### 5. Check whether 6^{n} can end with the digit 0 for any natural number n.

Here, n is a natural number and let 6^{n} end with digit 0.

6^{n} is divisible by 5.

But the prime factors of 6 are 2 and 3. i.e., 6 = 2 × 3

⇒ 6^{n} = (2 × 3)^{n}

i.e., In the prime factorization of 6^{n}, there is no factor 5.

So, by the fundamental theorem of Arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order in which the prime factorization occurs.

Our assumption that 6^{n} ends with digit 0, is wrong.

Thus, there does not exist any natural number n for which 6^{n} ends with zero.

### 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

We have

7 × 11 × 13 + 13 = 13((7 × 11) + 1) = 13(78), which cannot

be a prime number because it has factors 13 and 78.

Also, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= 5[7 × 6 × 4 × 3 × 2 × 1 + 1],

which is also not a prime number because it has a factor 5

Thus, 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

### 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?

Time taken by Sonia to drive one round of the field = 18 minutes

Time taken by Ravi to drive one round of the field = 12 minutes

LCM of 18 and 12 gives the exact number of minutes after which they meet again at the starting point.

Now, 18 = 2 × 3 × 3 and 12 = 2 × 2 × 3

LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36

Thus, they will meet again at the starting point after 36 minutes