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Reduced to single digit (Posted on 2019-09-06) Difficulty: 3 of 5
Suppose that we have two operations that we can perform on an integer:

Multiply it by any positive integer.
Delete the 0's in its decimal representation.

Beginning with any positive integer can we always obtain a single-digit number after a finite number of operations? For example, beginning with 7, we can multiply by 15 to obtain 105, delete the 0 to get 15, multiply by 2 to get 30, then delete the 0 to end with 3.

No Solution Yet Submitted by Danish Ahmed Khan    
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more thoughts | Comment 5 of 14 |
I have been thinking about numbers of the form 99, 999, 9999, etc.

Here is a contention: 

When such number is multiplied by any integer i that is n digits long
yielding a product j, j will _not_ have either of these properties:

1) j, a number that contains n embedded "0"s

2) j, number that contains n-1 embedded "0"s and, when they are stripped-out, is even, and when this is divided by 2, will be less than i.

e.g. 9999 x 9999 has 3 embedded "0"'s and comes out odd. 

I think this is true of all the other "nut" numbers I found (see below) like 11, 33, etc. I implied this below, but now I am wondering why this is so and how does such a number become predictably multiplied into a "non-nut" number? 

Edited on September 12, 2019, 2:02 am
  Posted by Steven Lord on 2019-09-12 01:02:32

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