No won won? Wone WonHappy days alphameticFrench alphametic (corrected)Multiplicative alphametic: This is too hardAlfred E. Neuman alphameticAn almost Shakespearian alphameticSquare dance alphameticBad grammar alphameticAlphametic between Kennedy and NixonCITEMAHPLA Reverse AlphameticIt is as simple as ABC
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No won won? Wone Won
Happy days alphameticFrench alphametic (corrected)Multiplicative alphametic: This is too hardAlfred E. Neuman alphameticAn almost Shakespearian alphameticSquare dance alphameticBad grammar alphameticAlphametic between Kennedy and NixonCITEMAHPLA Reverse AlphameticIt is as simple as ABC
$begingroup$
Find four different digits $ W, O, N, E $
that satisfies the equation
$$ overlineNO times overlineWON times overlineWON = overlineWONEWON $$
alphametic
edited May 9 at 18:11
PiIsNot3
4,258951
asked May 4 at 3:19
UvcUvc
51210
$endgroup$
add a comment |
$begingroup$
Find four different digits $ W, O, N, E $
that satisfies the equation
$$ overlineNO times overlineWON times overlineWON = overlineWONEWON $$
alphametic
edited May 9 at 18:11
PiIsNot3
4,258951
asked May 4 at 3:19
UvcUvc
51210
$endgroup$
1
$begingroup$
What's up with the dashes above the letters? Are we negating anything, or how should I interpret those?
$endgroup$
– Mast
May 4 at 19:11
$begingroup$
I have given some of earlier puzzles as straight combination...as NO..normally concatenation..somebody interpreted as multiplication..one of the editors modified to the current format..each letter stands for a digit..bar above signifies..it is one number
$endgroup$
– Uvc
May 4 at 19:45
add a comment |
$begingroup$
Find four different digits $ W, O, N, E $
that satisfies the equation
$$ overlineNO times overlineWON times overlineWON = overlineWONEWON $$
alphametic
edited May 9 at 18:11
PiIsNot3
4,258951
asked May 4 at 3:19
UvcUvc
51210
$endgroup$
Find four different digits $ W, O, N, E $
that satisfies the equation
$$ overlineNO times overlineWON times overlineWON = overlineWONEWON $$
alphametic
alphametic
edited May 9 at 18:11
PiIsNot3
4,258951
asked May 4 at 3:19
UvcUvc
51210
edited May 9 at 18:11
PiIsNot3
4,258951
asked May 4 at 3:19
UvcUvc
51210
edited May 9 at 18:11
PiIsNot3
4,258951
edited May 9 at 18:11
PiIsNot3
4,258951
edited May 9 at 18:11
PiIsNot3
4,258951
4,258951
asked May 4 at 3:19
UvcUvc
51210
asked May 4 at 3:19
UvcUvc
51210
asked May 4 at 3:19
UvcUvc
51210
51210
1
$begingroup$
What's up with the dashes above the letters? Are we negating anything, or how should I interpret those?
$endgroup$
– Mast
May 4 at 19:11
$begingroup$
I have given some of earlier puzzles as straight combination...as NO..normally concatenation..somebody interpreted as multiplication..one of the editors modified to the current format..each letter stands for a digit..bar above signifies..it is one number
$endgroup$
– Uvc
May 4 at 19:45
add a comment |
1
$begingroup$
What's up with the dashes above the letters? Are we negating anything, or how should I interpret those?
$endgroup$
– Mast
May 4 at 19:11
$begingroup$
I have given some of earlier puzzles as straight combination...as NO..normally concatenation..somebody interpreted as multiplication..one of the editors modified to the current format..each letter stands for a digit..bar above signifies..it is one number
$endgroup$
– Uvc
May 4 at 19:45
1
1
$begingroup$
What's up with the dashes above the letters? Are we negating anything, or how should I interpret those?
$endgroup$
– Mast
May 4 at 19:11
$begingroup$
What's up with the dashes above the letters? Are we negating anything, or how should I interpret those?
