Question

check whether 6n can end with the digit 0 for any natural number n.
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Answer 1

Given:

6^{n}

To find:

Check whether 6^{n} can end with the digit 0 for any natural number n

Solution :

To find whether the number 6^{n} ends with a digit 0 for any natural number n we first find the factorization of the value 6^{n} .

Now we know that the factors for 6=2 \times 3.

Putting 6^{n} we get 6^{n}=2^{n} \times 3^{n}

Thereby putting values like positive natural numbers in n = 1, 2, 3….. n but to get a 0 at the end the number has to be either multiplied by 0 or by 5, now the value of 2^{n} will never end with 0 and the value of 3^{n} can never end with 0 as well, the unit value of 2^{n} will always be even number except zero and unit value of 3^{n} will be 3, 9, 7, 1 thereby multiplying both factors will never have zero at the unit place of a

Answer 2

Step-by-step explanation:

if if the number 6 n, for any n is to end with 0 it should be divisible by 10 that is prime factorization of 6n should contain prime number 2 and 5

this this is not possible because 6n =(2*3)n , so the only prime factorization of 6n is 2 and 3

so the uniqueness of fundamental theorem of arithmetic guarantees that there is no other prime in factorization of 6n

so., there is no natural no. n for which 6n ends with 0