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Consider a number system which does not have the number 7 but has all other numbers. So the numbers are $1,2,3,4,5,6,8,9,10,11,12,13,14,15,16,18,...$. I want to find what is the $10^k$ number, where $k$ is an integer. For example $10^{th}$ number is $11$.

My try: I found that $10^k$ number is $11^k$ if $k \le5$. But for $k=6,11^6=1771561$ which contains $7$. So I am unable to solve this problem. If anyone can help it would be great. Thanks.

happymath
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    You are effectively working base $9$ - do you know how to convert from one base to another? – Mark Bennet Jun 18 '14 at 09:28
  • @MarkBennet yes I know how to convert from one base to another but I am sorry that I am unable to see the connection. So can you please elaborate – happymath Jun 18 '14 at 09:30

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Write your number as $a_0+a_1\cdot9^1+a_2\cdot9^2+a_3\cdot9^3+a_4\cdot9^4+a_5\cdot9^5+ \dots $ with $0\le a_i\lt 9$

Then if $0\le a_i\le 6$ set $b_i=a_i$ and for $7\le a_i\le 8$ set $b_i=a_i+1$. Then the expression you are looking for is $$b_rb_{r-1}\dots b_0$$

Mark Bennet
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  • I still do not get it,can you please outline the proof – happymath Jun 18 '14 at 10:05
  • @happymath You hev nine symbols. The $a_i$ are standard base $9$ digits $0-8$. The $b_i$ avoid the digit $7$. Have you yet tried converting your number to base $9$? – Mark Bennet Jun 18 '14 at 10:12