$endgroup$
– Mast
May 4 at 19:11
$begingroup$
I have given some of earlier puzzles as straight combination...as NO..normally concatenation..somebody interpreted as multiplication..one of the editors modified to the current format..each letter stands for a digit..bar above signifies..it is one number
$endgroup$
– Uvc
May 4 at 19:45
$begingroup$
I have given some of earlier puzzles as straight combination...as NO..normally concatenation..somebody interpreted as multiplication..one of the editors modified to the current format..each letter stands for a digit..bar above signifies..it is one number
$endgroup$
– Uvc
May 4 at 19:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The answer is
W = 1, O = 3, E = 0, N = 7
$73 * 137 * 137 = 1370137$
Method:
Divide both sides by $WON$
$NO * WON = 10001 + (E000 / WON)$
$ 0 <= E <= 9$
First consider cases when $E$ is not $0$ and
$(E000 / WON) = X $
$E000 = E * 2^3 * 5^3$
$N$ cannot be $0$ because $NO$ starts from it. So $WON$ cannot have both $2$s and $5$s as factors. If $WON$ does not have $5$s as factors it can be at most $9*2^3=72$ - not a three digit number. So $WON$ is an odd number divisible by $5$.
Now $X$ is obviously not divisible by $5$ because
$NO * WON = 10001 + X$
So $WON$ has to be divisible by $125$. That makes $X <=72$. Which contradicts $10001 + X$ being divisible by 125. So we proved that
$E = 0$ and
$NO * WON = 10001$
To get the last digit $1$ in the product with $N$ and $O$ being different digits we must have $NO = 37$ or $NO = 73$. $37$ is not a multiple of $10001$, so $NO = 73, WON = 137$
answered May 4 at 3:53
ppgdevppgdev
68538
$endgroup$
add a comment |
$begingroup$
W=1, O=3, N=7, E=0
works because
73 * 137 * 137 = 1370137.
Also,
W=O=N=E=0
works, but I don't think that's what you meant.
answered May 4 at 3:54
LarrySnyder610LarrySnyder610
23010
$endgroup$
1
$begingroup$
Welcome to Puzzling! (Take the Tour!) Regarding your "Also", the puzzle does say "four different digits" (emphasis added). :)
$endgroup$
– Rubio♦
May 4 at 5:44
$begingroup$
Thanks! And oops. :/
$endgroup$
– LarrySnyder610
May 4 at 10:56
add a comment |
$begingroup$
As this is not tagged "no-computers", I just used some Python :
Solution : 'W': 1, 'O': 3, 'N': 7, 'E': 0
Method (Python) :
def digit_sequence(string, digits): # string is sequence, digits are dictionairy with wone
result=0
for i in range(0, len(string)):
result+=digits[string[i]]*(10**(len(string)-i-1))
return result
for W in range(0, 10):
for O in range(0, 10):
if W == O:
continue
for N in range(0, 10):
if N == W or N == O:
continue
for E in range(0, 10):
if E == W or E == O or E == N:
continue
# Four different digits W, O, N, E
# Now check whether equation is fulfilled
digits="W": W, "O": O, "N": N, "E": E
if digit_sequence("NO", digits)*(digit_sequence("WON", digits)**2) == digit_sequence("WONEWON", digits):
print("Solution : "+str(digits))
edited May 4 at 13:19
I N T E R E S T I N G
3615
answered May 4 at 13:03
LMDLMD
23319
$endgroup$
1
$begingroup$
I thought it might be fun to put together a solution that does this with an SMT solver, too. I used cryptol as my front-end, and I think it turned out pretty beautiful. Here it is in a gist just six lines long; an excerpt of interest isisSolution w o n e = all isDigit [w,o,n,e] / no*won*won == wonewon / no != 0.
