Question

show that the number 4 power n Marion is a natural number cannot end with the digit 0 for any natural number N​

Answer 1

Proof by Contradiction :

Consider 4^n as the nth power of 4., where n ∈ N

Now , WE know that, 4 = 2^2,

=> 4^n = 2^(2n)

Since, (a^m)^n = a^(mn).

Now,

Let us assume 4^n ends with 0 for any one n ∈ N

If it ends with 0, Then it must be divisible by 2 and 5

To check it, Consider prime factorization of 4^n,

Which is,

2*2*2*2*2*2*2......................*2(For 2n times)

Clearly, it is divisible by 2, But

2 and 5 both being primes, 5 won't divide 2,

So, 5 doesn't divide 4^n => 10 doesn't divide 4^n => 4^n doesn't end with 0 for any n∈N.

Therefore, 4^n doesn't end with 0.

Proof by Symmetrical Repetition :

Since n ∈ N , Consider 4 powers,

They are : 4, 16 , 64 , 256 ..........

Clearly after every 2 numbers we get 4 as unit digit,

=> The only last digits are always 4 and 6, By using this symmetrical property we can clearly say never in 4^n where n ∈ N , Occur 0 as a unit digit.

Therefore once again, 4^n doesn't end with 0.

Hope you understand, Any query? Comment down,

Thanking you, Bunti 360 !