Question

In the number of digits in a natural number N is 4, then the number of digits in N^2 will be either.​

Answer 1

Step-by-step explanation:

For the number 4

n

to end with digit zero for any natural number n, it should be divisible by 5. This means that the prime factorisation of 4

n

should contain the prime number 5.But it is not possible because 4

n

=(2)

2n

so 2 is the only prime in the factorisation of 4

n

. Since 5 is not present in the prime factorization, so there is no natural number n for which 4

n

ends with the digit zero.