$endgroup$
– Daniel Wagner
May 4 at 17:29
$begingroup$
You could add this as answer, too, @DanielWagner
$endgroup$
– LMD
May 4 at 19:56
$begingroup$
(This isn't PPCG though. Different mechanical means of arriving at the same result don't really provide any new information here. If you want to elaborate on an innovative method as your answer that's one thing; but just providing code that brute-forces the solution, or uses language or library functionality that keeps the actual mechanism mostly in a black box, is little better than stating 'the answer is X because magic'—it provides the same answer someone else already has, and does nothing to explain how to reach it. A comment is fine... an additional code answer probably isn't.)
$endgroup$
– Rubio♦
May 4 at 23:40
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is
W = 1, O = 3, E = 0, N = 7
$73 * 137 * 137 = 1370137$
Method:
Divide both sides by $WON$
$NO * WON = 10001 + (E000 / WON)$
$ 0 <= E <= 9$
First consider cases when $E$ is not $0$ and
$(E000 / WON) = X $
$E000 = E * 2^3 * 5^3$
$N$ cannot be $0$ because $NO$ starts from it. So $WON$ cannot have both $2$s and $5$s as factors. If $WON$ does not have $5$s as factors it can be at most $9*2^3=72$ - not a three digit number. So $WON$ is an odd number divisible by $5$.
Now $X$ is obviously not divisible by $5$ because
$NO * WON = 10001 + X$
So $WON$ has to be divisible by $125$. That makes $X <=72$. Which contradicts $10001 + X$ being divisible by 125. So we proved that
$E = 0$ and
$NO * WON = 10001$
To get the last digit $1$ in the product with $N$ and $O$ being different digits we must have $NO = 37$ or $NO = 73$. $37$ is not a multiple of $10001$, so $NO = 73, WON = 137$
answered May 4 at 3:53
ppgdevppgdev
68538
$endgroup$
add a comment |
$begingroup$
The answer is
W = 1, O = 3, E = 0, N = 7
$73 * 137 * 137 = 1370137$
Method:
Divide both sides by $WON$
$NO * WON = 10001 + (E000 / WON)$
$ 0 <= E <= 9$
First consider cases when $E$ is not $0$ and
$(E000 / WON) = X $
$E000 = E * 2^3 * 5^3$
$N$ cannot be $0$ because $NO$ starts from it. So $WON$ cannot have both $2$s and $5$s as factors. If $WON$ does not have $5$s as factors it can be at most $9*2^3=72$ - not a three digit number. So $WON$ is an odd number divisible by $5$.
Now $X$ is obviously not divisible by $5$ because
$NO * WON = 10001 + X$
So $WON$ has to be divisible by $125$. That makes $X <=72$. Which contradicts $10001 + X$ being divisible by 125. So we proved that
$E = 0$ and
$NO * WON = 10001$
To get the last digit $1$ in the product with $N$ and $O$ being different digits we must have $NO = 37$ or $NO = 73$. $37$ is not a multiple of $10001$, so $NO = 73, WON = 137$
answered May 4 at 3:53
ppgdevppgdev
68538
$endgroup$
add a comment |
$begingroup$
The answer is
W = 1, O = 3, E = 0, N = 7
$73 * 137 * 137 = 1370137$
Method:
Divide both sides by $WON$
$NO * WON = 10001 + (E000 / WON)$
$ 0 <= E <= 9$
First consider cases when $E$ is not $0$ and
$(E000 / WON) = X $
$E000 = E * 2^3 * 5^3$
$N$ cannot be $0$ because $NO$ starts from it. So $WON$ cannot have both $2$s and $5$s as factors. If $WON$ does not have $5$s as factors it can be at most $9*2^3=72$ - not a three digit number. So $WON$ is an odd number divisible by $5$.
