Question

show that 4n can never end with the digit 2 for any natural number N

Answer 1

4n = (2×2)n and it is necessary to hv 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

Answer 2

Step-by-step 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 .