Starts in 2015 and Ends in 2015

$This\quad problem\quad can\quad be\quad solved\quad by\quad linear\quad congruent\quad Euqation.\\ \\ Let\quad x\quad be\quad the\quad whole\quad number\quad \\ which\quad yields\quad a\quad product\quad that\quad starts\quad and\quad ends\quad in\quad 2015\quad when\quad multiplied\quad by\quad 935.\\ \\ Step\quad 1)\quad Find\quad the\quad collection\quad of\quad x\quad that\quad ends\quad in\quad 2015\quad when\quad multiplied\quad by\quad 935\\ \\ 935x\equiv 2015\quad (mod\quad 10000)\\ 187x\equiv 403\quad (mod\quad 2000)\\ x\equiv 569\quad (mod\quad 2000)\\ \\ Therefore,\quad x=\quad 569\quad +2000n,\quad n\in \Re \\ \\ Step\quad 2)\quad Find\quad the\quad certain\quad n\quad that\quad make\quad the\quad product\quad starts\quad at\quad 2015\\ \\ First,\quad the\quad product\quad is\quad in\quad from\quad of\quad \\ 935(569+2000n)\\ =\quad 532015\quad +1870000n\\ =(53+187n)\times { 10 }^{ 4 }+2015\\ =\overline { (53+187n)2015 } \\ \\ Case\quad 1)\quad 53+187n=2015\quad \\ Impossible\quad because\quad n\quad is\quad not\quad an\quad integer.\\ \\ Case\quad 2)\quad 53+187n=\overline { 2015a } \\ \\ 53+187n=20150+a\\ 53\equiv 20150+a\quad (mod\quad 187)\\ a\equiv 99\quad (mod\quad 187)\\ Impossible\quad because\quad a\quad is\quad a\quad sigle-dight\quad number.\\ \\ Case\quad 3)\quad 53+187n=\overline { 2015bc } \\ \\ 53+187n=201500+\overline { bc } \\ 53\equiv 201500+\overline { bc } \quad (mod\quad 187)\\ \overline { bc } \equiv 139\quad (mod\quad 187)\\ Impossible\quad because\quad \overline { bc } \quad is\quad a\quad 2-dight\quad number.\\ \\ Case\quad 4)\quad 53+187n=\overline { 2015def } \\ \\ 53+187n=2015000+\overline { def } \\ 53\equiv 2015000+\overline { def } \quad (mod\quad 187)\\ \overline { def } \equiv 165\quad (mod\quad 187)\\ \\ Yes,\quad we\quad found\quad it!\quad So,\quad the\quad smallest\quad \overline { def } \quad is\quad 165.\\ \\ Therefore,\quad the\quad product\quad is\quad \boxed { 20151652015 } \quad and\quad the\quad corresponding\quad x\quad is\quad \boxed { 21552569 } .$