Now $X$ is obviously not divisible by $5$ because
$NO * WON = 10001 + X$
So $WON$ has to be divisible by $125$. That makes $X <=72$. Which contradicts $10001 + X$ being divisible by 125. So we proved that
$E = 0$ and
$NO * WON = 10001$
To get the last digit $1$ in the product with $N$ and $O$ being different digits we must have $NO = 37$ or $NO = 73$. $37$ is not a multiple of $10001$, so $NO = 73, WON = 137$
answered May 4 at 3:53
ppgdevppgdev
68538
$endgroup$
The answer is
W = 1, O = 3, E = 0, N = 7
$73 * 137 * 137 = 1370137$
Method:
Divide both sides by $WON$
$NO * WON = 10001 + (E000 / WON)$
$ 0 <= E <= 9$
First consider cases when $E$ is not $0$ and
$(E000 / WON) = X $
$E000 = E * 2^3 * 5^3$
$N$ cannot be $0$ because $NO$ starts from it. So $WON$ cannot have both $2$s and $5$s as factors. If $WON$ does not have $5$s as factors it can be at most $9*2^3=72$ - not a three digit number. So $WON$ is an odd number divisible by $5$.
Now $X$ is obviously not divisible by $5$ because
$NO * WON = 10001 + X$
So $WON$ has to be divisible by $125$. That makes $X <=72$. Which contradicts $10001 + X$ being divisible by 125. So we proved that
$E = 0$ and
$NO * WON = 10001$
To get the last digit $1$ in the product with $N$ and $O$ being different digits we must have $NO = 37$ or $NO = 73$. $37$ is not a multiple of $10001$, so $NO = 73, WON = 137$
answered May 4 at 3:53
ppgdevppgdev
68538
edited May 4 at 6:00
answered May 4 at 3:53
ppgdevppgdev
68538
answered May 4 at 3:53
ppgdevppgdev
68538
answered May 4 at 3:53
ppgdevppgdev
68538
68538
add a comment |
add a comment |
$begingroup$
W=1, O=3, N=7, E=0
works because
73 * 137 * 137 = 1370137.
Also,
W=O=N=E=0
works, but I don't think that's what you meant.
answered May 4 at 3:54
LarrySnyder610LarrySnyder610
23010
$endgroup$
1
$begingroup$
Welcome to Puzzling! (Take the Tour!) Regarding your "Also", the puzzle does say "four different digits" (emphasis added). :)
$endgroup$
– Rubio♦
May 4 at 5:44
$begingroup$
Thanks! And oops. :/
$endgroup$
– LarrySnyder610
May 4 at 10:56
add a comment |
$begingroup$
W=1, O=3, N=7, E=0
works because
73 * 137 * 137 = 1370137.
Also,
W=O=N=E=0
works, but I don't think that's what you meant.
answered May 4 at 3:54
LarrySnyder610LarrySnyder610
23010
$endgroup$
1
$begingroup$
Welcome to Puzzling! (Take the Tour!) Regarding your "Also", the puzzle does say "four different digits" (emphasis added). :)
$endgroup$
– Rubio♦
May 4 at 5:44
$begingroup$
Thanks! And oops. :/
$endgroup$
– LarrySnyder610
May 4 at 10:56
add a comment |
$begingroup$
W=1, O=3, N=7, E=0
works because
73 * 137 * 137 = 1370137.
Also,
W=O=N=E=0
works, but I don't think that's what you meant.
answered May 4 at 3:54
LarrySnyder610LarrySnyder610
23010
$endgroup$
W=1, O=3, N=7, E=0
works because
73 * 137 * 137 = 1370137.
Also,
W=O=N=E=0
works, but I don't think that's what you meant.
