# Thread: 8_0 eraser [tough and interesting!!!]

1. ## 8_0 eraser [tough and interesting!!!]

Start with arbitrary positive integer X(0), erasee all occurances of digit 8 and digit 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→1 22=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...

2. 1st 100 millions of the integers are verified