# Extremely difficult question from an IQ test...

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The plan is to start looking for pair correlations in the sequences I get from the stoicastic generation of solutions.

The spelling inconsistency is not specifically "blurred" by the method I used but the weight of each equation is. I could add some "blurr" for the different spellings but I am going to have to do a lot of re-writting for that

ro..ntgen or roentgen

plank or planck

block or bloch

cerenkov or cerenkov

lippman or lippmann

schrodinger or schro..dinger

michelson michaelson

what else is possible?

Can you spell einstein differently?

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The plan is to start looking for pair correlations in the sequences I get from the stoicastic generation of solutions.

The spelling inconsistency is not specifically "blurred" by the method I used but the weight of each equation is. I could add some "blurr" for the different spellings but I am going to have to do a lot of re-writting for that

ro..ntgen or roentgen

plank or planck

block or bloch

cerenkov or cerenkov

lippman or lippmann

schrodinger or schro..dinger

michelson michaelson

what else is possible?

Can you spell einstein differently?

Can we spell Feynman without the y...?

I think Einstein is the only way to spell it.

The following are just guesses (who knows how someone else may have (mis)-spelled a name)...

Lorentz or Lorenz

Kamerlingh seems like a candidate for a mispelling...

I don't know...

Sounds like you have a pretty good technique going on though.

Cheers,

w=f[z]

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If the name has German origin as many surnames of American Jewish do, it may be that Feynman was originally spelled Feinman, (fein man = roughly fine(as in good) man). It is quite a stretch though.

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Thanks for the suggestions, I will stick them in the "model" manually and see if any thing useful starts to come out.

It sounds so high fallooting to call it a model it is really a just a very messy spread sheet with a lot of obscure functions. All I do is calculate the square difference of each name/number from the given target value and try to minimise this value with solver. I cant help feeling that I am just ploughing futher and further away from the simple "trick" answer.

I introduce noise to the problem by randomly weighting the names

I need to adapt it to use a gaussian for the difference calculation to try and free up some of the "bad" equations more. My thinking is that if an equation is off a little bit then the value will need to stray quiet far, however this does mean I will start to get lots of local minima and I will need to explore them is some systemic way.

There must be a better tool for this than doing it on a spread sheet, but I will have to learn it. Ho Hum.

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Ok, first problem is that the system of linear equations is overdetermined and inconsistent as it stands in the question - so no programme will find a solution unless we somehow can eliminate the inconsistent equations. (The joys of mathematica - 7 lines of code and you have the matrix If anyone wants it just shout)

Now we need to work out which of the equations we can scrap and I'm guessing we need to scrap three of them to arrive at a unique solution. So i guess its a physics history lesson to work out which three physicists should not be on the list. I'm guessing its a theory, experiment division or something like that.

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;344375']I'll be darned, I found a pdf of the original test: [url']http://rooth.org/users/avfanatic/uber-IQ-test.pdf[/url]

The numbers presented in the OP don't match with the pdf-file:

roentgen 89 <-> 104

lorentz 91 <-> 102

curie 67 <-> 69

michelson 95 <-> 109

lippman 123 <-> 88

marconi 82 <-> 90

kamerlingh 127 <-> 120

planck 77 <-> 96

stark 66 <-> 60

einstein 109 <-> 94

bohr 43 <-> 47

millikan 135 <-> 103

siegbahn 114 <-> 87

perrin 78 <-> 98

richardson 115 <-> 155

heisenberg 122 <-> 118

schrodinger 122 <-> 168

anderson 86 <-> 127

davisson 121 <-> 103

fermi 62 <-> 57

stern 56 <-> 70

block 65 <-> 73

zernike 77 <-> 99

cherenkov 81 <-> 109

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The numbers presented in the OP don't match with the pdf-file:

Son of a gun...

I'll have to double check with my printed version (which I left home today ). I could have swore what I had matched my printed version....

Cheers

Edit: Okay... still have my printed version at home, but I am pretty darn sure "Heisenberg" and "Schrodinger" had the same value (122) because I tried to use that fact before. Just going from memory here... but something is very fishy about this d*mn problem now. That really pisses me off!

Hell... maybe with the "new" values, it works out nicer....

