Extremely difficult question from an IQ test...

[river_rat]

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.

 

Ok, but the original question (which i am still concerned about if you followed my posts) was inconsistent and overdetermined. IMHO the second question was someone's cop out the original problem, so we still need to solve the problem this thread is about.

 

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

 

For which problem, the first or second?

 

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

 

Why is that?

[cjohnso0]

river_rat,

 

I'm following you now, I was missing that you were referring to the original problem posted.

 

All of my work was for the .pdf version.

 

After playing with the original question, I cannot find a solution, at least not one that makes sense, yet.

 

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

by this I mean a mathematical solution for the .pdf question. If the original question posted is correct, there must be some obscure solution that's beyond me at this point.

[river_rat]

So i got Mathematica to churn through all the combinations and arrived at the following:

 

If you remove Marconi, Stern and either Davison or Cherenkov from the list you arrive at a consistent system with an integer solution (the same solution btw)

 

A -> 25

B -> 22

C -> 6

D -> 19

E -> 12

F -> 6

G -> 20

H -> 0

I -> 27

K -> 3

L -> 22

M -> 8

N -> 1

O -> 12

P -> 20

R -> 9

S -> 7

T -> 22

U -> 13

V -> 26

W ->4

Z -> 13

 

Some Trivia about each

 

Marconi - Received prize "in recognition of their contributions to the development of wireless telegraphy"

 

Davisson - Received prize "for their experimental discovery of the diffraction of electrons by crystals

 

Stern - Received prize "for his contribution to the development of the molecular ray method and his discovery of the magnetic moment of the proton"

 

Cherenkov - Received prize "for the discovery and the interpretation of the Cherenkov effect"

[doG]

Just an observation here. If this is really a question from an IQ test then it should be designed to be solvable mentally, without the use of Excel, Mathematica or even a calculator, within the time given for it on the test it's from. This suggests that there is some solution that the intelligent can deduce from the given information using the mind and only the mind. Am I missing something here?

[w=f[z]]

Hi doG,

Take a look at the other questions on the test (a link to the pdf was provided). Normally I'd agree with you, but this test had no time limit (unlike most internet IQ tests) and the other questions are pretty tough too.

 

My take, is that the folks that would do well on this test would be able to teach themselves the necessary skills needed to answer the questions. Just a thought.

 

Cheers

[river_rat]

Adding to what w=f[z] said, the resulting systems here are sparse so you could do them by hand if you had accountant like accuracy. Being able to accurately copy down 550 odd numbers has never been a strong point in my mathematical career though lol.

[doG]

;346584']Hi doG' date='

Take a look at the other questions on the test (a link to the pdf was provided). Normally I'd agree with you, but this test had no time limit (unlike most internet IQ tests) and the other questions are pretty tough too.[/quote']

 

It looks to me like you should be able to do it by hand in the space provided. I certainly wouldn't expect it to be OK to do any analysis with a computer for any of the questions.

[w=f[z]]

These were the instructions on my version (annoyingly enough... not on the pdf link). I'll type in what I have so forgive any typos.

 

We have designed this high-ceiling test to be so challenging that gifted individuals will get average scores, and only geniuses will be able to achieve the highest scores. This test is deceptively difficult and is, therefore, only recommended for persons whose IQ's are in the top one percent and above. Considering that the average person who successfully completes this test answers less than two questions correctly, and that the average professional mathematician of physics professor would barely be able to answer half of them correctly, you can see the extreme difficult nature of this test. Of the hundreds of people who complete this test each week, nobody has been able to achieve a perfect score. All of the Society's tests tend to be more difficult than the average IQ test, but this one in particular has the unique ability to determine IQ scores in the extreme high-end of the spectrum and has one of the highest ceilings of any test currently available.

 

Although a mastery of advanced mathematics is not required, proficiency in general areas of math will be beneficial in answering many of the questions. Understanding the wording of the problem, which is occasionally rather intricate, and the ability to deduce various complex concepts, are also essential. This test offers a wide variety of different types of questions and is well suited to the assessment of general intelligence.

 

The use of reference materials, books, calculators, and computers are permitted. We ask that you do not share your answers in any public forum in order to maintain the integrity of the test. There is no penalty for wrong answers so it is to your advantage to guess whenever you are unsure of an answer.

 

This test has been developed and copyrighted by Nathan Haselbauer and Mike Dickheiser and is the exclusive property of teh International High IQ society. The accuracy of both the questions and the answers has been verified by members of the International Intelligence Testing Committee. The test results are constantly reviewed by the Society to ensure the most accurate score possible.

 

The test is also available in a printed version for your convenience.

 

 

A few comments;

This test is no longer on their site, so I thought posting this problem here was fair game now.

