For 12 you can basically make it a large set of linear equations and then solve them.
Since each letter appears once in the solution, make 2 variables for all the letters of the alphabet, one with value equal to position, the other indicating whether it's present (A = value, a = present)
1) Alpha != Beta for all non zero (alpha, beta) pairs.
2) Alpha + Beta + Gamma + Delta = Xi, where Xi is one of the clues
3) alpha + beta + gamma + delta = xi, where xi is one of the clues
4) the increasing sequence spells a word with some mathematical meaning
5) Alpha <= 10 for all Alpha in the alphabet.
6) sum(lowercase) = 10
eg. for a), you have
S+I+D+E=17
s+i+d+e=3
F+A+C+E = 12
f+a+c+e = 2
D+I+S+K = 10
d+i+s+k = 2
C+O+N+E = 23
c+o+n+e = 3
R+E+C+T = 17
r+e+c+t = 2
R+A+Y+S = 9
r+a+y+s = 2
L+U+M+P = 13
l+u+m+p = 3
M+A+T+H = 11
m+a+t+h = 2
One assumption that may prove useful is that all 10 letters in the eventual solution are present in these clues (not stated in the question, though). This would reduce the full alphabet to
acdefhiklmnoprstuy
Only 18 letters.
We have 8 words with a total of 19 letters present in the final solution, meaning that there are 9 "repeats" of letters.
Letter frequencies are:
a3 c3 d2 e4 f1 h1 i2 k1 l1 m2 n1 o1 p1 r2 s3 t2 u1 y1
With the 8 word equations and the sum(alpha) = 10, we only have 9 equations and 18 unknowns so a unique solution is not guaranteed. But since they're binary variables it's easy to construct a tree of possibilities, starting with the most common letters and eliminating possibilities.
A simple contradiction can be found if we assume 'e' is not present:
s+i+d+e = 3, d+i+s+k = 2, e = 0 => k = -1
ie.
e = 1 -> s+i+d=2, f+a+c=1, c+o+n=2, r+c+t = 1, k = 0
a = 1 -> f = 0, c = 0, r+y+s = 1, m+t+h = 1, o = 1, n = 1, r+t = 1
s = 1 -> i+d = 1, r = 0, y = 0, t = 1, m = 0, h = 0, l = 1, u = 1, p = 1
This gives 2 possibilities
(e,a,o,n,s,t,l,u,p,i)
(e,a,o,n,s,t,l,u,p,d)
Neither of these find quick anagrams so replace one assumption
s = 0 -> i = 1, d = 1, r+y = 1
r = 1 -> y = 0, t = 0, m + h = 1, l + u + m + p = 3
We have 6 so far, so the only possibility that gives 10 letters is h, l, u, p = 1
(e,a,o,n,i,d,r,h,l,u,p) Edit: This line is proof that Stereo can't count to 10. x_x
Again, no quick anagrams, so step back to the assumption a = 1.
a = 0 -> f+c = 1, r+y+s = 2, m+t+h = 2
c = 1 -> f = 0, o+n = 1, r = 0, t = 0, y = 1, s = 1, m = 1, h = 1, i+d = 1, l+u+p = 2
This gives a whole bunch of valid possibilities:
(e, c, y, s, m, h) + o/n, i/d, 2 of (l, u, p)
Ex. ecysmhoilu, ecysmhodlu, etc. Unfortunately, none of these seem to be anagrams of actual words.
c = 0 -> f = 1, o = 1, n = 1, r+t = 1
This always results in 11 letters or more in the solution so it's invalid.
Take one of those 'potential solutions' and find positions
ECYSMHODLU
S+D+E = 20
C+E = 12
D+S = 10
C+O+E = 23
contradiction - O = 11
And so forth. Obviously doing all of these is kinda a waste of time - so try starting with just the "known" formulas and see what introduces contradictions.
