End in 2015

We can get the solution from working from the last digit for the product $117n = p = \overline{A2015}$ . It is obvious that the last digit of $n$ must be $5$ . And the rest the digits are found as follows:

$\begin{array} {rr} & 117 \\ \times & 5 \\ \hline & 585 \\ & \color{#D61F06}{63} \space \space \\ \hline & \color{#D61F06}{1}5 \end{array} \quad \Rightarrow \quad \begin{array} {rr} & 117 \\ \times & 95 \\ \hline & 11115 \\ & \color{#D61F06}{49} \quad \\ \hline & \color{#D61F06}{0}15 \end{array} \quad \Rightarrow \quad \begin{array} {rr} & 117 \\ \times & 795 \\ \hline & 93015 \\ & \color{#D61F06}{49} \quad \space \space \\ \hline & \color{#D61F06}{2}015 \end{array} \quad \Rightarrow \quad \begin{array} {rr} & 117 \\ \times & 7795 \\ \hline & 91\color{#3D99F6}{2015} \end{array}$

$\Rightarrow n = \boxed{7795}$

$This\quad problem\quad can\quad be\quad solved\quad by\quad linear\quad congruent\quad Euqation.\\ \\ Let\quad x\quad be\quad the\quad whole\quad number\quad \\ which\quad yields\quad a\quad product\quad that\quad ends\quad in\quad 2015\quad when\quad multiplied\quad by\quad 117.\\ \\ 117x\equiv 2015\quad (mod\quad 10000)\\ By\quad Extended\quad Euclidean\quad Algorithm,\quad 1453\quad is\quad the\quad multiplicative\quad inverse\quad of\quad 117.\\ So,\quad x\equiv 2015\times 1453\quad (mod\quad 10000)\\ \quad \quad \quad x\equiv 7795\quad (mod\quad 10000)\\ \\ Therefore,\quad \boxed { 7795 } \quad is\quad the\quad smallest\quad whole\quad number\quad with\quad the\quad required\quad property.$

2015=403 . 5=13 . 31 .5 117= 13 . 9 We are looking for a number x that multiplied by 117 results in a y that end in 2015 How 117 is a multiple of 9 and 13,y must be a multiple of 9 and 13. To be a multiple of 9 the sum of digits of a number must be a multiple of 9. 2+0+1+5=8.the next multiple of 9 is 9 but we can just ad 1 to 2015 because 12015 is not a multiple of 13 and 2015 is and 10000 is not.so we need to get a y which has the sun of digits equal to 18 and the numbers added before 2015 must be a multiple of 13 .so we get in 912015.dividing it by 117 we found 7795