# How is this done?

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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name and page number of the book please. – Arjang Feb 16 '13 at 12:49
R S Agarwal Aptitude Test Book, Eg-5 in Number System – bad_keypoints Feb 16 '13 at 12:58

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.