Number made up of 1's

First, in order for f(n) to be divisible by 9, n has to be multiple of 9 by the divisibility rule. Now, rewrite f(n) as following: $f(n)=111...11=\frac { 999...99 }{ 9 } =\frac { { 10 }^{ n }-1 }{ 9 }$ Since f(n) also has to be divisible by 13 and 17, $f(n)=\frac { { 10 }^{ n }-1 }{ 9 } \equiv 0\quad (mod\quad 13)\rightarrow { 10 }^{ n }\equiv 1\quad (mod\quad 13)\\ Now,\quad the\quad Fermat's\quad Little\quad Theorem\quad states\quad that\\ { a }^{ p }\equiv a\quad (mod\quad p)\quad or\quad { a }^{ p-1 }\equiv 1\quad (mod\quad p)\quad where\quad p\quad is\quad prime.\\ Since\quad 13\quad is\quad prime,\quad { 10 }^{ 13-1 }\equiv 1\quad (mod\quad 13).\\ \therefore \quad n\quad has\quad to\quad be\quad a\quad multiple\quad of\quad 12\\ Similarly,\quad since\quad 17\quad is\quad also\quad prime,\quad { 10 }^{ 17-1 }\equiv 1\quad (mod\quad 17)\\ \therefore \quad n\quad has\quad to\quad be\quad a\quad multiple\quad of\quad 16$

In conclusion, n has to be a multiple of 9, 12, and 16. Therefore, the minimum value of n is the least common multiple of 9, 12, and 16, which is 144.