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8_0 eraser [tough and interesting!!!]
04-06-2010, 06:25 PM
Post: #1
8_0 eraser [tough and interesting!!!]
Start with arbitrary positive integer X(0), erasee all occurances of digit 8 and 0 in decimal form of 8X.
If all digits of 8X are gone, we are done, otherwise we obtain a positive integer X(1) by the remaining digits in their original order. We call it X(1) and we can do the same thing to X(1) as we did for X(0) and possibly can obtain X(2) that way...

Prove that for any initial X(0), the above procedure can only produce an finite sequence. In other words, all digits will be erased within finite iterations.

For example, let X(0) = 2, we have

X(0) → 16 = X(1) → 128 → 12 = X(2) → 96 = X(3) → 768 → 76 = X(4) → 608 → 6 = X(5)→48→4 = X(6)→32 = X(7)→256 = X(8)→2048→24 = X(9)→192 = X(10)→1536 = X(11)→12288→122 = X(12)→976 = X(13)→7808→7 = X(14)→56 = X(15)→448→44 = X(16)→352 = X(17)→2816→216 = X(18)→1728→172 = X(19)→1376 = X(20)→11008→11 = X(21)→88→※

You then see the problem is not trivial!! And actually I'm not sure the statement is ture. Maybe start with some big number, it'll get into some cyclic iteration chains...
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We can re-state the problem in more mathematical way:
Define \(f_d: \mathbf{N} \to \mathbf{N}\) as: \(f_d(n)\) is \(0\) if the decimal representation of \(nd\) contains only digits \(0\) or \(d\), otherwise \(f_d(n)\) equals the integer obtained by remove all digits \(0\) and \(d\) from the decimal representation of \(nd\)

Now the problems is to prove that \(\forall n \in \mathbf{N}\quad \exists N(n) \in \mathbf{N} (f_8^{N(n)}(n) = 0)\)
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04-08-2010, 02:18 PM
Post: #2
RE: 8_0 eraser [tough and interesting!!!]
Remarks
(1) Not all digits have the same behavior.
\(\quad\) (a) If we replace 8 by 2 or 5, we can easily show that the similar statements are ture.
\(\quad\) (b) If we replace 8 with 3, we get \(f_3^3(15)=15\)
(2) Let \(A=\left\{1,2,3,4,5,6,7,9\right \}=\left\{0,1,2,3,4,5,6,7,8,9\right \}\setminus \left\{0,8\right \}\), then \(f_8 (A^*) = A^*\)
\(\quad\)where \(A^* = \left\{\overline{a_1\cdots a_m}=\sum_{k=1}^m 10^{m-k}a_k \quad|\quad a_k \in A, \quad m \in \mathbf{N}\right\} \)
\(\quad\)thus we have \(\forall m \in \mathbf{N} , \exists n\in \mathbf{N} \left(\left|\left\{f^k(n) |f^k(n) \neq 0, k\in \mathbf{N}\right\}\right| = m\right) \)
\(\quad\) This means that there exists non-duplicating sequence \(f(n),f^2(n),\cdots,f^m(n)\) of any length \(m\)
\(\quad\) Proof. Let \((A_1,A_2,A_3,A_4,A_5,A_6,A_7,A_9)=(135,35,375,5,735,75,975,1135)\)
\(\quad\) then for any \(a_i \in A\) we have \( f_8(\overline{A_1\cdots A_k})= \overline{a_1\cdots a_k}\)
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