Saved Bookmarks
| 1. |
Show that 12^n constant end with digit 0 for any natural number n |
| Answer» If a number ends with zero then it would be divisible by 5 that is it\'s prime factorization must be 2*5. But this is not possible with 12n because its prime factorization is 2*2*3.It cannot be end with zero because its prime factors don\'t contain pair of 2*5 | |