Saved Bookmarks
| 1. |
Show that 12n cannot end with digit 0 or 5 for any natural n |
| Answer» 12 = 22{tex}\\times{/tex} 3{tex}\\therefore 12^n=(2^2 \\times3)^n=(2^2)^n \\times3^n{/tex}\xa0So, only primes in the factorisation of 12n are 2 and 3 and, not 5.Hence, 12n cannot end with digit 0 or 5. | |