#3 came from the back of Omni magazine in the late 80's. I presented it to the head of the math dept. at my university, and it stumped him good.

Isn't it redundant to say "fawn obsequiously"?

It is a very stupid mode of reclutament.
The google bussines not is an inteligence question is other things

Do all of these actually have answers? I'm neither a math guy nor a programmer, so most of them just blew me away.

Not all icosahedra have 20-sides.
They should have specified regular
icosahedron. Sloppy....

http://mathworld.wolfram.com/Icosahedron.html

Those math problems are of no more than moderate difficulty. Sure, their wording is sloppy, but I thought it was apparent what they were looking for. The stuff about Google, though, who knows. I think it was an interesting recruiting strategy, got me interested.

Brad W. and I solved questions 16 and 17.

Question 4 has several typographical errors --
A - should probably have another comma before the last clause
C - should probably be "MPoRPG" instead of "MMoRPG",
E - definitely needs another comma before the last clause, and the word "a" before "whole"

Question 10 is not really completely specified -- it really should state that the resistors are the connections between the lattice points.

Yoshi, were you able to get an explicit formula for f in question 17, or did you just evaluate it at key values?

I'll do a write up describing our solution(s) to problem 17 and I'll post a link to it here.

Concering question 4, the proper vernacular is actually massively multiplayer online role playing game, or MMORPG. Ack, it appears on second reading that you were right but swapped your terms.

SPOLER:

one
one one
two ones
one two one one
one one one two two ones

D@MNIT!!! I read the spoiler before trying to solve the riddle, i assumed it was a BS question with no answer until i kept reading through. many are simple but vague. I liked #10.

Spoiler:

Resistors in parallel have final resistance R=1/(1/R1) + (1/R2) + ... + (1/Rn); which for infinite n = 0!!! so unless the resistors arent connected there is no resistance.

The most beautiful math equation would have to be the equation for total energy. I'm leaving a few out but E = mgh + (1/2)kX^2 + (1/2)mv^2
E = mc^2 may be incorrect and is currently thought of as E=MC^2(1/((1-(V/C)2)0.5)-1) which i believe engineers made up so they wouldnt be pressured to make a rocket go the speed of light. "WE CAN'T DO IT SO IT MUST BE IMPOSSIBLE!!!"(if v=c e=infinity)

The most beautiful math equation would have to be the equation for total energy.

Uh, no. That's a physics equation! They were talking about math equations!

The most beautiful maths equation would be:

e^(pi * i) - 1 = 0

The most beautiful math equation ever would definnitely have to be 1=1, or 0=0, or just X=X; true beauty.

it happens that one of my friend's dream is to work in google, makasih ya discanin hehe...

312211

I think the correct equation is

e^(pi * i) + 1 = 0

(not -1)

> should probably be "MPoRPG" instead of "MMoRPG"

MMoRPG means "massively multiplayer online role playing game. No typo there.

But it says "MPoRPG" already.

the most beautiful math equation?

S.M.I^2L.E.=infinity

The most beautiful math equation out there is 2+2=4.

"beautiful math equation" is an oxymoron :S

Nearly all mathematical equations are beautiful. Mathematics is filled with beauty. This is definitely *not* an oxymoron.

Question 4 answer (C) already says "MPoRPG"

And yes, the most beautiful mathematical equation ever written would have to be

exp(i * pi) + 1 = 0

or

exp(i * pi) = -1

(where i = sqrt(-1))

karma police, arrest this girl-- her hitler hairdo is making me feel ill

"It is a very stupid mode of reclutament.
The google bussines not is an inteligence question is other things

Posted by: Excatedra"

What does this mean, reclutment?

For old dog. "reclutment" sounds like baby talk for recruitment. How profound can you get? :-)

Well, Boys and Girls!
Of course Google is not recruiting by means of that so called test. It is marketing, they are using a very imaginative and clever way to "segment" their target markets--I mean us.

What magazines did you find these GLAT tests in? I know that there is also other recruiting methods that google uses. I found the GLAT test in MIT's Technology Review magazine.

This test has also been in Scientific American.

Q.12 is 2 easy, the answer is 2 simple, it's actually a riddle, heres how you tell.

