Question

Show that 45ⁿ can not end with the digit 0, n being a natural number. Write the prime number ‘a’ which on multiplying with 45ⁿ makes the product end with the digit 0.

Answer 1

The prime factorization of 45ⁿ

= (3 × 3 × 5)ⁿ

= (3² × 5)ⁿ

= 3²ⁿ × 5ⁿ

We have only prime factors 3 and 5.

If 45ⁿ is ends with zero, then we need prime factors 2 and 5 both or divisible by 10 but the prime factor 2 is absent.

So, 45ⁿ cannot end with the digit 0.

The required prime number is 2 for 45ⁿ makes the product and with the digit 0.

Therefore, a is 2.