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Upon Reflection (Posted on 2013-11-22) Difficulty: 3 of 5
The Alphametics:
WED + TAB + PIN = X j k Y
DEW + BAT + NIP = Y k j X
..... when considered concurrently have 864 solutions.

In the totals X=1 and Y=2 or X=2 and Y=1, and j and k are distinct integers represented in the set of variables.

Why can X and Y never equate to W, T, P, D, B or N? The digits 0-9 are all available but there are no leading zeroes.

See The Solution Submitted by brianjn    
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Some Thoughts re(3): Disappointed! - I AM HAPPY- now resolved | Comment 7 of 13 |
(In reply to re(2): Disappointed! - I AM by brianjn)

I am very pleased that meanwhile I found the solution which in 

my opinion provides the explanation to what has happened.

I would not call it a " slanging" match but rather presenting two
views, one based on trying to solve the problem, the other not
addressing the puzzle and referring to unrelated history and
plainly suggesting " Forget it!"

Seeing , how deeply Brian is convinced that the puzzle is impeccable, I therefore assumed that it either comes from reliable source (and perhaps something was lost or modified) or was created by brianjn, fully debugged and helas worded ambiguously (imho) while being absolutely true if interpreted as Charlie found only after unrestricted computer search.

I went back to my process of solution revised it and quickly found the following:
1. if we take the XandY
wording meaning" both not,but one of them yes" and the 1st statement as  1**2 ; and discount  permutations there is only one set of letters to be used i.e. 023456789, = all except 1.
2. Showing that would constitute an answer.

I  also found a solution, proved that 1232/2321 is the only couple eligible to represent the sums, checked the # of possible ways to permute this particular solution got 864 ways,( with AIE=(803)) and was about to post it and then saw Ch's generic solution with AIE =(704) and a quoted count 864.
- Thus my perception was that something is wrong in the text.

The truth is that there are  6*6*6*2=432 permutations for each generic solution (The 1st line with 1232)  and not 864 as I wrongly counted inspired by the data given and Ch's sample.
The permutation for a given AIE are:
6 for the order of the sumands
6 for the order of WTP
6 for the order of DBN
2 for the order two of the vowels(AIE) , one must be zero

432 per each  generic solution , 864 in all as stated.

I erroneously "counted twice" (added interchanging 1232 with 2321, which clearly is counted within the letter permutations)- the rest is history.

My conclusions ( I  must add a disclaimer - it represents only my views):
1. It is a very nice puzzle
2. It would be much better to ask : prove that a digit 1 cannot appear in the solution set.  Same goal, no ambiguity
3, "No leading zeroes". is redundant, it shows immediately
4. Same for "0-9 are available". It is a tautology.
4. The number 864, although true is  optional(i.e. redundant, not needed to solve).
5. Last , but not the least : I did nothing wrong.
    I was discouraged to present my reservations on the board, which will not preclude my further active involvement in the future.
When I quoted Seneca's  "Errare humanum est" I was fully aware that it is also applicable to my humble self.

REM (2015: reread & fixed typo :Last , but not the least )

Edited on July 18, 2015, 2:51 am
  Posted by Ady TZIDON on 2013-11-25 05:10:50

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