Edit #2: It seems these guys were using the same values I originally posted too...

http://episteme.arstechnica.com/eve/forums/a/tpc/f/6330927813/m/6740938424/p/1

Edit #3: Looking at the date-stamp on the pdf link I gave for the puzzle, I see that it is 11/22/05 - which is about 3 years newer than what I was originally working from. In light of that, it might be a good idea to work form those numbers in case there indeed was a mistake that the puzzle makers corrected. Sheesh! Sorry guys for the values I supplied, but I was only posting what I had in front of me and what I originally used....

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I think the puzzle was reset to make it a new puzzle for people that had got the old one. Perhaps the old "answer" got out into the public domain and the authors had to do this. If you read the site they discourage solving the problem on forums as it invalidates the test, something I have a great deal of sympathy with

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I just double checked with my original printed version. What I have matches the OP. But I would still suggest working this with the new values.

I think the puzzle was reset to make it a new puzzle for people that had got the old one. Perhaps the old "answer" got out into the public domain and the authors had to do this. If you read the site they discourage solving the problem on forums as it invalidates the test, something I have a great deal of sympathy with

Yes, I saw that too. They have taken this problem off their website so I figured it was fair game now. Plus, I've had this problem in the back of my brain for so long now, that for my own "well being," I really want to see the solution.

Cheers,

w=f[z]

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if they have with drawn it it may be because they consider it to be too flawed to continue using it, I am going to take a break from the problem. I am sure you have found that it helps to come back to a problem after a period of rest and some fresh ideas.

On this occasion I have learnt that I need to be able to add more flexibility into an approach and I will try to do this from the beginning next time.

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I think the puzzle was reset to make it a new puzzle for people that had got the old one. Perhaps the old "answer" got out into the public domain and the authors had to do this. If you read the site they discourage solving the problem on forums as it invalidates the test, something I have a great deal of sympathy with

Yes, very likely, but they didn't change the spelling...

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Just chiming in here, i've been thinking / working on this one for a week now, and finally i've come to the conclusion that the answer is inderminate. Given the problem specifications, and using the data from the .pdf linked above, the letter Y can have any value.

Methodology was to make a huge matrix and reduce until I had some equations I could solve, at that point I saw the problem. Only 1 occurance of the letter Y.

Here's values I've gotten for all the others:

A 6

B 11

C 20

D 24

E 7

F 16

G 12

H 5

I 9

K 13

L 21

M 2

N 22

O 8

P 14

R 23

S 15

T 3

U 10

V 4

W 17

Z 18

Using these values, all names add up correctly to the .pdf values, and Y can have any value.

Thinking about it, I would be not be surprised if the answer is something like this, as they can't expect you to solve 23 equations at once, high IQ or not.

Anyhow, that's just my take, hopefully there's not too many errors...

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Whoa -

Just looked at my data in my last post and the numbers go 2-24, omiiting 19.

If Y = 19, then Feynman = 94

thoughts?

M 2

T 3

V 4

H 5

A 6

E 7

O 8

I 9

U 10

B 11

G 12

K 13

P 14

S 15

F 16

W 17

Z 18

Y 19

C 20

L 21

N 22

R 23

D 24

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Whoa -

Just looked at my data in my last post and the numbers go 2-24, omiiting 19.

If Y = 19, then Feynman = 94

thoughts?

M 2

T 3

V 4

H 5

A 6

E 7

O 8

I 9

U 10

B 11

G 12

K 13

P 14

S 15

F 16

W 17

Z 18

Y 19

C 20

L 21

N 22

R 23

D 24

!!!!!

Your values indeed work! (At least for the named I tried.)

There is still the matter of the missing 4 letters/values, j, q, x, & y.

I got so excited that I had to post a message before thinking about how y might fit in.

How exactly did you solve it? Good old linear algebra? If so, which names did you exclude/include?

Great job!

Cheers,

w=f[z]

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If there is a definite answer, then we would have to account for the letters j, q, x, and y. And we would also have to determine if the values start at 1 and go to 26, or start at 2 and go to 27. My guess would be that they start at 1.

Seems like in order to maximize our confidence that we have figured out the correct value for y, we should also figure out the values for j, q, & x.

My thought is that there should be a pattern to how the letters are assigned values. Determining what that pattern is should be the next step.