answered May 4 at 3:54
LarrySnyder610LarrySnyder610
23010
answered May 4 at 3:54
LarrySnyder610LarrySnyder610
23010
answered May 4 at 3:54
LarrySnyder610LarrySnyder610
23010
answered May 4 at 3:54
LarrySnyder610LarrySnyder610
23010
23010
1
$begingroup$
Welcome to Puzzling! (Take the Tour!) Regarding your "Also", the puzzle does say "four different digits" (emphasis added). :)
$endgroup$
– Rubio♦
May 4 at 5:44
$begingroup$
Thanks! And oops. :/
$endgroup$
– LarrySnyder610
May 4 at 10:56
add a comment |
1
$begingroup$
Welcome to Puzzling! (Take the Tour!) Regarding your "Also", the puzzle does say "four different digits" (emphasis added). :)
$endgroup$
– Rubio♦
May 4 at 5:44
$begingroup$
Thanks! And oops. :/
$endgroup$
– LarrySnyder610
May 4 at 10:56
1
1
$begingroup$
Welcome to Puzzling! (Take the Tour!) Regarding your "Also", the puzzle does say "four different digits" (emphasis added). :)
$endgroup$
– Rubio♦
May 4 at 5:44
$begingroup$
Welcome to Puzzling! (Take the Tour!) Regarding your "Also", the puzzle does say "four different digits" (emphasis added). :)
$endgroup$
– Rubio♦
May 4 at 5:44
$begingroup$
Thanks! And oops. :/
$endgroup$
– LarrySnyder610
May 4 at 10:56
$begingroup$
Thanks! And oops. :/
$endgroup$
– LarrySnyder610
May 4 at 10:56
add a comment |
$begingroup$
As this is not tagged "no-computers", I just used some Python :
Solution : 'W': 1, 'O': 3, 'N': 7, 'E': 0
Method (Python) :
def digit_sequence(string, digits): # string is sequence, digits are dictionairy with wone
result=0
for i in range(0, len(string)):
result+=digits[string[i]]*(10**(len(string)-i-1))
return result
for W in range(0, 10):
for O in range(0, 10):
if W == O:
continue
for N in range(0, 10):
if N == W or N == O:
continue
for E in range(0, 10):
if E == W or E == O or E == N:
continue
# Four different digits W, O, N, E
# Now check whether equation is fulfilled
digits="W": W, "O": O, "N": N, "E": E
if digit_sequence("NO", digits)*(digit_sequence("WON", digits)**2) == digit_sequence("WONEWON", digits):
print("Solution : "+str(digits))
edited May 4 at 13:19
I N T E R E S T I N G
3615
answered May 4 at 13:03
LMDLMD
23319
$endgroup$
1
$begingroup$
I thought it might be fun to put together a solution that does this with an SMT solver, too. I used cryptol as my front-end, and I think it turned out pretty beautiful. Here it is in a gist just six lines long; an excerpt of interest isisSolution w o n e = all isDigit [w,o,n,e] / no*won*won == wonewon / no != 0.
$endgroup$
– Daniel Wagner
May 4 at 17:29
$begingroup$
You could add this as answer, too, @DanielWagner
$endgroup$
– LMD
May 4 at 19:56
$begingroup$
(This isn't PPCG though. Different mechanical means of arriving at the same result don't really provide any new information here. If you want to elaborate on an innovative method as your answer that's one thing; but just providing code that brute-forces the solution, or uses language or library functionality that keeps the actual mechanism mostly in a black box, is little better than stating 'the answer is X because magic'—it provides the same answer someone else already has, and does nothing to explain how to reach it. A comment is fine... an additional code answer probably isn't.)
$endgroup$
– Rubio♦
May 4 at 23:40
add a comment |
$begingroup$
As this is not tagged "no-computers", I just used some Python :
Solution : 'W': 1, 'O': 3, 'N': 7, 'E': 0
Method (Python) :
def digit_sequence(string, digits): # string is sequence, digits are dictionairy with wone
result=0
for i in range(0, len(string)):
result+=digits[string[i]]*(10**(len(string)-i-1))
return result
for W in range(0, 10):
for O in range(0, 10):
if W == O:
continue
for N in range(0, 10):
if N == W or N == O:
continue
for E in range(0, 10):
if E == W or E == O or E == N:
continue
# Four different digits W, O, N, E
# Now check whether equation is fulfilled
digits="W": W, "O": O, "N": N, "E": E
if digit_sequence("NO", digits)*(digit_sequence("WON", digits)**2) == digit_sequence("WONEWON", digits):
print("Solution : "+str(digits))
edited May 4 at 13:19
I N T E R E S T I N G
3615
answered May 4 at 13:03
LMDLMD
23319
$endgroup$
1
$begingroup$
I thought it might be fun to put together a solution that does this with an SMT solver, too. I used cryptol as my front-end, and I think it turned out pretty beautiful. Here it is in a gist just six lines long; an excerpt of interest isisSolution w o n e = all isDigit [w,o,n,e] / no*won*won == wonewon / no != 0.
