Re: Mini-Challenge: Zip Code Message #18 Posted by Thomas Klemm on 27 Sept 2010, 2:59 p.m., in response to message #1 by Thomas Klemm
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The following challenge is based on an arithmetic problem I found in a mathematics book for elementary school.
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I will add my solution for the HP-15C here within a couple of days.
Initialization:
10 STO 0
001 - 42,21,11 LBL A 022 - 30 -
002 - 32 0 GSB 0 023 - 9 9
003 - 32 0 GSB 0 024 - 0 0
004 - 32 0 GSB 0 025 - 20 ×
005 - 43,30, 7 TEST 7 026 - 34 x<>y
006 - 34 x<>y 027 - 43 33 R-^
007 - 33 R-v 028 - 30 -
008 - 43,30, 7 TEST 7 029 - 9 9
009 - 34 x<>y 030 - 9 9
010 - 33 R-v 031 - 9 9
011 - 43,30, 7 TEST 7 032 - 20 ×
012 - 34 x<>y 033 - 40 +
013 - 43 33 R-^ 034 - 43 32 RTN
014 - 43,30, 7 TEST 7 035 - 42,21, 0 LBL 0
015 - 34 x<>y 036 - 45,10, 0 RCL ÷ 0
016 - 43 33 R-^ 037 - 43 44 INT
017 - 43,30, 7 TEST 7 038 - 43 36 LASTx
018 - 34 x<>y 039 - 42 44 FRAC
019 - 33 R-v 040 - 45,20, 0 RCL × 0
020 - 43,30, 7 TEST 7 041 - 34 x<>y
021 - 34 x<>y 042 - 43 32 RTN
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What is the behavior if you iterate this? Make a speculation.
Most numbers will end up with 6174. Only numbers composed of the same digit will end up with 0.
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Try to proof your assumption.
My observation was the same as described in Mysterious number 6174. (cf. "Which way to 6174?")
So I guess I don't have to repeat that proof here. However I didn't ignore possible duplicates of the 55 numbers that remain after the first iteration.
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Can your calculator be of any help?
Here's a small program that produces the 55 numbers left after the first iteration:
Initialization:
9 STO 1
EEX 3 / STO 2
043 - 42,21,12 LBL B 055 - 42, 6, 2 ISG 2
044 - 45 1 RCL 1 056 - 43 32 RTN
045 - 9 9 057 - 42, 5, 1 DSE 1
046 - 9 9 058 - 43 20 x=0
047 - 9 9 059 - 43 32 RTN
048 - 20 × 060 - 45 1 RCL 1
049 - 45 2 RCL 2 061 - 26 EEX
050 - 43 44 INT 062 - 3 3
051 - 9 9 063 - 10 ÷
052 - 0 0 064 - 44 2 STO 2
053 - 20 × 065 - 33 R-v
054 - 40 + 066 - 43 32 RTN
Now you can switch between the two programs A and B:
B: 8991
A: 8082
A: 8532
A: 6174
B: 9081
A: 9621
A: 8352
A: 6174
(...)
B: 0
A: 0
The data was then used to create this graph.
There's a direct proof in the same article as well. (cf. "Only 6174?")
Again the HP-15C can be used to solve the system of linear equations:
| 1 0 1 -1 | | a | | 10 |
| | | | | |
| 1 1 -1 0 | | b | | 9 |
| | | | = | |
| 0 1 -1 -1 | | c | | 1 |
| | | | | |
| 1 -1 0 -1 | | d | | 0 |
The solution is: [ a b c d ] = [ 7 6 4 1 ]
I must admit that I wasn't aware of Kaprekar's constant. I just saw this arithmetic problem in that book which made me wonder why. So thanks a lot to Paul Dale for pointing this out.
I know it wasn't a hard problem but still an occasion to use your best loved machines. Nevertheless I hope you enjoyed the contest.
Cheers and thanks for your contributions
Thomas
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PS: And where will all this lead us?
6174 Sörenberg
Edited: 27 Sept 2010, 3:43 p.m.
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