I'm so excited I'm about to pee my pants!

Cheers

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My methodology started out poorly...

1st up was to boot up excel, my tried and true numerical analysis program. Then I simply laid out a matrix of all 23 used letters, each row being an equation for one of the names. Then I attempted gaussian elimination. This is when I saw that Y can equate to anything I choose, without effecting the rest of equations. At this point, my answers were way off base, so I tried using the solver tool in excel on the first 23 names (23 becuase I only had 23 potential letters) (got the method from googling "Excel Simultaneous Equations", initial guess was 1 for all letters)

Bammo! I got my answers in seconds

So I tried the linear algebra, but failed, then used my easy tool to solve.

Given the problem definition, I'm still not convinced that there is a definate solution, or that the missing letters are of any consequence. We'll need to see if anyone can find a pattern in this.

Not sure if I can, but I can post the excel file if anyone wants, or e-mail, as that will work.

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Excel... damn, I was trying Maple and Mathematica when good ol' Excel would have done the job.

So... if the letters go from 1 to 26, then j, q, x, and y can have possible values of 1, 19, 25, & 26. Which is which is yet to be determined.

I am going to give Excel a shot when I get time.

Thanks!

Edit:

It seems like there are only three possibilities;

1. the values were assigned randomly, or

2. there is a pattern in the numbers themselves, or

3. the numbers were mapped to each name/physicist.

If 1, then there is no hope of finding y except for guessing (1 out of 4 chance). If 3, then one would need the "key" to decipher the code. My initial guess is that it is 2 - as that would be the easiest to implement (a formula so to speak). The fact that they redid the question (i.e., new values to the names compared to what I originally worked from) makes me even that more confident that it is possibility 2.

Cheers

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if you change the values of the names you don't have to exclude any names as you can always make a system A X = B consistent by juggling the values of B. So i guess someone made the problem easier by doing that (but doesn't tell you how to assign the other letters as i cannot see a pattern there - if you can please enlighten me) , which leaves the original problem wide open. My money is still on finding the three physicists which don't fit (which fits into the general knowledge section of these uber IQ tests) and arriving at a 22 x 22 system which you can solve for integer values.

w=f[z], do you want my mathematica notebook?

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w=f[z], do you want my mathematica notebook?

Hi RR,

Thanks, but I think cjohnso0 found the correct values. They seem to work (at least for all the names I tried...). So I probably don't need your mathematica notebook. Thanks though!

I think it's a matter of finding the pattern (if there is one...). Nothing jumps out at me yet.

Cheers,

w=f[z]

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My money is still on finding the three physicists which don't fit

Do you think it will make any difference which three you remove ?

Why don't you try it, remove three randomly, check the values and then remove three other.

I doubt you will get different values.

EDIT 1:

I think the numbers are picked randomly -> No pattern, except for the gap in the line where Y fits nicely.

(The example with Glaser is not changed.)

EDIT2:

I forgot, well done cjohnso0, nice work! All the numbers fit.

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Do you think it will make any difference which three you remove ?

Why don't you try it, remove three randomly, check the values and then remove three other.

I doubt you will get different values.

You are kidding right? Excluding equations changes your solutions, and there are 2300 different solution sets for the original problem.

For example, consider the system

$x + y = 2$ (1)

$x - y = 0$ (2)

$x + y = 1$ (3)

Removing equation 1 gives you the solution set $(x, y) = (\frac{1}{2}, \frac{1}{2})$ while excluding equation 2 still leaves an inconsistent system. Removing equation 3 gives the solution set $(x, y) = (1, 1)$.

So bar solving the 2300 systems and picking out the integer solutions and going from there as a brute force method i don't see that as a "fair" attempt at the solution.

EDIT 1:

I think the numbers are picked randomly -> No pattern, except for the gap in the line where Y fits nicely.

(The example with Glaser is not changed.)

Huh? Gap in the line?

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river_rat,

In your example, your system of equations does not have a solution. If you were to start with a solution, you could then build any number of equations which would satisfy it:

x,y,z = 1,1,1

x + y = 2

x - z = 0

y + 2z = 3

2x - 5y + z = -2

Given these 4 equations, you can remove any one and still get the same solution, assuming the equations you choose have at least one occurence of each variable.