Most beautiful equation will obviously result in the most beautiful number, look of most beautiful number in google and Voila, PHI, 1.618......an common and curious equation to get phi, can be calculated with all fives, therefore we, get, 5 ^ .5 * .5 + .5 = Φ

Bingo.

I chose bright pink, lime green and light purple for the colours :)

My name is "shoji".came from Japan.I am writing PUICP-blog
I searched "GLAT" with Google-Japan and visited here.
I seldom understand English.
Although the track back was carried out, since it is Japanese, it cannot display.
Please delete, if convenience is bad.

the most beautiful math equation to me is a^2 + b^2 = c^2

for my three colors I would pick green

based on how the Army does things I wouldn't go over 11 for a project team. However, based on my software experience I wouldn't go over 5 typically - and oftentimes I prefer 1 :O)

too bad question 6 doesn't have the option - not be impressed and continue on as if the person is anyone else I work with..

and for what I'd do in the bay area on a beautful sunday afternoon - spend it with my family.

#20

96!

ten
nine
sixty
ninety
seventy
sixtysix

So.

ninetysix (96)!

For only post my correct URL.

the most beautiful math equation to me is : -0

A hint for 17.

10% of all the digits to be counted are 1's.

Spoiler: pof ujnft sfo sp sif sfosi qmvt uxp
(roll every letter back by one)

#20
ninetysix or
tengoogol

About the resistor problem (#10)...

Consider applying 1V across any two points.
No more than 1Amp will go through any single
resistor right? Now consider that any point
has exactly 4 resistors coming to it. So an
upper bound of current going to any point is
4Amps. Doesn't that mean a lower bound for
resistance between any two points is 1/4 Ohms?
How could the answer be 0?

dbozy, the net current at any point in the lattice is always 0 (Kirchoff's current law). You cannot have 4 amps coming into one point -- where would it go?

That said I don't know the answer. I calculated the resistance looking at all possible paths of length 3 (3 of them) and of length 5 (found 11 of them) and got 0.313 ohms. The real answer must be less than that, considering all the longer paths that the current can take.

I meant the 2 points where the circuit starts and ends. So if you had a 1V battery and hooked the + to one point, and the - to the second point. Shouldn't the current flowing out of the second point into the battery be bounded by 4Amps? Otherwise that means one of the four resistors going to it is carrying more than 1Amp, which implies that the voltage across it greater than 1V, which doesn't make sense.

About your calculation - how are you handling overlapping paths?

> Spoiler: pof ujnft sfo sp sif sfosi qmvt uxp
> (roll every letter back by one)

You meant: pof ujnft ufo up uif ufoui qmvt uxp.
And that's not the right answer either.

Whats the answer to #3? Is it a variation of the Fibonacci sequence?

Think i got it nevermind

I think the answer for coloring icosahedron would 3^20, as there are three colors and 20 different sides to color. Hence, the number of different ways would be 3*3*3*...20 times.

I think the answer for coloring icosahedron would 3^20, as there are three colors and 20 different sides to color. Hence, the number of different ways would be 3*3*3*...20 times.

#17 spoiler

the number is 199981

what happens to f(i)/i when i ---> infinity? I got it by brute force, but I cannot go much higher than 1999999 or maybe 19999999 (these are the function peaks)

the icosohedron problem is not nearly 3^20 because of rotation equivalence

the f(i)=i problem. did you really count the ones up to 199981?:
that number seems way too low. did anyone else come up with a number?

Yes, I did too... but I guess cheated. I used a computer program.

I mean, I also got 199981.

#16: Seems too simple, but what about an equilateral triangle. Use compass and straight edge to find perpendicular bisectors of each side. Bisectors meet at point P. All resulting triangles would have equal perimeters. The question didn't say the point needed to lie outside ABC.
Another solution would be a right angle isosceles triangle with AC as the long side. Construct a perpendicular bisector of AC. Duplicate the distance of the midpoint to B along the bisector as segment midpoint to P. AC and BP are diagonals of a square. The resulting triangles have equal perimeters.

anybody get 19?

