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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Part 2: Guess? Spoiler? | Comment 3 of 8 |
3 digits  D(88) = 109
4 digits  D(978) = 1099
5 digits  D(9878) = 10099
6 digits  D(99778) = 100999

OK, so it's not a total guess, but I haven't checked this answer.  (Sorry, no time)
Consider the 6 digit case.
999001 - 100999 seems (without too much thought) to be the biggest increase possible with 6 digits, and (999001 - 100999)/9 = 99778.

  Posted by Steve Herman on 2010-09-18 12:07:42
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