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

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

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

If n is any natural number, then 9^(n)-5^(n) ends with

Assertion (A): 6^(n) ends with the digit zero where n is a natural number. Reason (R):Any number ends with digit zero, if its prime factor is of the form 2^(m)xx5^(n) , where m,n are natural numbers.

prove that 2nlt(n+2)! for all natural numbers n.

Can the number 6^(n),n being a natural number, end with the digit 5? Give reason.

If 6^n - 5^n always end with a digit, where n is any natural number, then the desired digit is

Prove that the sum of first n natural numbers (n>2) cannot be prime.