Find all three-digit numbers $\overline{abc}$ such that 6003 digit number $\overline{abcabcabc.....abc}$ is divisible by 91? Find all three-digit numbers $\overline{abc}$ such that $6003$ digit number $\overline{abcabcabc......abc}$ is divisible by 91?Here $\overline{abc}$ occurs $2001$ times.I know the divisibility rule for 91 which states to subtract $9$ times the last digit from the rest and, for large numbers,to form the alternating sum of blocks of three numbers from right to left. However, I am not able to see how I could apply this rule to determine the numbers $\overline{abc}$. How can I solve this?
 A: Note that:


*

*$91=7\cdot13$

*$7\mid\overline{abc\ldots abc}\iff7\mid(\overline{abc}-\overline{abc}+\overline{abc}-\ldots+\overline{abc})\iff7\mid\overline{abc}$

*$13\mid\overline{abc\ldots abc}\iff13\mid(\overline{abc}-\overline{abc}+\overline{abc}-\ldots+\overline{abc})\iff13\mid\overline{abc}$


Therefore $91\mid\overline{abc\ldots abc}\iff91\mid\overline{abc}$, which holds in either one of the following cases:


*

*$\overline{abc}=091$

*$\overline{abc}=182$

*$\overline{abc}=273$

*$\overline{abc}=364$

*$\overline{abc}=455$

*$\overline{abc}=546$

*$\overline{abc}=637$

*$\overline{abc}=728$

*$\overline{abc}=819$

*$\overline{abc}=910$

A: The given number can be written as follows,
$abc(1+10^3+10^6+\cdots+10^{6000})$
Now, $91|1001=1+10^3$ . The sum $S=1+10^3+10^6+\cdots+10^{6000}$ has $2001$ terms,  therefore, $91$ and $(1+10^3)+10^6(1+10^3)+\cdots+10^{1999}(1+10^3)+10^{6000}$ are relatively prime $\implies$ $abc$ is a multiple of 91. 
Therefore, the required numbers are $91\times n$ , where $n={2,3,4,5,6,7,8,9}$ i.e., the required numbers are :
$182,273,364,455,546,637,728,819$ and $910$
