Remember
“Unique and restricted” ? ,b (pid=13696)
There I have asked for a restricted answer to an alphametic puzzle and got a set of many words.
Now I have fiddled with a similar equation and again will allow only answers not using any of the letters appearing in “TWELVE”.
TWELVE + TWELVE = (Oompha, grubby, payoff, droppy ….et al)
Your task is to find an answer to my puzzle such that adding the numerical values of all 6 letters in the word chosen by you (a long list of candidate solutions) will be closest to 24.
Start your chase.
Good luck!
Summary: no solution when pod = 24
I have scanned, but not fully read the most recent several posts, and I realized that I was placing a restriction such that all 10 digits must be used. The Other Word need not have 5 distinct letters, it might have fewer. I was working with letter patterns and number patterns but had not done a number search including all the restrictions.
So I went through all the possible numbers, N, that might represent the Other Word including all restrictions:
definitions: N is the number on the RHS and H = N/2 = TWELVE
N is even;
sod(N) = 24
N has less than 6 distinct digits (TWELVE is already using 5 letters);
H has 5 distinct digits with the 3rd and 6th digit the same;
None of the digits in H appear in N.
There are only two such potential solutions for the RHS, 373092 and 973032:
TWELVE OtherNumber Example
186546 373092 ABACUS
486516 973032 CHAZAN
But in each case, the alphametic solution is not unique.
Thus there can be no solution with sod = 24.
... which I suspect Charlie already proved, now I'll go read the prior posts.
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sod_of_RHS = 24
rightnumbers = []
for n in range(200000, 1000000, 2):
if sod(n) != sod_of_RHS:
continue
strn = str(n)
if len(set(strn)) == 6:
continue
str12 = str(int(n/2))
if len(set(str12)) != 5:
continue
if str12[2] != str12[5]:
continue
hasTWELV = False
for letr in str12:
if letr in strn:
hasTWELV = True
if hasTWELV:
continue
rightnumbers.append(n)
print(rightnumbers)
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Posted by Larry
on 2024-04-27 10:47:30 |
re 24
