Question

show that 21 power n cannot end with the digits 0,2,4,6,8 for any natural number n

Answer 1

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

Consider  $21^{1}$ = 2 x (10) + 1
                $21^{2}$ = 441 = 2 x (220) + 1
                $21^{3}$ = 9261 = 2 x (4630) + 1
                .
                .
                .
                .
                .
                $21^{n}$ =  2 ( k ) + 1

Thus  $21^{n}$ for n= 1,2,3,.....n It always takes a form that has 2 parts: - 

1) 2 x 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, In other words  $21^{n}$ will always generate an odd integer

Answer 2

Solution :-

21ⁿ can be written as (3 × 7)ⁿ

We know that for any power of 3, we cannot get even number as its unit digit or end digit. Same thing is with 7 also that we cannot get even number as its end digit.

Therefore,

(21)ⁿ cannot end with digits 0, 2, 4, 6 or 8 for any natural number n.


It can be checked with these examples:-

If n = 2

⇒ (21)² = 21*21 = 441 (End digit is 1 and not 0, 2, 4, 6 or 8)

⇒ (21)³ = 21*21*21 = 9261 (End digit is 1 and not 0, 2, 4, 6 or 8)

⇒ (21)⁴ = 21*21*21*21 = 194481 (End digit is 1 and not 0, 2, 4, 6 or 8)

⇒ (21)⁵ = 21*21*21*21*21 = 4084101 (End digit is 1 and not 0, 2, 4, 6 or 8)

⇒ (21)⁶ = 21*21*21*21*21*21 = 85766121 (End digit is 1 and not 0, 2, 4, 6 or

8)

⇒ (21)¹⁰ = 21*21*21*21*21*21*21*21*21*21 = 16679880978201 (End digit is

1 and not 0, 2, 4, 6 or 8)

Now, we can say that (21)ⁿ cannot end with the digits 0, 2, 4, 6, 8 for any natural number n.

Proved.