1
$\begingroup$

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?

$\endgroup$
3
  • 2
    $\begingroup$ That's true for every combination $\overline{abc}$ which is divisible by $91$, since $\overline{abc}-\overline{abc}+\overline{abc}-\dots+\overline{abc}=\overline{abc}$. $\endgroup$ – barak manos Jan 18 '17 at 9:09
  • $\begingroup$ Why did you add an overline to $abc$? $\endgroup$ – mvw Jan 18 '17 at 11:00
  • $\begingroup$ @mvw: To emphasize that we're dealing with a concatenation of these digits, and not with their product (which is $a\cdot b\cdot c$). $\endgroup$ – barak manos Jan 18 '17 at 12:57
4
$\begingroup$

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:

  • $\overline{abc}=091$
  • $\overline{abc}=182$
  • $\overline{abc}=273$
  • $\overline{abc}=364$
  • $\overline{abc}=455$
  • $\overline{abc}=546$
  • $\overline{abc}=637$
  • $\overline{abc}=728$
  • $\overline{abc}=819$
  • $\overline{abc}=910$
$\endgroup$
3
  • $\begingroup$ How did you arrive at those alternating sums? $\endgroup$ – mvw Jan 18 '17 at 12:02
  • 1
    $\begingroup$ @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: $n\mid1\underbrace{0\dots0}_{k\text{ times}}1\iff n\mid\text{ the alternating sum, where the length of each term is }k+1\text{ digits}$. $\endgroup$ – barak manos Jan 18 '17 at 12:47
  • 1
    $\begingroup$ @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. $\endgroup$ – barak manos Jan 18 '17 at 12:53
3
$\begingroup$

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 $910$

$\endgroup$
8
  • $\begingroup$ You forgot $091$ $\endgroup$ – barak manos Jan 18 '17 at 9:17
  • $\begingroup$ With a heavy heart, I have made an edit :) $\endgroup$ – user356774 Jan 18 '17 at 9:21
  • $\begingroup$ 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. $\endgroup$ – barak manos Jan 18 '17 at 9:21
  • 1
    $\begingroup$ I think we should use the normal representation without leading zero digits, if nothing else is specified. $\endgroup$ – mvw Jan 18 '17 at 12:00
  • 1
    $\begingroup$ @barak manos : I think I should go with my original answer. $\endgroup$ – user356774 Jan 18 '17 at 12:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.