# 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?

• That's true for every combination $\overline{abc}$ which is divisible by $91$, since $\overline{abc}-\overline{abc}+\overline{abc}-\dots+\overline{abc}=\overline{abc}$. – barak manos Jan 18 '17 at 9:09
• Why did you add an overline to $abc$? – mvw Jan 18 '17 at 11:00
• @mvw: To emphasize that we're dealing with a concatenation of these digits, and not with their product (which is $a\cdot b\cdot c$). – barak manos Jan 18 '17 at 12:57

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:

• $\overline{abc}=091$
• $\overline{abc}=182$
• $\overline{abc}=273$
• $\overline{abc}=364$
• $\overline{abc}=455$
• $\overline{abc}=546$
• $\overline{abc}=637$
• $\overline{abc}=728$
• $\overline{abc}=819$
• $\overline{abc}=910$
• How did you arrive at those alternating sums? – mvw Jan 18 '17 at 12:02
• @mvw: It's a known "divisibility thumb rule" for $7$, $11$ and $13$. The reason behind this is that each one of them divides $1001$. The more general rule is: $n\mid1\underbrace{0\dots0}_{k\text{ times}}1\iff n\mid\text{ the alternating sum, where the length of each term is }k+1\text{ digits}$. – barak manos Jan 18 '17 at 12:47
• @mvw: For example: $17$ divides $100000001$, therefore $17$ divides $k$ if and only if it divides the alternating digit-sum of $k$, where the length of each term is $8$ digits. – barak manos Jan 18 '17 at 12:53

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 $910$

• You forgot $091$ – barak manos Jan 18 '17 at 9:17
• With a heavy heart, I have made an edit :) – user356774 Jan 18 '17 at 9:21
• Well, it really depends on OP's definition. Indeed, with $\overline{abc}=091$, the number $\overline{abc\ldots abc}$ will consist of $6002$ digits, not $6003$... unless we allow leading zeros... So like I said, it's up to the definition within the question. But I think that the intention is clear enough whether or not we state that $091$ is a solution. – barak manos Jan 18 '17 at 9:21
• I think we should use the normal representation without leading zero digits, if nothing else is specified. – mvw Jan 18 '17 at 12:00
• @barak manos : I think I should go with my original answer. – user356774 Jan 18 '17 at 12:42