For 19:
Mearly let N be an empty set. Then all K (null) is contained in the set of yours (null). Since a null set is contained in every set this is the simpleist way to answer the question. However, is null less than 1/2 null? No for 0=1/2*0. Hmm this is odd.

damn tough! or making fun or us!

what does hair? color has to do with personal information and that too with (*) sign?

Hey,

Google is a very different kind of company. I like it! Google on!

Later_

does 17 = 3^20/2/10? This would divide out bi-symetry and the rotation equivalencies...

#16 has an analytical solution for any acute isosceles triangle. For triangle with sides measuring a, a, b, the distance y from the vertex is b(2a-b)/[2(a-b+L)], where L is the altitude sqrt(a^2 + (b/2)^2). For a 17-17-16 triangle, altitude is 15 and y=9. The perimiters of the 3 new triangles are 36.

Just don't know how to do it geometrically. Sure, 17-(16/2)=9, but consider a 13-13-10 triangle. L=12 and y=5.33_ P=26.66_.

Spoilers:
an icosahedron always has 20 sides (see http://mathworld.wolfram.com/Icosahedron.html) but all sides are not necessarily of equal shape or size, so no rotation equivolency or bi symmetry except in the special case of a regular icosahedron. The answer is 3^20, or (3^20)/20 for the special case of a regular icosahedron considering rotation equivolency. For the triangle problem, it is easy to do for the special case of an equilateral triangle, but you must be able to do it for the more general case of any triangle for which it is possible, and prove it is impossible for other triangles. For resistors, lack of a current does not make the resistors dissapear. The greater the path length you consider, the greater the resistance of each, but the greater the number of paths as well, you need to construct an equation using equations for parallel resistors and knowledge of how many paths exist with any given length(this is the tough part), and it will approach some number as the length gets large. Most of the others have me stumped, great test!

the most beautiful math equation to me is a^2 + b^2 = c^2

this isn't an equation, it is a theorem.

#16: For an arbitrary right triangle, treat the hypotenuse as the diagonal of a rectangle. The 4th corner is the solution point.

Most beautiful equation question(#12) 36-24-36 of course!

Most beautiful equation question(#12) 36-24-36 of course!

You are all morons.

some of u guys are v. clever and the most beautiful equation is 1233*A*-14567=1 MILLION HARHAR HAR

Google is the best company in the world. I cant wait for them to take it over. Imagine a Google OS. Or even google food!? That would be the most efficient food ever concieved. Google cars that take you wherever you want to go in 0.092 secs. I want to work for Google :'(.............

Darn, I got #3 wrong. I saw it as the sums being a fibonacci sequence with a single one starting on alternating left/right side of the line followed by twos, followed by ones. My answer was: 12222211

I figure it out a tad bit late (the billboard puzzel)... just looked at their first problem dedicated at E (base of the natural logarithm). Its a dam sequence that repeates itself.

2.718281828 and the pattern repeats till it gets to the last 4 number and continues on to infinity. O'well. Its been years since I did any real math.

This would be the answer to phase 2 of their original math posting. You are going to miss out on a few good people.

Maybe not as a replacement but as a complement.

Coolest hack I have done would have to be opening th refrigerator without the key....was really cool

wow that sounds fun but im a good boy so i dont do stuff like that can some1 post an answer to 19 thanks

no9 answer please

i noe all the answers..9)type f^*$ as fast as u can 10^99 times.
4)E
6)A
rule 1:these answers are all right
rule 2: if they're wrong, you may read this over again or repeat your answers while commiting suicide.

OK here goes:
1. 777589-188103=589486

3. 312211

4. b ( not d because i am a fighter and want to find what the gnomes are hiding rather than finding an exit path)

6. e (obviously)

8. (3^20)/20 for regular

12. most beautiful equation would've to be that of a hyperbola... u can c the essential's of a woman in that can't you ?

16. The answer applies to all triangles in which a solution is possible. Keep a compass and an edge ready. Draw tri(ABC) such that AB>AC>BC (keep slightly different to attain solution). Mark X on AC such that AB=AX. mark Z such that in between C and X (keep CZ as small as possible). Mark Z' between A and X such that XZ'=CZ. Now mark Z" on AB such that BZ"=CZ'. Mark Y on BC such that CY=CA. With compass on A and measure BY mark Y' on AC and Y'on AB. NOW, draw perpendicular bisectors of Z"Y" and ZY'. These 2 meet in P.
Read above explanation 4 times to get the logic. took me 2hrs to reach solution.. that's normal for me.