ECYSMH:
E+C = 12 (FACE = 12)
E+C = 17 (RECT = 17)
EAONSTLUP:
A+E = 12
O+N+E = 23
E+T = 17
A+S = 9
L+U+P = 13
A+T = 11
E = 10 (as before), A = 2, thus contradiction: T = 7 and 9.
That leaves
EAONIDRHLUP
I+D+E = 20
A+E = 12
D+I = 10
O+N+E = 23
R+E = 17
R+A = 9
L+U+P = 13
A+H = 11
E = 10, A = 2, R = 7, H = 9
I+D = 10, O+N = 13, L+U+P = 13
These need to fit into (1, 3, 4, 5, 6, 8). Obviously 1 is L, U, or P, leaving the other 2 equal 12. So 10, 12, 13 -> 3 doesn't fit into any of these.
Since I've kinda eliminated all possibilities I guess I made a mistake.
On the other hand, I've got a better handle on the question.
E = 10, is in the solution. K is not in the solution.
so clues are reduced to
SID, 2, 10
FAC, 1, 2
CON, 2, 13
RCT, 1, 7
This leads to elimination of C (can't be both 2 and 7), so
O and N are in the solution, and their positions total to 13
FA, 1, 2
RT, 1, 7
RAYS, 2, 9
By inspection, if A is in RAYS, then another letter has 7. If R is in rays, then another letter has 2. This is only true if
a) RA are both in the solution, with R = 7, A = 2. FT are not in the solution.
b) FT are both in the solution, with F = 2, T = 7. AR are not in the solution.
a)
_A____R__E, plus O and N in (4, 9) or (5, 8)
_A_(ON)__R_(NO)E
_A__(ON)_R(NO)_E
b)
This implies MH, 1, 4. -> one or the other is in the solution.
Which in turn implies that O and N are in (5, 8)
_F_(MH)(ON)_T(NO)_E
At this point, SID, 2, 10 must be (1, 9) because (3,6) don't add to 10. SF, DF don't make sense.
IF_(MH)(ON)_T(NO)(SD)E
LUMP can only have 3 letters if M is present, and requires that the other 2 letters add to 9 and are in (LUP)
IF(LUP)M(ON)(UPL)T(NO)(SD)E
Assuming this word is constructed with vowels in typical positions
IFUMN(PL)TO(SD)E
Doesn't seem to have any good possibilities. So eliminate option b).
Going to take a second look at a)
_A_(ON)__R_(NO)E
_A__(ON)_R(NO)_E
Plus remaining useful clues:
SID, 2, 10
LUMP, 3, 13
MTH, 1, 9 - this eliminates one of the options, again.
_A__(ON)_R(NO)(MTH)E
1 3 4 6 are left. Needs to fit 2 of SID and 3 of LUMP into those 4 spots. So it's kinda gotta be an M.
_A__(ON)_R(NO)ME
4+6 for SID, 1+3 for LUP
(LUP)A(UPL)(SID)N(SID)ROME
I would say the answer is:
PALINDROME
This is just 12.a)
12. d)
HALF 3 22
ONES 3 11
TRIO 3 18
DUOS 3 15
TWOS 2 7
FOUR 2 11
FIVE 2 11
PLUG 2 12
O must be in the solution, otherwise it has 11 letters.
F must not be in the solution, otherwise FIVE / FOUR share 2 letters in the same spots (due to being 2,11)
HALO is in the solution.
NES 2
TRI 2
DUS 2
TWS 1
UR 1
IVE 2
PLUG 2
Assume S is in the solution (random guess)
-> TW are not in the solution
-> RI are in the solution
-> U is not in the solution
-> D is in the solution
NE 1
VE 1
PG 1
Since we've got HALOSRID already, N/V cannot be in the solution (too many letters)
HALOSRIDEP
HALOSRIDEG
Interestingly, an anagram of HALOSRIDEP is SPHEROIDAL
Checking the equations this works.
