – q²) is a composite number.

📝 Solution:

Given 43/2⁴×5³ is in the lowest form and power of 2 = 4, Power of 5 = 3, Max. {4, 3} = 4

∴ 43/2⁴×5³ will terminate after 4 places of decimal.

📝 Solution:

Smallest prime number = 2 = 21

Smallest composite number = 4 = 22

H.C.F. (2, 4) = 21 = 2