Question

Show that 21^n cannot end with digits 0,2,4,6 or 8 for any natural number n.

Answer 1

If any of the numbers 21^n ends in 0,2,4,6,8 that means that 21^n has 2 in its prime factorization. But 21^n = 3^n * 7^n and each number can have only one unique prime factorization. So there can't be 2 in the prime factorization of 21^n.

Answer 2

Answer:

To show : $21^n$ cannot end with digits 0,2,4,6 or 8 for any natural number n?

Solution :

$21^n$ is always going to end with an odd integer.

Let, For values of n=1,2,...k

$21^1=21=2\times 10+1$

$21^2=441=2\times 220+1$

$21^3=9261=2\times 4630+1$

..

$21^k=2\times k+1$

So, $21^n$ for n= 1,2,3,.....n.

It always takes a form that has 2 parts :

1) $2\times k$  which will always be an even integer

2) +1 

Adding one to an even number always gives an odd integer.

Thus, $21^n$  will never generate a 2,4,6,8 at its unit's place.