$endgroup$
– Daniel Wagner
May 4 at 17:29
$begingroup$
You could add this as answer, too, @DanielWagner
$endgroup$
– LMD
May 4 at 19:56
$begingroup$
(This isn't PPCG though. Different mechanical means of arriving at the same result don't really provide any new information here. If you want to elaborate on an innovative method as your answer that's one thing; but just providing code that brute-forces the solution, or uses language or library functionality that keeps the actual mechanism mostly in a black box, is little better than stating 'the answer is X because magic'—it provides the same answer someone else already has, and does nothing to explain how to reach it. A comment is fine... an additional code answer probably isn't.)
$endgroup$
– Rubio♦
May 4 at 23:40
add a comment |
$begingroup$
As this is not tagged "no-computers", I just used some Python :
Solution : 'W': 1, 'O': 3, 'N': 7, 'E': 0
Method (Python) :
def digit_sequence(string, digits): # string is sequence, digits are dictionairy with wone
result=0
for i in range(0, len(string)):
result+=digits[string[i]]*(10**(len(string)-i-1))
return result
for W in range(0, 10):
for O in range(0, 10):
if W == O:
continue
for N in range(0, 10):
if N == W or N == O:
continue
for E in range(0, 10):
if E == W or E == O or E == N:
continue
# Four different digits W, O, N, E
# Now check whether equation is fulfilled
digits="W": W, "O": O, "N": N, "E": E
if digit_sequence("NO", digits)*(digit_sequence("WON", digits)**2) == digit_sequence("WONEWON", digits):
print("Solution : "+str(digits))
edited May 4 at 13:19
I N T E R E S T I N G
3615
answered May 4 at 13:03
LMDLMD
23319
$endgroup$
As this is not tagged "no-computers", I just used some Python :
Solution : 'W': 1, 'O': 3, 'N': 7, 'E': 0
Method (Python) :
def digit_sequence(string, digits): # string is sequence, digits are dictionairy with wone
result=0
for i in range(0, len(string)):
result+=digits[string[i]]*(10**(len(string)-i-1))
return result
for W in range(0, 10):
for O in range(0, 10):
if W == O:
continue
for N in range(0, 10):
if N == W or N == O:
continue
for E in range(0, 10):
if E == W or E == O or E == N:
continue
# Four different digits W, O, N, E
# Now check whether equation is fulfilled
digits="W": W, "O": O, "N": N, "E": E
if digit_sequence("NO", digits)*(digit_sequence("WON", digits)**2) == digit_sequence("WONEWON", digits):
print("Solution : "+str(digits))
edited May 4 at 13:19
I N T E R E S T I N G
3615
answered May 4 at 13:03
LMDLMD
23319
edited May 4 at 13:19
I N T E R E S T I N G
3615
edited May 4 at 13:19
I N T E R E S T I N G
3615
edited May 4 at 13:19
I N T E R E S T I N G
3615
3615
answered May 4 at 13:03
LMDLMD
23319
answered May 4 at 13:03
LMDLMD
23319
answered May 4 at 13:03
LMDLMD
23319
23319
1
$begingroup$
I thought it might be fun to put together a solution that does this with an SMT solver, too. I used cryptol as my front-end, and I think it turned out pretty beautiful. Here it is in a gist just six lines long; an excerpt of interest isisSolution w o n e = all isDigit [w,o,n,e] / no*won*won == wonewon / no != 0.
