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Monotonous integers (Posted on 2015-12-30) Difficulty: 3 of 5
For how many integers having between 1 and 10 digits (base 10) are all of their digits when read from left to right monotonically increasing? In other words, every digit is less than or equal to all of those to its right. For example, 244467889 is one of them, and 0 is another, but there are more.

See The Solution Submitted by Steve Herman    
Rating: 1.0000 (1 votes)

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Some Thoughts My mapping Comment 9 of 9 |

 -   found my write-up on the mapping -


I have finally succeeded to define the mapping of all n-digit increasing numbers into a binary word of n+9 bits.


Lemma:
Every binary string of length n+9 bits  (n ones, 9 zeroes) represents an  n-digit decimal number written in  “all leading zeroes included” notation i.e. all numbers from 0 to 10^n-1.


Take all 2 digit increasing numbers:
  00000000011…..…99 
  00000000101……89       
  00000000110……88         
  00000001001……79         
  00000001010……78         
  00000001100……77 
  ……………………
  00010010000……35         
  ……………………          
  00011000000……33         
  ……………………          
  10000001000…….07         
  ……………………           
  11000000000……..00           


The first 1 - just  write the digit equal to the number of zeroes preceding it, the second 1 transforms into   the digit equal to the first augmented  by the number of zeroes between the ones.


There are  COMB(11,2) =55 increasing numbers below 100   and  COMB(19,10) =92378 increasing numbers below 10^10.

    One can prove that   the sum for comb(8+m,8) ,  (m   from 0 to m) equals comb(9+m,m).
  
 


 


 


 


 


 



Edited on January 7, 2016, 1:34 am
  Posted by Ady TZIDON on 2016-01-06 08:30:56

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