Question

show that 9^n cannot end with digit 0 for any natural number

Answer 1

because the prime factorization of 9 doesn't comprise 2

Answer 2

Hii. ....friends,

The answer is here,

$= > \: {9}^{1} = \: units \: digit \: 9$

$= > \: {9}^{2} = units \: digit \: 1$

$= > \: {9}^{3} = units \: digit \: 9$

$= > \: {9}^{4} = units \: digit \: 1$

And so on

So, We can conclude that , For all natural numbers, 9^n has units digit as 1 or 9.

Hence 9^n doesn't ends with 0 for any natural number.

we can also prove it in another way that,

Prime factorization of 9 = 3×3.

So, to get 0 as a units digit there must be 5 and 2 as factor.

But there is no 5 & 2 . Hence 9^n doesnt end with 0.

:-)Hope it helps u.