$endgroup$
– Daniel Wagner
May 4 at 17:29
$begingroup$
You could add this as answer, too, @DanielWagner
$endgroup$
– LMD
May 4 at 19:56
$begingroup$
(This isn't PPCG though. Different mechanical means of arriving at the same result don't really provide any new information here. If you want to elaborate on an innovative method as your answer that's one thing; but just providing code that brute-forces the solution, or uses language or library functionality that keeps the actual mechanism mostly in a black box, is little better than stating 'the answer is X because magic'—it provides the same answer someone else already has, and does nothing to explain how to reach it. A comment is fine... an additional code answer probably isn't.)
$endgroup$
– Rubio♦
May 4 at 23:40
add a comment |
1
$begingroup$
I thought it might be fun to put together a solution that does this with an SMT solver, too. I used cryptol as my front-end, and I think it turned out pretty beautiful. Here it is in a gist just six lines long; an excerpt of interest isisSolution w o n e = all isDigit [w,o,n,e] / no*won*won == wonewon / no != 0.
$endgroup$
– Daniel Wagner
May 4 at 17:29
$begingroup$
You could add this as answer, too, @DanielWagner
$endgroup$
– LMD
May 4 at 19:56
$begingroup$
(This isn't PPCG though. Different mechanical means of arriving at the same result don't really provide any new information here. If you want to elaborate on an innovative method as your answer that's one thing; but just providing code that brute-forces the solution, or uses language or library functionality that keeps the actual mechanism mostly in a black box, is little better than stating 'the answer is X because magic'—it provides the same answer someone else already has, and does nothing to explain how to reach it. A comment is fine... an additional code answer probably isn't.)
$endgroup$
– Rubio♦
May 4 at 23:40
1
1
$begingroup$
I thought it might be fun to put together a solution that does this with an SMT solver, too. I used cryptol as my front-end, and I think it turned out pretty beautiful. Here it is in a gist just six lines long; an excerpt of interest is
isSolution w o n e = all isDigit [w,o,n,e] / no*won*won == wonewon / no != 0.$endgroup$
– Daniel Wagner
May 4 at 17:29
$begingroup$
I thought it might be fun to put together a solution that does this with an SMT solver, too. I used cryptol as my front-end, and I think it turned out pretty beautiful. Here it is in a gist just six lines long; an excerpt of interest is
isSolution w o n e = all isDigit [w,o,n,e] / no*won*won == wonewon / no != 0.$endgroup$
– Daniel Wagner
May 4 at 17:29
$begingroup$
You could add this as answer, too, @DanielWagner
$endgroup$
– LMD
May 4 at 19:56
$begingroup$
You could add this as answer, too, @DanielWagner
$endgroup$
– LMD
May 4 at 19:56
$begingroup$
(This isn't PPCG though. Different mechanical means of arriving at the same result don't really provide any new information here. If you want to elaborate on an innovative method as your answer that's one thing; but just providing code that brute-forces the solution, or uses language or library functionality that keeps the actual mechanism mostly in a black box, is little better than stating 'the answer is X because magic'—it provides the same answer someone else already has, and does nothing to explain how to reach it. A comment is fine... an additional code answer probably isn't.)
$endgroup$
– Rubio♦
May 4 at 23:40
$begingroup$
(This isn't PPCG though. Different mechanical means of arriving at the same result don't really provide any new information here. If you want to elaborate on an innovative method as your answer that's one thing; but just providing code that brute-forces the solution, or uses language or library functionality that keeps the actual mechanism mostly in a black box, is little better than stating 'the answer is X because magic'—it provides the same answer someone else already has, and does nothing to explain how to reach it. A comment is fine... an additional code answer probably isn't.)
$endgroup$
– Rubio♦
May 4 at 23:40
add a comment |
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$begingroup$
What's up with the dashes above the letters? Are we negating anything, or how should I interpret those?
$endgroup$
– Mast
May 4 at 19:11
$begingroup$
I have given some of earlier puzzles as straight combination...as NO..normally concatenation..somebody interpreted as multiplication..one of the editors modified to the current format..each letter stands for a digit..bar above signifies..it is one number
$endgroup$
– Uvc
May 4 at 19:45