12. d) SPHEROIDAL.
Since each letter appears once in the solution, make 2 variables for all the letters of the alphabet, one with value equal to position, the other indicating whether it's present (A = value, a = present)
1) Alpha != Beta for all non zero (alpha, beta) pairs.
2) Alpha + Beta + Gamma + Delta = Xi, where Xi is one of the clues
3) alpha + beta + gamma + delta = xi, where xi is one of the clues
4) the increasing sequence spells a word with some mathematical meaning
5) Alpha <= 10 for all Alpha in the alphabet.
6) sum(lowercase) = 10
eg. for a), you have
S+I+D+E=17
s+i+d+e=3
F+A+C+E = 12
f+a+c+e = 2
D+I+S+K = 10
d+i+s+k = 2
C+O+N+E = 23
c+o+n+e = 3
R+E+C+T = 17
r+e+c+t = 2
R+A+Y+S = 9
r+a+y+s = 2
L+U+M+P = 13
l+u+m+p = 3
M+A+T+H = 11
m+a+t+h = 2
One assumption that may prove useful is that all 10 letters in the eventual solution are present in these clues (not stated in the question, though). This would reduce the full alphabet to
acdefhiklmnoprstuy
Only 18 letters.
We have 8 words with a total of 19 letters present in the final solution, meaning that there are 9 "repeats" of letters.
Letter frequencies are:
a3 c3 d2 e4 f1 h1 i2 k1 l1 m2 n1 o1 p1 r2 s3 t2 u1 y1
With the 8 word equations and the sum(alpha) = 10, we only have 9 equations and 18 unknowns so a unique solution is not guaranteed. But since they're binary variables it's easy to construct a tree of possibilities, starting with the most common letters and eliminating possibilities.
A simple contradiction can be found if we assume 'e' is not present:
s+i+d+e = 3, d+i+s+k = 2, e = 0 => k = -1
ie.
e = 1 -> s+i+d=2, f+a+c=1, c+o+n=2, r+c+t = 1, k = 0
a = 1 -> f = 0, c = 0, r+y+s = 1, m+t+h = 1, o = 1, n = 1, r+t = 1
s = 1 -> i+d = 1, r = 0, y = 0, t = 1, m = 0, h = 0, l = 1, u = 1, p = 1
This gives 2 possibilities
(e,a,o,n,s,t,l,u,p,i)
(e,a,o,n,s,t,l,u,p,d)
Neither of these find quick anagrams so replace one assumption
s = 0 -> i = 1, d = 1, r+y = 1
r = 1 -> y = 0, t = 0, m + h = 1, l + u + m + p = 3
We have 6 so far, so the only possibility that gives 10 letters is h, l, u, p = 1
(e,a,o,n,i,d,r,h,l,u,p) Edit: This line is proof that Stereo can't count to 10. x_x
Again, no quick anagrams, so step back to the assumption a = 1.
a = 0 -> f+c = 1, r+y+s = 2, m+t+h = 2
c = 1 -> f = 0, o+n = 1, r = 0, t = 0, y = 1, s = 1, m = 1, h = 1, i+d = 1, l+u+p = 2
This gives a whole bunch of valid possibilities:
(e, c, y, s, m, h) + o/n, i/d, 2 of (l, u, p)
Ex. ecysmhoilu, ecysmhodlu, etc. Unfortunately, none of these seem to be anagrams of actual words.
c = 0 -> f = 1, o = 1, n = 1, r+t = 1
This always results in 11 letters or more in the solution so it's invalid.
Take one of those 'potential solutions' and find positions
ECYSMHODLU
S+D+E = 20
C+E = 12
D+S = 10
C+O+E = 23
contradiction - O = 11
And so forth. Obviously doing all of these is kinda a waste of time - so try starting with just the "known" formulas and see what introduces contradictions.
