Question

prove that there is no natural no. for which 4^nends with the digit zero..
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Answer 1

Answer:

any no. which have 2 and 5 as a prime factors are terminating and 4 have a power n can be expressed in 2 so it cannot be end with digit 0

Answer 2

Answer:

No

Step-by-step explanation:

We know that any positive integer ending with the digit zero is divisible by 5 and so its prime factorization must contain the prime 5

We have

4 n =2

2n

⇒ The only prime in the factorization of 4 n

is 2.

⇒ There is no other primes in the factorization of 4 n =2

2n

[By uniqueness of the Fundamental theorem of Arithmetic]

⇒ 5 does not occur in the prime factorization of 4

n

for any n.

⇒ 4

n

does not end with the digit zero for any natural n.

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