Question

Can the number 4^n n being a natural number end with the digit 0

Answer 1

Answer: No

Step-by-step explanation:

To prove this is wrong , first consider that the case is correct. Hence for some value of n being a natural number, $4^{n}$ has a value that ends with 0. In that case , the factors of the number must be an integral multiple of 4. In simple terms , all the factors of that value must be only 4.

But , let us see if 4 satisfies that case.

$4^{1}$ = 4

4² = 16

4³ = 64

$4^{4}$ = 256

We can find that the digits are repeating, so even if this series go on to infinity, we won't have a number such that $4^{n}$ end with 0.

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