Question

4. Check whether 4^n where n is a natural number, can end with digit 0 for any natural number n.

Class 10

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Answer 1

Heya!

Here is yr answer.....


Given,

4^n , where n is any natural no.

If any no. is going to end with '0' then the no. is divisible by 2 and 5.

Therefore, in its prime factorization 2 and 5 should occur.

But, 4^n = (2²)^n = 2^2n


Here, prime factor 5 is not there.

Hence, 4^n never ends with '0' for any natural no.


Hope it hlpz...

Answer 2

we know that any positive integer ending with zero is divisible by 5 and so it's prime factorisation must contain the prime number 5
There is no another primes in the factorisation of 4^n 5 doesn't occur in the prime factorisation of 4^n.so 4^n does not end with digit 0 for any natural number n