[4" ,where "n" is a natural number.Check whe "],[" ends with the digit zero."]

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

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

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.

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 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.

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

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

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

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