Write out the positive integers consisting of only $1$s, $6$s, and $9$s in ascending order as in: $1,6,9,11,16,\dots$. a. Find the order of $1996$ in the sequence. b. Find the $1996$th term in the sequence.
Problem
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Tags: geometry, geometric transformation
18.01.2013 16:39
a)
b)
08.10.2024 22:08
$\textbf{a)}$ Answer: $\boxed{65}$ Let the number of $k$-digit numbers be $n_k$. We obtain the relation $n_1=3$, $n_k=3n_{k-1}$,$n_k=3^k$. Let $M_k$ represent the number of numbers with up to $k$ digits. $M_1 = 3, \quad M_2 = 3^2 + 3 = 12, \quad M_3 = 3^3 + 12 = 39.$ The number 1999 is the 66th number, which is given by $M_3 + n_3 = 39 + 27 = 66.$ Therefore, 1996 is the 65th number. $\textbf{b)}$ Answer: $\boxed{6191661}$ Since: $M_6 = 3^6 + 3^5 + 3^4 + 3^3 + 3^2 + 3 = \frac{3^7 - 3}{3 - 1} = 1092$, As $1996-1092=904$, we will find the 904th 7-digit number. At this stage, let's make the following transformations: $1\to 0$, $6\to 1$, $9\to 2$. We are trying to find the 904th element in the sequence: $0000000$, $0000001$, $0000002$, $0000010$, $\ldots$, $2222222$, $\ldots$ This is the number $(903)_{10}=(1020110)_3$. Reversing the transformation, we get $1020110 \to 6191661$.