Q) Prove that 4n can 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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