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This problem has been vexing me all day. It comes from an old IQ test. Note that while the test says not to discuss the problems on it in order to maintain its integrity, the test itself closed years ago so this shouldn't matter now. I have managed to narrow it down decisively to 4 possible solutions.

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I've gotten what I thought was the hard part out of the way; that is, I did some matrix witchcraft using an online calculator (this is just a massive linear systems of equations problem) to find the value of all the letters used here... except for "y", which appears only in "Feynman." As it turns out, finding y is the real challenge here.

Each letter has a different value. Here are the letters assigned to each value, in order of value: $$ \begin{array}{cc} \text{Value}&\text{Letter}\\\hline 1& \\ 2& M\\ 3& T\\ 4& V\\ 5& H\\ 6& A\\ 7& E\\ 8& O\\ 9& I\\ \end{array}\hspace{10mm} \begin{array}{cc} \text{Value}&\text{Letter}\\\hline 10& U\\ 11& B\\ 12& G\\ 13& K\\ 14& P\\ 15& S\\ 16& F\\ 17& W\\ 18& Z \end{array}\hspace{10mm} \begin{array}{cc} \text{Value}&\text{Letter}\\\hline 19& \\ 20& C\\ 21& L\\ 22& N\\ 23& R\\ 24& D\\ 25& \\ 26& \\ &\\ \end{array} $$ There are known letters with values 2-24, with one exception (19). I think it's safe to assume that the values start at 1 and end at 26, because there are 26 letters in the alphabet. Since every letter has a different value, I suspect that Y is either 1, 19, 25, or 26.

I've read through every message board on the internet that has this problem. None of them have explained the solution, but one of them seemed to have a user who knew the solution but refused to divulge it. He said the following:

"Besides,are you sure there isn't room for something like TWO independent (but consistent) systems of equations with juggled integers values regarding alphabetical order ? Fact that choice of value for "Y" isn't logically clear / unique is good enough sign for suspicious minds."

In another post:

"There are mathematically consistent system & subsystems of lin. eq.+ logically implemented subsystem to the puzzle. Altogether they generate a unique solution, which isn't 94. One side of the puzzle people here realized fine, but it seems that the other people didn't."

I don't exactly follow him. From what I know of matrices, there is either exactly one solution, infinitely many solutions, or no solution to large systems of equations like these. So the concept of this equation having another unique set of solutions doesn't make sense in my opinion.

One more thing: I don't think it's without significance that when the letters and their values are lined up in order of value, all five vowels appear in a row. Y is "sometimes" a vowel. Is this a clue?

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  • $\begingroup$ To me, it looks like you could solve the problem using a system of linear equations then allow for the math to figure itself out. I say this, because there are 25 equations with answers and less than 25 characters used (e.g. no J and no Q). For each letter, let it be X_1 thru X_26, then set the coeff to be cardinality of how many times the letter appears in the name. Mathematica can do this pretty easily... just takes time to type the equations... after you figure out the value for each letter, writing the last should be easy. After you see there is an answer, you can look for an easier way. $\endgroup$ – VenomFangs Oct 7 '15 at 22:16
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    $\begingroup$ Yes, @VenomFangs, that's pretty much what I did with matrices. The issue was not finding a value for each letter, the issue was finding a value for Y specifically, which only appears in one of the words (Feynman) and therefore cannot be solved using a system of equations. Y must be found as part of a pattern, and it's the pattern that I can't find. $\endgroup$ – Jacob Crofts Oct 7 '15 at 22:42
  • $\begingroup$ The only pattern I see is 6-10 are the vowels: a, e, o, i, u (and they aren't even alphabetically ordered). I haven't been able to find any pattern by plotting the values vs. letters. $\endgroup$ – user1717828 Oct 8 '15 at 12:31
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    $\begingroup$ Maybe the question was originally multiple-choice, and of the four possible values, only one was listed. $\endgroup$ – user3294068 Oct 8 '15 at 13:57
  • $\begingroup$ Perhaps the "second part" is that when you put the scientists in numerical order by score, they are also in some other order (alpha by first name, date of birth, alpha by middle initial, chrono by date of I-don't-know-what" and Therefore Feynman belongs between Someone and AnotherOne and so Y must be whatever-value-would-fit. $\endgroup$ – Kate Gregory Apr 8 '17 at 17:54
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Buried on page 4 of this message board discussion is a post from a member of the High IQ Society stating that the correct answer (according to the person who set the puzzle) is:

FEYNMAN=101 (because Y=26)

However he also includes this:

"note that no two physicists are equally valued"

which is incorrect as

Davisson and Millikan are both 103.

He doesn't explain why the answer is as stated and suggests contacting the puzzle creator. According to the High IQ Society web page, Nathan is still a member, so the stated email address (nathan@highiqsociety.org) should still be valid.

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  • $\begingroup$ @JacobCrofts are you going to follow up on this or are you still looking for more information on the puzzle? $\endgroup$ – user1717828 Oct 12 '15 at 1:07
  • $\begingroup$ @user1717828 I'm trying to figure it out myself. I'll definitely post an update if I figure it out. $\endgroup$ – Jacob Crofts Oct 30 '15 at 4:16
  • $\begingroup$ @user1717828 I tried sending an email to the email address mentioned above, and got the following response from the server: 550 sorry, no mailbox here by that name.. So it looks like that email address is no longer valid. $\endgroup$ – GentlePurpleRain Nov 20 '15 at 13:09
  • $\begingroup$ Seeking further clarification from Nathan Haselbauer seems unlikely to be fruitful: dailycamera.com/boulder-county-news/ci_28721132/… $\endgroup$ – Gareth McCaughan Aug 26 '16 at 13:13
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This puzzle is way above my IQ, but I had fun thinking about it:

After solving the 1st 25 equations, we are left with finding 75 + f('Y'). It is reasonable to assume that the letters form an ordered set (but it's a bit unreasonable to assume that it must start at 'one' instead of 'zero'). Four letters are undefined ('J','Q','X','Y') so 'Y' is not forced unique yet.

What other pattern determines f('Y')? The names are a partial list of 20th century Nobel winners in physics, sorted by Nobel year. Note that 'Block' is an aberrant spelling of 'Bloch' (is that a clue?). We might suspect that external data is needed (but not too obscure) such as the initial letter of the person's first name. That would only introduce the letter 'J'.

It is noteworthy that the vowels are contiguous in the solved variables, but 'O' and 'I' are not in alphabetical order. The font in the puzzle allows them to be interpreted as 'zero' and 'one'. Is that a clue (about parity)? (i.e., 'O' was forced to even and 'I' to odd)

If the variables are sequential (reasonable), they could be numbered 1..26 or 0..25. In either case, only one missing value is even (0 or 26). Is there a reason why 'Y' should be that even value?

If we look at the parity of each sum (odd or even), we get 13 even and 12 odd. Is that enough reason to make 'Feynman' odd? If so, how do we choose between answers 75+0 and 75+26?

Note that Lord Rayleigh (not listed) won in 1904. His partial sum is 83 so he would have been just as valid an entry as Feynman for solving 'Y' so far, but his initial is ambiguous. Is that a clue?

My tentative answer is 101, but I can't exclude 75. Are my parity observations a red herring?

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Ok just a thought, I have no idea, but what if the letters didn't have one value each?

If you could find two values for each letter, then when trying to find the value for $y$, you would have two references for each of the four possibilities and also two for each of the other letters in the word and you could add up those two sets of values.

I personally find it hard to even add things so this is just something that popped into my head and might be totally meaningless and wrong. But thought I'd write it anyway in case it was useful.

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