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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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Thoughts | Comment 1 of 14
Most, but certainly not all numbers have a multiple containing a zero within their first 10 multiples.  If they all did, we would be done because could would shrink at every step.  

Even numbers when multiplied by 5 end in a zero.  Deleting this zero halves the original number.  So if the process works for all odd numbers, it works for all evens as well.

Likewise for numbers ending in 5.  They can be divided by 5.

I found solutions for all two-digit numbers ending in 1,3,7,9 (and so all two digit numbers).

Sometimes you need a higher multiple to see a zero, but multiples of this new number eventually bring it back down.  
Example: 83 (the first two steps)
83*13=1079
179*6=1074
174/2=87
87*7=609
69*3=27
27*4=108
18/2=9

One possible proof would be that all numbers do this.  Longer numbers tend to have more zeros and should be able to shrink more quickly.

  Posted by Jer on 2019-09-10 11:13:47
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