 

Also notice their mention of "advanced mathematics" and "reference materials."

 

And here's a minor nitpick, but they encourage "guessing." While guessing might increase your IQ "score" (if you guess correctly)... a correct answer by guessing I don't think is an accurate reflection of one's real IQ (whatever that is...).

 

 

Okay, so there are the rules that came with the test. The use of computers was fair game.

 

Cheers

[psi20]

Spyman, when you combined the equations like in step II, did you subtract?

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

If you did, shouldn't the L have a negative sign in front?

 

This is too hard for me. I'm just going to accept that I'm a person of inferior intelligence.

[Spyman]

Spyman, when you combined the equations like in step II, did you subtract?

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

If you did, shouldn't the L have a negative sign in front?

 

This is too hard for me. I'm just going to accept that I'm a person of inferior intelligence.

Well, I think your question might be better answered by the people in the math section, but here goes:

 

The equal sign, "=", is like the center or axis of a weighing scale or a seesaw for children, and it says that the weights of both sides are equal, they are in balance.

 

Weighing scale -> http://en.wikipedia.org/wiki/Weighing_scale

Seesaw -> http://en.wikipedia.org/wiki/Seesaw

 

In math you are allowed to change the weights as long as they always are in balance...

...you must change both sides with the same amount:

If X=5 then 1+X=5+1 or X-5=5-5 or X-X=5-X or 2*X=5*2 or X/2=5/2 and so on...

You are also allowed to flip the left side and the right side, if X=5 then 5=X too.

 

OK, now lets make an easy example, where I think you can easily guess what A, B and C are:

A+B=3, A+C=4, B+C=5

 

B+C=5 is equal to -C+B+C=5-C which is B=5-C.

A+C=4 is equal to -C+A+C=4-C which is A=4-C.

 

A+B=3 but we now know that A=4-C so (4-C)+B=3 and since B=5-C it's also true that (4-C)+(5-C)=3.

 

If (4-C)+(5-C)=3 then 4-C+5-C=3 and 4+5=9 so 9-C-C=3 and 2*-C=-C-C so 9-2C=3.

 

Balancing the scales: -3+9-2C=3-3 gives 6-2C=0 and 2C+6-2C=0+2C gives 6=2C and 6/2=2C/2 which gives 3=C.

 

And since A=4-C it's also A=4-(3) so A=1, and B=5-C is also B=5-(3) so B=2.

Solved, A=1, B=2 and C=3.

 

Now to my combinations of equations, if A+B=3 then 3=A+B so I can use it on A+C=4 in different ways:

If A+C=4 then (3)+A+C=4+(3) or (A+B)+A+C=4+(A+B) or (A+B)+A+C=4+(3) or (3)+A+C=4+(A+B).

 

If I wish to remove A, I choose the last one, (3)+A+C=4+(A+B).

 

Balancing the scales: 3+A+C=4+A+B, -3+3+A+C=4+A+B-3 -> A+C=1+A+B and -A+A+C=1+A+B-A -> C=1+B

 

 

L+O+R+E+N+T+Z=102 and Z+E+R+N+I+K+E=99:

I want to remove Z so I need them to be on opposite sides to cancel out:

Z+E+R+N+I+K+E=99 so 99=Z+E+R+N+I+K+E.

L+O+R+E+N+T+Z=102 so (99)+L+O+R+E+N+T+Z=102+(Z+E+R+N+I+K+E)

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

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

-Z+L+O+R+E+N+T+Z=3+Z+E+R+N+I+K+E-Z -> L+O+R+E+N+T=3+E+R+N+I+K+E

-R+L+O+R+E+N+T=3+E+R+N+I+K+E-R -> L+O+E+N+T=3+E+N+I+K+E

-E+L+O+E+N+T=3+E+N+I+K+E-E -> L+O+N+T=3+N+I+K+E

-N+L+O+N+T=3+N+I+K+E-N -> L+O+T=3+I+K+E

Then I only rearranged them in alphabetical order, 3+I+K+E is equal to E+I+K+3 so L+O+T=E+I+K+3.

 

The math is not that hard, the "intelligence" part is finding the shortest route to the answer.

(Which BTW, I'm pretty sure I failed.)

 

But if all equations are true any route will give the correct answer.

 

The first question asked in the OP is a harder nut to crack, since all equations can't be true.

 

Personally I think it has a typo somewhere inside it, the only other option is that they changed it fundamentally.

[w=f[z]]

Just so we're all on the same page (including myself...) -

So the status of this problem is as follows:

 

The original problem (let's call "P1") still remains unsolved as far as the values of the letters (unless river-rat's are correct). If RR's values are correct, then the pattern seems tougher than the second version of this problem.

