Question

Show that the number 8n can never end with digit zero for any natural number n

Answer 1

For a number to end with the digit 0 it's prime factorization should have 2 and 5 as a common factor.

here 8^n = (2*4)^n doesn't have 5 in its prime factorization.

Therefore 8^n cannot end with the digit 0.

Hope this helps!

Answer 2

Answer:

It’s not possible for $8^n$ to be end with zero having n as any natural number

Solution:

For any number to have zero at the end, it should be multiplied with 5, 10 or by the multiples of 10.

In case of terms like $x^n$, the value x must contain zero or zeroes at the end otherwise it must have the factor as 5 or 10.

Here in 8, the factors of 8 are 2 alone

It doesn’t have 5 to make the two as 10.  

$\begin{array} { c } { 8 = 2 \times 2 \times 2 } \\\\ { 8 ^ { 1 } = 8 } \\\\ { 8 ^ { 2 } = 64 } \\\\ { 8 ^ { 3 } = 512 } \\\\ { 8 ^ { 4 } = 4096 } \\\\ { 8 ^ { 5 } = 32768 } \end{array}$

Therefore, for $8^n$ cannot end with the digit zero for n having any natural number.