20. C ninetysix or onegoogel ( yea its one not ten count properly)

also
9. convert the rectangle into an empty rectangular carton without a lid by drawing 3 easy lines... and wallah you have something more than an empty area.. an empty volume

other answers coming shortly

people from google please feel free to clarify anything.

sorry missed again
8. colors to be chosen have to be RGB cause these are the only colors... all others are combinations of these

ok seems answer to 8 is wrong

ok seems answer to 8 is wrong

holy shit look what I found when my karma was killed by my dogma..

Thank God I finished the test before I found this site!! It's no fun when you see a spoiler.

It took me a month, but I solved every one except #10.
I think #19 was the hardest, and I don't think making "n" equal to null or zero is a valid answer.

Overall, a very fun and challenging test! I hope they continue the campaign.

Thank you very much,I like this website! Do it good.

Regarding the resistance problem, you are all wrong! The equivalent resistance doesnt have a value! Instead, it varies, depending on the potencial between the 2 points. y? Because R=U/I and if u apply a really small voltage, the current will be very small. In the limit, it can only consist of a single electron passing by. In that case, it will follow the least resistent path, which is 3 ohm. So, the resistance will vary between 3 ohm and a lower limit value, when the number of electrons won't matter and everything can be treated as a continuos function. The lower limit is 1/2, and i wont explain how. If you are curious, just ask by e-mail :)

GLAT: Google Labs Aptitude Test
When you flip through Dr. Dobbs you may run across the green Google Labs Aptitude Test.

I wonder if the like of Cedric, Adam Bosworth, Josh Block, and Neal Gafter applied in this way? :)

Some of the questions are hilarious:

Solve this cryptic equation: WWWDOT - GOOGLE = DOTCOM
Write a haiku describing possible methods for predicting search traffic smoothly
1
11
21
1211
111221
What is the next line?
What's broken with Unix? How would you fix it?
This space left intentionally blank. Please fill it with something that improves upon emptiness
On an infinite, two-dimensional, rectangular lattic of 1-ohm resistors, what is the resistance between two nodes that are a knight's move away?
In your opinion, what is the most beautiful math equation ever derived?
A fun place to work?

Number 8
http://mathworld.wolfram.com/PolyhedronColoring.html gives all the info shown below:
First you have to define if the coloring is "proper" or not. If so, no two faces with a common edge can be the same color.
The answer in this case is 144. Find this in the book "Mathematical Recreations and Essays"(W.W.R. Ball and H.S.M. Coxeter), pp. 239-242.
If you have no such restrictions for adjacent faces, then:
1- find the graph automorphisms for the skeleton of the icosahedron
2-remove all symmetries that invert a face, leaving pure rotational symmetries only
3-find the induced symmetry group for the faces
4- apply the Polya enumeration theorem, an extension of Burnside's theorem.
The resultant polynomial is 2/5n^4 + 1/3n^8 +1/4n^10 +1/60n^20, which for n=3 is 58130055

the answer to number 3 is 312211. Read the lines, they are descriptions of the lines before.
1
11 "one one"
21 "two ones"
1211 "one two, one one"
111221 "one one, one two, two ones"

so it follows that the next one is 312211 "three ones, two twos, one one"

Hello. I just wanted to give a quick greeting and tell you I enjoyed reading your material.

x²-x-1=0

The most beatifull math equation may be x²-x-1=0, since x yields the Phi/phi number, the Gold number or Divine Proportion. Everything that is beatifull have that proportion, everything. From clothes design to music.

x= square root of 5 -/+ 1 all divided by 2.

The most beautiful math equation is undisputably euler's equation because it contains all the essential terms you could need in math...e, pi, i (imaginary number), and both identities 0 and 1.

e^(pi * i) + 1 = 0.

Eulers equation is definately in a league of its own.

Add this to the test:
16
17
20
22
24
31
100
121
?

I think you guys have made a mistake on question number 8. it needs to be 20*19*18, due to the fact that you must apply three colors to the surfaces. thus, the answer is 6840 ways.