 

The second (pdf) version (let's call "P2") has been solved for the letters (except j, q, x, & y) by cjohnso0 (and spyman). The pattern between the letters remains unsolved.

 

P2 was solved by linear-algebra techniques (& excel) omitting any of the names (so that we get an augmented matrix-type problem), but P1 requires the omission of only certain names - which seems to require more knowledge in order to determine which ones to omit. Once that is done, then linear-algebra techniques give the values (such as the ones river-rat got).

 

 

Did I get all that right?

 

 

River-rat, did you get your values by reducing an augmented matrix omitting only certain names? I'm not sure I caught exactly how you got your values (i.e., the technique).

 

Cheers,

w=f[z]

[river_rat]

Hi w=f[z]

 

I got those solutions by just kicking out the equations for Marconi, Stern and Davison and solving the resulting system and then by kicking out the equations for Marconi, Stern and Cherenkov and solving the resulting linear system.

[mikeiq]

"S" in

...

 

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

Now when it is practically solved I must inform you that chioce Y=19

is not the most logical one.

In that case would be FEYNMAN=94 ,but take a look how all the others are valued.

Do you see / note that no two physicists are equally valued ?

Somebody already got 94 there

 

FEYNMAN=101 is correct answer.

Ask Nathan H. to explain you why Y=26 (he created that puzzle ,not me).

 

Contact email:

nathan@highiqsociety.org

president@highiqsociety.org

[mikeiq]

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.

I repeat that FEYNMAN=101 is the ONLY correct solution to the puzzle!

 

Bye,

 

Mike

[aliasgherman]

This is a very good thread indeed. I would like to point out that Fenyman can get 94, because BOTH michelson and cherenkov have got the same value(109).

[foodchain]

Well, the longest amount of letters in a name is 11, I think the shortest is four. The max value assigned that is visible does not exceed 130, while the min does not go below 40. This should be able to suggest a range of values to use, which I imagine happen to be only whole numbers, though that is not specified. I also think that would be the best place to start, as it not simply trying to crack the code via iterations with different values, but rather attempting to subtract whole numbers that simply cannot fall in range with the letters used. I would think working from the upper and lower bounds to a mean would be a good function with different sets of whole numbers on a range of say 1 through 20.

[mikeiq]

@aliasgherman;

 

There are mathematically consistent system & subsystems of lin. eq.+

logically implemented subsystem to the puzzle.

Alltogather they generate unique solution ,which isn't 94.

One side of the puzzle people here realized fine,but another part seems they didn't.

Instead of giving you complete solution for this case I'll show you one more puzzle which belongs to the same cathegory of the puzzles (courtesy of IHIQS):

 

 

----------------------------------------|

BARROW =71

NEWTON =70

WHISTON =104

SAUNDERSON=129

COLSON = 51

WARING =92

MILNER =58

WOODHOUSE=108

TURTON =80

AIRY =46

BABBAGE =84

KING =45

LARMOR =58

DIRAC =52

LIGHTHILL =130

HAWKING =____?

-----------------------------------|

 

 

 

Text is the same as before and your objective is to figure out

professor's Hawking numerical value .

 

The situation here is a little bit different becouse now you have 19

unknowns but "only" 15 simultaneous equations at first glance.

Brute force approach will not work again.

However,just like in the previous puzzle the solution is unique.

 

Have fun.

 

cheers

[BenTheMan]

So, if you assume that the thing is unique, just put it into a system of equations and solve for x.

 

you should be able to take the inverse of a 26x26 matrix with pencil and paper:)

[Raffael]

 

 

----------------------------------------|

BARROW =71

NEWTON =70

WHISTON =104

SAUNDERSON=129

COLSON = 51

WARING =92

MILNER =58

WOODHOUSE=108

TURTON =80

AIRY =46

BABBAGE =84

KING =45

LARMOR =58

DIRAC =52

LIGHTHILL =130

HAWKING =____?

-----------------------------------|

 

 

 

Text is the same as before and your objective is to figure out

professor's Hawking numerical value .

 

The situation here is a little bit different becouse now you have 19

unknowns but "only" 15 simultaneous equations at first glance.

Brute force approach will not work again.

However,just like in the previous puzzle the solution is unique.

 

 

A=17,B=5,C=9,D=1,E=13,G=22,H=26,I=11,K=2,L=6,M=4,N=10,O=3,R=14,S=20,T=16,U=21,W=18,Y=4

 

Is there room for another solution under assumption that values were taken from set {1,2,3,...,26}?

 

 

 

 

:-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :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.

 

In the "Feynman" puzzle I'd rather go with Y=26.

You have a neat regularity(pattern) in ascribing the values to the letters you're supposed to find.

A little help-notice how grouped the vowel's values are : A=6,E=7,I=9,O=8,U=10.

Coincidence?I don't think so...