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p(Ascending Number) closest to .01 (Posted on 2020-03-16) Difficulty: 2 of 5
An Ascending Number has been defined by the following 2 properties:

- The number has at least 2 digits.

- All digits of the number are in strictly ascending order, in base 10.

For completeness, let's also add: no leading zeros or duplicate digits.

For what value of N is the probability that a randomly chosen N-digit number is also an ascending number closest to .01, i.e. a 1% chance?

For this value of N, what is the exact probability the random number N digits long is ascending?

No Solution Yet Submitted by Larry    
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Solution Solution | Comment 1 of 3
Start with N=2.  There are 90 two digit numbers.  Consider the last digit (L) of each, and how often that last digit could be part of an ascending number (A):
L: 0 1 2 3 4 5 6 7 8 9
A: 0 0 1 2 3 4 5 6 7 8 
A total of 36 ascending numbers, so probability = 36/90 = 40%.
Now N=3.  There are 900 of these.  Once again consider the last digit of each, and realize that we already counted the possible sets of preceding two digits above, i.e. the number of three-digit ascending numbers ending in X is the total of all two-digit ascending numbers ending in a digit < X.  Therefore for N=3 we get:
L: 0 1 2 3 4 5  6  7  8  9
A: 0 0 0 1 3 6 10 15 21 28
For a total of 84 ascending numbers.  84/900 = 9.33%.  Closer but not there yet. 
Similar logic applies as N increases.  The number of four-digit ascending numbers ending in X is the number of three-digit ascending numbers ending < X.  So for N=4:
L: 0 1 2 3 4 5  6  7  8  9
A: 0 0 0 0 1 4 10 20 35 56
That makes 126/9000 = 1.4%.  Clearly N=5 and above will be far less than 1%, so this is the solution.  


  Posted by tomarken on 2020-03-16 13:24:04
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