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


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.

(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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