The most beautiful math equation is "answer to life the universe and everything = 42" because if you google the first clause the google calculator spits out 42... Allusion to Hitchhikers' Guide to the Galaxy

i think next line is
12212111

#1: 222110 - 111199 = 110911

#15.)
Marginal Revenue = Marginal Cost

To understand this concept read this:
http://ingrimayne.saintjoe.edu/econ/MakeProfit/OptimalInputs.html

About the resistor question, how about we think about it _conceptually_, because I have not yet seen a correct answer to it (not even you, Miguel). Ben has been the closest so far, with zero resistance.

The lattice is _infinite_, which means that there is no definition as to when it ends, and there are no more resistors. If you drew a picture and stared at it a while, you would see that there are many, many branches between the two points (infinite, in fact), and even resistors that transfer between them, which reduce the total resistance further.

Each parallel circuit (branch, if you want to call it that) reduces the total resistance to less than the resistance of either of the branches. If we imagine an infinite set of 1-ohm branches, we see that the first branch is 1 ohm; add the second branch, and (using the previously stated formulas) we get 1/2 ohms; add the third, and we get 1/3 ohms, and so on. This number gets _infinitely_ small.

So, when Ben said that the total resistance was zero, he was very close. In fact, he was infinitely close. The total resistance between any two points on an _infinite_ two-dimensional lattice is always infinitesimal.

I found this interesting....

e^(pi*i)+1=0

AND

e^(phi*i)-1=0

I love math equations. That question #3 is a neat one to introduce to people at a restaurant while you are waiting for food. That's where I came across it anyways :P

This is really neat:

http://porocrom.blogspot.com

#15 is clearly 3, since you are tripling the project team, and definitely won't get anywhere near that in return. 2 would be the best from 1, since you are only doubling the team, but likely tripling the throughput.

imho

For solving problem 17:

#include
#include

int main() {
unsigned long int n;
unsigned long int ac=1;
char str[10];

for (n=2; n int len;
char* i;
len = (int)floorf(log10(n)+1);

sprintf(str, "%d", n);
for (i=str; i if (*i=='1') ac++;
if (ac==n) {
printf("%d\n",n);
break;
}
}
return 0;
}

Mathworld has posted solutions to nearly the entire test:

http://mathworld.wolfram.com/news/2004-10-13/google/

#20 is b = a googleplex

i believe that the case of the resistors, q10 - electricity takes the path of least resistance, so the infinite lattice can be ignored. all that is important is the set of resistors directly connected in the L-shape of the knight's move. So, there are 2 paths to be considered (draw it out). They both take in 3 resistors of 1-ohm each. this constitutes a 2-path parallel resistor circuit with 3 ohms each. R=1/(1/R1) + (1/R2) which is 1/(1/3 + 1/3) or 1/(2/3) or 3/2 so,

R=1.5

I havn't been convinced on anybody's answer for number 10...so I'm giving it my best shot.

I wish I could post a diagram...Hopefully my asci-art will show up. (^w^ and z are resistors)
Pick a random point on this infinite lattice and call it A and consider A to be the beginning of your circuit. The 'end' of your circuit is at infinity B(yes, some may say that you can't have a single point that denotes the end because there are an infinite number of ends...to that I say it doesn't matter, bear with me or picture the entire sheet to be on the surface of an infinite sphere and point B is the point directy on the other side). Apply a volage between A and and B...I do this because it makes sense to most people that we need a 'source' and 'drain' for a standard electric circuit.
You'll notice that A has 4 neighbours each one resistor away...call these points 1,2,3,4 respectively. see diagram...

