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The question here: A number when successively divided by 9, 11 and 13 ... I found the answer to it in a book and this was the answer:

The least number that satisfies the condition= 8 + (9×9) + (8×9×11) = 8 + 81 + 
792 = 881

I brute forced the solution like this when I didn't get the author's solution.

N = 9a+8.

And we have a = 11b+9, and b=13c+8 (Successively it's said in the question, no?)

So N = 99b+89 = 99(13c+8)+89 = 1287c+792+89 = 1287c+881.

So lowest value of N = 881 (c = 0)

I know I'm probably missing something really easy here, I'm not understanding the author's calculations.

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    $\begingroup$ name and page number of the book please. $\endgroup$
    – jimjim
    Feb 16, 2013 at 12:49
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    $\begingroup$ R S Agarwal Aptitude Test Book, Eg-5 in Number System $\endgroup$
    – bad_keypoints
    Feb 16, 2013 at 12:58

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Note that in your calculation:$$89=9\times9+8$$$$792=8\times9\times11$$ so you calculate successively$$N=8+9a=8+9\times9+(9\times11)b=8+9\times9+9\times11\times8+[9\times11\times13c]$$ where the last term vanishes when you put $c=0$

The successive remainders are multiplied by $1, 9, 9\times11, [9\times11\times13]$ - so you can see the pattern. The author has just used this pattern to write down the answer immediately.

NOTE: successive division in this case is not the same as modular arithmetic, where the chinese remainder theorem comes into play (eg 881 leaves remainder 1 when divided by 11). It can be thought of as a number system where the base varies by place - so in this system of 9,11,13 we could write 898 instead of the conventional 881 which would be equivalent to using 10,10,10.

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  • $\begingroup$ Ah, I see, the language is a bit imprecise, so a few folks misinterpreted it as CRT problem in the cited prior duplicate, and others ran with this interpretation. I've update the prior question. Thanks for making that clear. $\endgroup$
    – Math Gems
    Feb 16, 2013 at 17:45
  • $\begingroup$ @MathGems That's fine. The first answer I started writing used CRT until I realised that 881 obviously doesn't work mod 11. I've always wondered about mixed bases for number systems, though I've never taken the thought very far. $\endgroup$
    – Mark Bennet
    Feb 16, 2013 at 21:11
  • $\begingroup$ +1 for taking care to find the correct interpretation when so many others missed it! (to be fair, the prior question had neither the worked example nor citation, so it was easier to misinterpret). The varying bases rep is called mixed radix rep. Iirc Knuth discusses it in TaoCP 2: Seminumerical Algorithms. $\endgroup$
    – Math Gems
    Feb 16, 2013 at 21:29

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