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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 4: More guessing. Spoiler? | Comment 7 of 8 |
Infinite examples of D(n) < 9n

109= D(89) < 9*89
1099 = D(979) < 9*979
10099 = D(9889) < 9*9889
100999 = D(99789) < 9*99789
etc.

Infinite examples of D(n) > 9n

120 = D(10) > 9*10
1200 = D(100) > 9*100
12000 = D(1000) > 9*1000
120000 = D(10000) > 9*10000
etc.

And, unless I miss my guess (as long as I'm guessing),
10, 100, 1000 look like the smallest n for which D(n) have 3, 4, 5, ... digits respectively 

Edited on September 18, 2010, 4:16 pm
  Posted by Steve Herman on 2010-09-18 15:31:29

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