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show that 4^n cannot end with the digit 0 for any number.
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4^n = (2×2)^n
Since, it is necessary to have 2 and 5 as a factor to end with digit 0 . And according to fundamental theorem of arithmatic its prime factorization is unique and thus it cant end with digit 0.
Since, it is necessary to have 2 and 5 as a factor to end with digit 0 . And according to fundamental theorem of arithmatic its prime factorization is unique and thus it cant end with digit 0.
Answered by
0
We have 4n where n = 1, 2, 3, 4.......
if n = 1 then 4n = 41 = 4
if n = 2 then 4n = 42 = 16 and so on
If a number ends with zero then it is divisible by 5.
Here, 4 and 16 are not dividible by 5.
Therefore, 4n can never end with zero.
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