Check whether `6^n` can end with the digit '0' for any natural numbers n.

Show that 12^(n) cannot end with the digit 0 or 5 for any natural number ‘n’.

Show that 3^nxx4^m cannot end with the digit 0 or 5 for any natural numbers 'n' and 'm'.

6^n can not end with……….

If the mean of first n natural numbers is 11 then n =

Mode of first 'n' natural numbers is ........

Write the negation of the following statements and check whether the resulting statements are true, Every natural number is greater than 0.

Sum of first 'n' natural numbers is, ________

Prove that 6^(2n)-35n-1 is divisible by 1225 for all natural numbers of n.

Consider the number 4^n where n is a natural number. Check whether there is any value of n which 4^n ends with the digit zero?

Is the units digit in the value of 6^n where 'n' is a natural number is zero? Justify your answer.