**Q) **Prove that 4^{n }can never end with digit 0, where n is a natural number.

**Ans: **

Let’s assume that 4^{n } ends with 0.

Since now it ends with zero, it is multiple of 10 and hence, it must be divisible by 2 and 5 both.

This clearly means, that the factors of 4^{n } should include 2 and 5 both.

But since 4 = 2 x 2 has only 2 is included as its factor, not 5.

Now , since 5 is not the factor of 4^{n }, therefore it will not become multiple of 10,

**Hence, there is no natural number n, for which value of 4 ^{n } will end with 0.**

**Check:**

*(On rough work, you can check this number as below.. This is only for your understanding, DO NOT submit below working as solution).*

*Check the possible value of n for which value of 4 ^{n} will have 0 at the end….*

*for n = 1: 4 ^{n} = 4^{1} = 4*

*for n = 2: 4*

^{n}= 4^{2}= 16*for n = 3: 4*

^{n}= 4^{3}= 64*for n = 4: 4*

^{n}= 4^{4}= 256…. and so on.

*Therefore, for different values of n, value of 4 ^{n}ends with either 4 or 6, not with 0.*

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