Observations: consider wed+tab+pin=1**2
If we add up both alphametics we must get a number ending by 3
(13,23,33,43).
It is quite easy to understand that this number is either 23 or 33, since a sum
of 6 biggest digits is at most 39
-less than 43 , and the sum of six smallest, zero included is 15, more
than 13.
1. W+D+T+B+P+N=23 OR 33
2. 23 subdivides into 11+12
so both the 1st sum and the second will begin by 1
3. So W+D+T+B+P+N= 33
4. The subdivision must be
D+ B+N=22 and W+ T+ P=11, leaving A+E+I =10 or 11
5. Trying 1hh2 , using the
same digit and trying leading zeros nulls immediately those options, - no need
to disclaim them a priori.
6. Same thing with VOWELS A+E+I=10 cannot produce a sum and its
reversal.
7. So be D+B+N=22 and W+ T+
P=11, leaving A+E+I = 11
8. Total of WEB PANDIT IS 44 i.e. DIGIT 1 NOT USED
It boils down to solving the basic set as follows:
wed+tab+pin=1232
dew+bat+nip=2321
W+D+T+B+P+N= 33, remember the subdivision
I=0; ARBITRARY CHOICE
A+E =11.
9. Trying 2+9, 3+8, 4+7, 5+6 for A+E we get generic solutions in no time –
presented in my previous post,-
Chose two generic solutions , and remember
that each represents 864 permuted versions.
<o:p> </o:p>
So the answer to a modified version should
refer to this number: 1728<o:p></o:p>
I
honestly apologize for the condensed explanation- I am sure that whoever
tried to solve this puzzle gets a coherent
expose.
It is significantly easier to solve than to provide nicely structured
explanation.
I will sum up, how in my eyes this puzzle
should be rectified and
I
suggest you try to solve it from this text:
<begin <d3
A set of alphametics
WED + TAB + PIN = 1##2 or 2##1
DEW + BAT + NIP = 1##2 or 2##1,
in which "#" may represent any digit, has numerous
solutions.
Find , how many distinct solutions are there.
<end><o:p></o:p>
Optional
Rem: No need to solve the alphametics<o:p></o:p>
Another
wording for puzzle's question" Using each of
the distinct solutions of the above set - in how many different ways can
you code " WEB PANDIT"? <o:p></o:p>
<o:p></o:p>
Ans:
1728<o:p></o:p>
WEBPANDIT
:(SET1) &(SET2)- compare my prev post.<o:p></o:p>
Again
, excuse my shorthand style, using both capital and regular letters for the
same<o:p></o:p>
Item,
not editing for punctuation etc.<o:p></o:p>
<o:p> </o:p>
<o:p> Errrare humanum est, perseverare diabolicum.</o:p>
<o:p> </o:p>