Question

Show that 9n can never ends with unit digit zero.

Answer 1

If a number ends with the digit zero then it will be divisible by 2 and 5 both. So here first write the given number as the product of prime factors and then check this prime factorization contains 2 and 5 or not.If it contains 2 and 5 both, then given number ends with the digit 0, otherwise not.

SOLUTION:
Given, n is a natural number.
Let 9ⁿ ends with 0, then 9ⁿ is divisible by 2 and 5. But prime factors of 9 are 3×3

9ⁿ= (3×3)ⁿ = 3²ⁿ
Thus, prime factorization of 9ⁿ does not contain 2 and 5. so the uniqueness of the fundamental theorem of arithmetic guarantees that there is no other primes in the factorization of 9ⁿ. So 2 and 5 does not occur in the prime factorization of 9ⁿ. Hence 9ⁿ can never ends with digit 0 for any natural number.

HOPE THIS WILL HELP YOU ..

Answer 2

If 9^n ended with the digit 0 then it would be divisible by 5 . This means that the prime factorization of 9^n would contain the prime factor 5.
But 9^n = 3(3^n) . This means that 9^n contains only the prime factor three which is guaranteed by the uniqueness of the fundamental theorem of arithmetic .
Hence since 9^n does not have 5 as a factor it cannot end with the digit 0.