I just ran the numbers using 3 different sets of 23 from the available 25, and the solution came up the same. I'm not going to try all 2300

I'm still not sure that there will be a solution to the initial problem.

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You are kidding right? Excluding equations changes your solutions, and there are 2300 different solution sets for the original problem.

Well, I solved it manually and I got the same answers as cjohnso0 using different equations.

Huh? Gap in the line?

:-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :confused:

"-""M""T" "V""H" "A""E" "O""I" "U""B" "G" "K""P" "S""F""W""Z"GAP "C""L" "N""R" "D""-" "-"

All other available values are outside of the span of values, so in that case Y could have any value.

Since it should be solvable and there is only one gap, I guess thats where Y belong.

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OK, here is the question copied from the PDF-file and my solution...

(Hope it don't contains to many typos.)

Each letter has an associated numerical value attached to it, and the total of all the letters equals the physicist's total value. For example, if the letters G, L, A, S, E, and R had the values 12, 7, 9, 14, 21, and 5, respectively, American physicist Glaser would have a numerical value of 68.

Your objective is to figure out what the last physicist -- Feynman -- should be valued.

+------------------+------------------+

! ROENTGEN=104_____! PERRIN=98________!

! LORENTZ=102______! RICHARDSON=155___!

! CURIE=69_________! HEISENBERG=118___!

! MICHELSON=109____! SCHRODINGER=168__!

! MARCONI=90_______! ANDERSON=127_____!

! KAMERLINGH=120___! DAVISSON=103_____!

! PLANCK=96________! FERMI=57_________!

! STARK=60_________! STERN=70_________!

! EINSTEIN=94______! BLOCK=73_________!

! BOHR=47__________! ZERNIKE=99_______!

! MILLIKAN=103_____! CHERENKOV=109____!

! SIEGBAHN=87______! FEYNMAN=?________!

+------------------+------------------+

Without "J", "Q" and "X" there are 26 equations with 23 unknowns.

Step I:

First I save the following equations to last, they are the only ones with one letter in.

"U" C+U+R+I+E=69

"W" C+H+A+D+W+I+C+K=114

"F" F+E+R+M+I=57

"Y" F+E+Y+N+M+A+N=?

(With Feyman removed Fermi goes too.)

Now I have 22 equations with 19 unknowns.

Step II:

There are only two "Z" and "V" so I rule them out by pairing them up.

"Z" L+O+R+E+N+T+Z+99=Z+E+R+N+I+K+E+102 => L+O+T=E+I+K+3

"V" C+H+E+R+E+N+K+O+V+103=D+A+V+I+S+S+O+N+109 => C+2E+H+K+R=A+D+I+2S+6

There are only four "D" and "B", they are removed now.

"D" A+N+D+E+R+S+O+N+C+2E+H+K+R=A+D+I+2S+6+127 => C+3E+H+K+2N+O+2R=I+S+133

R+I+C+H+A+R+D+S+O+N+168=S+C+H+R+O+D+I+N+G+E+R+155 => A+13=E+G

"B" H+E+I+S+E+N+B+E+R+G+87=S+I+E+G+B+A+H+N+118 => 2E+R=A+31

B+O+H+R+73=B+L+O+C+K+47 => H+R+26=C+K+L

If I save Kamerlingh to last, there is only two "G" left.

"G" K+A+M+E+R+L+I+N+G+H=120

Remove the last two "G" by pairing them up.

A+13+R+O+E+N+T+G+E+N=104+E+G => A+E+2N+O+R+T=91

Now there are 14 equations with 14 unknowns.

Step III:

I have been trying to find more shortcuts but failed, so lets do the messy part...

(But since this is a test for uber-IQ, there might still be more.)

"O" L+O+T=E+I+K+3 => O=E+I+K+3-L-T Insert "E+I+K+3-L-T" for "O" in the other equations.

C+3E+H+K+2N+O+2R=I+S+133 => C+4E+H+2K+2N+2R=L+S+T+130

A+E+2N+O+R+T=91 => A+2E+I+K+2N+R=L+88

M+I+C+H+E+L+S+O+N=109 => C+2E+H+2I+K+M+N+S=T+106

M+A+R+C+O+N+I=90 => A+C+E+2I+K+M+N+R=L+T+87

"C" H+R+26=C+K+L => C=H+R+26-K-L Insert "H+R+26-K-L" for "C" in the other equations.

