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Which multiple of 9? (Posted on 2010-09-17) Difficulty: 3 of 5
It is a well known fact that if you permute the digits of a number the difference will be a multiple of 9.

Define the sequence D, where D(n) is the smallest positive value that can be increased by 9n through a permutation of its digits. No leading zeroes are allowed so the first term is D(1)=12 not 10

1) Find the next 14 terms of D.

2) Note D(8) is the greatest n with two digits. What is the greatest n with 3, 4, 5, ... digits?

3) There are some numbers a, b such that a≠b but D(a)=D(b). Prove there are infinitely many such pairs.

4) Sometimes D(n)>9n and sometimes D(n)<9n. Prove that both cases happen an infinity of times.
5) Are there any values of n such that D(n)=9n?

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

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Some Thoughts Part 2: Revised guess. Spoiler? | Comment 5 of 8 |
Oh, right, Charlie.  Thanks.

999001 - 100999 is not the biggest possible increase by rearranging 6 digits.

999100 - 100999 = 9*99789 is bigger

So, revised guesses (with a little more confidence) are:

3 digits  D(    89) =       109
4 digits  D(  979) =      1099
5 digits  D( 9889) =   10099
6 digits  D(99789) = 100999

  Posted by Steve Herman on 2010-09-18 15:07:23
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