Show that any number of the form 4^n, n∈N can never end with the digit 0.
(Class 10 Maths Sample Question Paper)
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Given, n is a natural number.
Let 4ⁿ ends with 0, thus 4ⁿ is divisible by 2 and 5. But the prime factors of 4 are 2×2 .
4ⁿ = (2×2)ⁿ = 2²ⁿ
Thus, prime factorization of 4ⁿ does not contain 5.Show the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 4ⁿ.
Hence, there is no natural number n for which 4ⁿ ends with digit 0.
HOPE THIS WILL HELP YOU...
Let 4ⁿ ends with 0, thus 4ⁿ is divisible by 2 and 5. But the prime factors of 4 are 2×2 .
4ⁿ = (2×2)ⁿ = 2²ⁿ
Thus, prime factorization of 4ⁿ does not contain 5.Show the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 4ⁿ.
Hence, there is no natural number n for which 4ⁿ ends with digit 0.
HOPE THIS WILL HELP YOU...
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