C+4E+H+2K+2N+2R=L+S+T+130 => 4E+2H+K+2N+3R=2L+S+T+104

C+2E+H+2I+K+M+N+S=T+106 => 2E+2H+2I+M+N+R+S=L+T+80

A+C+E+2I+K+M+N+R=L+T+87 => A+E+H+2I+M+N+2R=2L+T+61

P+L+A+N+C+K=96 => A+H+N+P+R=70

"A" 2E+R=A+31 => A=2E+R-31 Insert "2E+R-31" for "A" in the other equations.

A+2E+I+K+2N+R=L+88 => 4E+I+K+2N+2R=L+119

A+E+H+2I+M+N+2R=2L+T+61 => 3E+H+2I+M+N+3R=2L+T+92

A+H+N+P+R=70 => 2E+H+N+P+2R=101

L+I+P+P+M+A+N=88 => 2E+I+L+M+N+2P+R=119

S+T+A+R+K=60 => 2E+K+2R+S+T=91

M+I+L+L+I+K+A+N=103 => 2E+2I+K+2L+M+N+R=134

"L" 4E+I+K+2N+2R=L+119 => L=4E+I+K+2N+2R-119 Insert "L" in the other equations.

4E+2H+K+2N+3R=2L+S+T+104 => 2H+134=4E+2I+K+2N+R+S+T

2E+2H+2I+M+N+R+S=L+T+80 => 2H+I+M+S+39=2E+K+N+R+T

3E+H+2I+M+N+3R=2L+T+92 => H+M+146=5E+2K+3N+R+T

2E+I+L+M+N+2P+R=119 => 6E+2I+K+M+3N+2P+3R=238

2E+2I+K+2L+M+N+R=134 => 10E+4I+3K+M+5N+5R=372

"M" H+M+146=5E+2K+3N+R+T => M=5E+2K+3N+R+T-H-146 Insert "M" in the other equations.

2H+I+M+S+39=2E+K+N+R+T => 3E+H+I+K+2N+S=107

6E+2I+K+M+3N+2P+3R=238 => 11E+2I+3K+6N+2P+4R+T=H+384

10E+4I+3K+M+5N+5R=372 => 15E+4I+5K+8N+6R+T=H+518

"H" 15E+4I+5K+8N+6R+T=H+518 => H=15E+4I+5K+8N+6R+T-518 Insert "H" in the other equations.

2H+134=4E+2I+K+2N+R+S+T => 26E+6I+9K+14N+11R+T=S+902

3E+H+I+K+2N+S=107 => 18E+5I+6K+10N+6R+S+T=625

2E+H+N+P+2R=101 => 17E+4I+5K+9N+P+8R+T=619

11E+2I+3K+6N+2P+4R+T=H+384 => 2P+134=4E+2I+2K+2N+2R

"P" P+E+R+R+I+N=98 => P=98-E-I-N-2R Insert "P" in the other equations.

17E+4I+5K+9N+P+8R+T=619 => 16E+3I+5K+8N+6R+T=521

2P+134=4E+2I+2K+2N+2R => 330=6E+4I+2K+4N+6R

"K" 2E+K+2R+S+T=91 => K=91-2E-2R-S-T Insert "K" in the other equations.

26E+6I+9K+14N+11R+T=S+902 => 8E+6I+14N=7R+10S+8T+83

16E+3I+5K+8N+6R+T=521 => 6E+3I+8N=4R+5S+4T+66

18E+5I+6K+10N+6R+S+T=625 => 6E+5I+10N=6R+5S+5T+79

330=6E+4I+2K+4N+6R => 2S+2T+148=2E+4I+4N+2R

"R" S+T+E+R+N=70 => R=70-E-N-S-T Insert "R" in the other equations.

8E+6I+14N=7R+10S+8T+83 => 15E+6I+21N=3S+T+573

6E+3I+8N=4R+5S+4T+66 => 10E+3I+12N=S+346

6E+5I+10N=6R+5S+5T+79 => 12E+5I+16N+S+T=499

2S+2T+148=2E+4I+4N+2R => 4S+4T+8=4I+2N

"S" E+I+N+S+T+E+I+N=94 => S=94-2E-2I-2N-T Insert "S" in the other equations.

