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

Ans:

Let’s assume that 4n  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 4n  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 4n , therefore it will not become multiple of 10,

Hence, there is no natural number n, for which value of 4n  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 4n will have 0 at the end….

for n = 1: 4n = 41 =     4
for n = 2: 4n = 42 =   16
for n = 3: 4n= 43  =   64
for n = 4: 4n = 44 = 256

…. and so on.

Therefore, for different values of n, value of 4nends with either 4 or 6, not with 0.

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