Show that any number of the form 4n nen can never end with the digit 0
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➡4^n can not end with the Digit Zero because in their prime factoraisation 5 is not contain
➡4^n = (2^2)^n
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Explanation:
→ No, 4ⁿ can never end with the digit 0 for any natural number n .
→ If 4ⁿ ends with 0 then it must have 5 as a factor .
But, 4ⁿ = ( 2² )ⁿ = 2²ⁿ .
→ It shows that 2 is the only prime factor of 4ⁿ .
Also, we know from the fundamental theorem of airthematic that the prime factorisation of each number is unique .
So, 5 is not a factor of 4ⁿ .
Hence, 4ⁿ can never end with the digit 0 .
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