15E+6I+21N=3S+T+573 => 21E+12I+27N+2T=855

10E+3I+12N=S+346 => 12E+5I+14N+T=440

12E+5I+16N+S+T=499 => 10E+3I+14N=405

4S+4T+8=4I+2N => 384=8E+12I+10N

"T" 12E+5I+14N+T=440 => T=440-12E-5I-14N Insert "T" in the other equation.

21E+12I+27N+2T=855 => 2I+25=3E+N

"N" 2I+25=3E+N => N=2I+25-3E Insert "N" in the other two equations.

10E+3I+14N=405 => 31I=32E+55

384=8E+12I+10N => 22E+134=32I

"I" 31I=32E+55 => I=(32E+55)/31 Insert "I" in the last equation.

22E+134=32I => 2394=342E

"E" 2394=342E => E=2394/342=7

Step IV:

Solving the other equations backwards with accumulating aquired values.

I=(32E+55)/31=(32*7+55)/31=9

N=2I+25-3E=2*9+25-3*7=22

T=440-12E-5I-14N=440-12*7-5*9-14*22=3

S=94-2E-2I-2N-T=94-2*7-2*9-2*22-3=15

R=70-E-N-S-T=70-7-22-15-3=23

K=91-2E-2R-S-T=91-2*7-2*23-15-3=13

P=98-E-I-N-2R=98-7-9-22-2*23=14

H=15E+4I+5K+8N+6R+T-518=15*7+4*9+5*13+8*22+6*23+3-518=5

M=5E+2K+3N+R+T-H-146=5*7+2*13+3*22+23+3-5-146=2

L=4E+I+K+2N+2R-119=4*7+9+13+2*22+2*23-119=21

A=2E+R-31=2*7+23-31=6

C=H+R+26-K-L=5+23+26-13-21=20

O=E+I+K+3-L-T=7+9+13+3-21-3=8

Step V:

Use the saved formulas to find the last unknowns.

K+A+M+E+R+L+I+N+G=120 => G=120-K-A-M-E-R-L-I-N-H

G=120-K-A-M-E-R-L-I-N-H=120-13-6-2-7-23-21-9-22-5=12

B+O+H+R=47 => B=47-O-H-R

B=47-O-H-R=47-8-5-23=11

A+N+D+E+R+S+O+N=127 => D=127-A-N-E-R-S-O-N

D=127-A-N-E-R-S-O-N=127-6-22-7-23-15-8-22=24

D+A+V+I+S+S+O+N=103 => V=103-D-A-I-S-S-O-N

V=103-D-A-I-S-S-O-N=103-24-6-9-15-15-8-22=4

L+O+R+E+N+T+Z=102 => Z=102-L-O-R-E-N-T

Z=102-L-O-R-E-N-T=102-21-8-23-7-22-3=18

C+U+R+I+E=69 => U=69-C-R-I-E

U=69-C-R-I-E=69-20-23-9-7=10

C+H+A+D+W+I+C+K=114 => W=114-C-H-A-D-I-C-K

W=114-C-H-A-D-I-C-K=114-20-5-6-24-9-20-13=17

F+E+R+M+I=57 => F=57-E-R-M-I

F=57-E-R-M-I=57-7-23-2-9=16

And last finding the missing value for Y:

A= 6 - 01=?

B=11 - 02=M

C=20 - 03=T

D=24 - 04=V

E= 7 - 05=H

F=16 - 06=A

G=12 - 07=E

H= 5 - 08=O

I= 9 - 09=I

J= ? - 10=U

K=13 - 11=B

L=21 - 12=G

M= 2 - 13=K

N=22 - 14=P

O= 8 - 15=S

P=14 - 16=F

Q= ? - 17=W

R=23 - 18=Z

S=15 - 19=? -> Y

T= 3 - 20=C

U=10 - 21=L

V= 4 - 22=N

W=17 - 23=R

X= ? - 24=D

Y= ? - 25=?

Z=18 - 26=?

All the values are individual and in incremental steps, except for the gap at 19, so my logical conclusion is Y=19.

=> F+E+Y+N+M+A+N=16+7+19+22+2+6+22=94

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