Zero is a natural number.

Consider the numbers 4^n , where n is a natural number. Cheek whether there is any value of n for which 4^n ends with the digit zero.

A positive interber is a natural number also.

Prove that there is no natural number for which 4^n ends with the digit zero.

If n=2^3xx3^4xx5^4xx7 , then the number of consecutive zeros in n , where n is a natural number, is (a) 2 (b) 3 (c) 4 (d) 7

The number of natural numbers n for which (15n^2+8n+6)/n is a natural number is:

If the mean of the squares of first n natural numbers is 105 , then find the median of the first n natural numbers .

The variance of first n natural number is:

If n is a natural number, then sqrt(n) is:

The standard deviation for first 10 natural number is

The S.D. of the first n natural numbers is