ECYSMH:
E+C = 12 (FACE = 12)
E+C = 17 (RECT = 17)
EAONSTLUP:
A+E = 12
O+N+E = 23
E+T = 17
A+S = 9
L+U+P = 13
A+T = 11
E = 10 (as before), A = 2, thus contradiction: T = 7 and 9.
That leaves
EAONIDRHLUP
I+D+E = 20
A+E = 12
D+I = 10
O+N+E = 23
R+E = 17
R+A = 9
L+U+P = 13
A+H = 11
E = 10, A = 2, R = 7, H = 9
I+D = 10, O+N = 13, L+U+P = 13
These need to fit into (1, 3, 4, 5, 6, 8). Obviously 1 is L, U, or P, leaving the other 2 equal 12. So 10, 12, 13 -> 3 doesn't fit into any of these.
Since I've kinda eliminated all possibilities I guess I made a mistake.
On the other hand, I've got a better handle on the question.
E = 10, is in the solution. K is not in the solution.
so clues are reduced to
SID, 2, 10
FAC, 1, 2
CON, 2, 13
RCT, 1, 7
This leads to elimination of C (can't be both 2 and 7), so
O and N are in the solution, and their positions total to 13
FA, 1, 2
RT, 1, 7
RAYS, 2, 9
By inspection, if A is in RAYS, then another letter has 7. If R is in rays, then another letter has 2. This is only true if
a) RA are both in the solution, with R = 7, A = 2. FT are not in the solution.
b) FT are both in the solution, with F = 2, T = 7. AR are not in the solution.
a)
_A____R__E, plus O and N in (4, 9) or (5, 8)
_A_(ON)__R_(NO)E
_A__(ON)_R(NO)_E
b)
This implies MH, 1, 4. -> one or the other is in the solution.
Which in turn implies that O and N are in (5, 8)
_F_(MH)(ON)_T(NO)_E
At this point, SID, 2, 10 must be (1, 9) because (3,6) don't add to 10. SF, DF don't make sense.
IF_(MH)(ON)_T(NO)(SD)E
LUMP can only have 3 letters if M is present, and requires that the other 2 letters add to 9 and are in (LUP)
IF(LUP)M(ON)(UPL)T(NO)(SD)E
Assuming this word is constructed with vowels in typical positions
IFUMN(PL)TO(SD)E
Doesn't seem to have any good possibilities. So eliminate option b).
Going to take a second look at a)
_A_(ON)__R_(NO)E
_A__(ON)_R(NO)_E
Plus remaining useful clues:
SID, 2, 10
LUMP, 3, 13
MTH, 1, 9 - this eliminates one of the options, again.
_A__(ON)_R(NO)(MTH)E
1 3 4 6 are left. Needs to fit 2 of SID and 3 of LUMP into those 4 spots. So it's kinda gotta be an M.
_A__(ON)_R(NO)ME
4+6 for SID, 1+3 for LUP
(LUP)A(UPL)(SID)N(SID)ROME
I would say the answer is:
PALINDROME
This is just 12.a)
12. d)
HALF 3 22
ONES 3 11
TRIO 3 18
DUOS 3 15
TWOS 2 7
FOUR 2 11
FIVE 2 11
PLUG 2 12
O must be in the solution, otherwise it has 11 letters.
F must not be in the solution, otherwise FIVE / FOUR share 2 letters in the same spots (due to being 2,11)
HALO is in the solution.
NES 2
TRI 2
DUS 2
TWS 1
UR 1
IVE 2
PLUG 2
Assume S is in the solution (random guess)
-> TW are not in the solution
-> RI are in the solution
-> U is not in the solution
-> D is in the solution
NE 1
VE 1
PG 1
Since we've got HALOSRID already, N/V cannot be in the solution (too many letters)
HALOSRIDEP
HALOSRIDEG
Interestingly, an anagram of HALOSRIDEP is SPHEROIDAL
Checking the equations this works.
12. d) SPHEROIDAL.