In the number of digits in a natural number N is 4, then the number of

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

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

Answer»

For the NUMBER 4

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

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

=(2)

so 2 is the only prime in the factorisation of 4

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

ends with the digit zero.