o--^w^--o--^w^-13--^w^--o--^w^--o
|              |               |              |              |
z              z                z              z              z
z              z                z              z              z
|              |                |              |              |
o--^w^-14--^w^--5--^w^-15--^w^--o
|              |                |              |              |
z              z                z              z              z
z              z                z              z              z
|              |                |              |              |
16-^w^--6--^w^--1--^w^--7--^w^-17
|              |                |              |              |
z              z                z              z              z
z              z                z              z              z
|              |                |              |              |
8--^w^--2--^w^--A--^w^--3--^w^--9--^w^--19
|              |                |              |              |
z              z                z              z              z
z              z                z              z              z
|              |                |              |              |
20-^w^-10--^w^--4-^w^-11--^w^--21
|              |                |              |              |
z              z                z              z              z
z              z                z              z              z
|              |                |              |              |
0--^w^-22--^w^-12-^w^-23-^w^--o
                                  |
                                  z
                                  |
                                  24
Because of symmetry, we can safely say that the potential drop in each of the 4 directions is equal...therefore each of point 1,2,3,4 are at the same potential with respect to B (ground). Taking one step further, using symmetry once again, points 5,6,7,8,9,10,11,12 are also at equal potential with respect to ground. To get 'a knight move away', we'll do this once more...points 13-24 will be at an equivalent potential (note point 18 is off to the left of point 8).
No that we've 'hit' the knight move and we've determined which points are at equivalent potential, we can re-draw the circuit...
                                                             __ww__
                                                           |__ww__|
                                                           |__ww__|
                                                           |__ww__|
                                                           |__ww__|
                                                           |__ww__|
                                     __ww__      |__ww__|
                                  |__ww__|      |__ww__|
          __ww__        |__ww__|      |__ww__|
        |                |      |__ww__|      |__ww__|
        |__ww__|      |__ww__|      |                |
        |                |      |__ww__|      |__ww__|
----o                   o---|__ww__o---o__ww__|---o
        |__ww__|      |__ww__|      |                |
        |                |      |__ww__|      |__ww__|
        |__ww__|      |__ww__|      |__ww__|
                                |__ww__|      |__ww__|
                                |__ww__|      |__ww__|
                                                         |__ww__|
                                                         |__ww__|
                                                         |__ww__|
                                                         |__ww__|

Using simple series/parallel formulas yields:
R/4 + R/12 + R/20
== R*23/60

Note that the number in parellel grows according to the formula 8*n-4 where n is the number of steps...1 being the first. So the general formula for the resistance of the circuit would be:
R/4 + R/12 + R/20 + ... + R/(8*n-4)
where n is the number of 'steps' away from A. Also note that for no steps away, the resistance is zero.

Clearly, as n->infinity, this is no better than 1/8th of the harmonic series which is divergent...therefore the resistance of the entire circuit is infinite.

Thoughts?

The most beautiful math equation would be the polar equation for a cardioid. r = a(1-cos(theta)) Its the graph of a heart.

your test sucks, i thought you were smarter

mensa sent the GLAT to members
most members were pretty stumped

not all isohedron have 20 sides

number one is quite simple

555378
177104
------
378274

for question 10 and other questions:
http://mathworld.wolfram.com/news/2004-10-13/google/

Hey, the answer to the resistor question is 1/infinity...

Because the resistance of parallel resistors is product over sum...

TTYL
D

Red, green, and blue are not the colors that create all others, Vijay. On can eisily see when they buy printer cartridges(or take a high school art class) that the true primary colors are cyan, magenta, and yellow. Red it magenta darkened with black, green is yellow and blue, and blue is cyan darkened with black.
The most beautiful equation: a/b=b/c where b+c=a or Phi. Phi is the golden ratio found in almost all elements of nature.

I believe that RGB are indeed the primaries as magenta is created from red/blue mixtures, cyan from green/blue and yellow from red/green.

On second look, Cuule, after some research you both appear to be correct - http://home.att.net/~B-P.TRUSCIO/COLOR.htm RGB are primary colors for light, and CMY are the primaries for paint.

this is amazing. some of the questions do not have definite answers and are opinion. others are concrete fact. i learned how to answer 17 in ninth grade AP algebra.

Question 10:
Since it's a knight's move away, it's strongly implied the sides as well as legs of each ladder step has the resistor. Since normal resistors have 5% tolerance, you can stop after calculating for the first 20 most significant resistors. The infinite ends of the lattice become quickly insignificant anyway. It seems clear 18 will be plenty. You can set the rest of the lattice to 2 ohms and then 3 ohms at any point to get a min and max for the rest of the lattice (open circuit and then short circuit the rest of it). Each leg of the lattice as it splits off is 2.730 +/- 0.003 ohms after just doing 6 resistors. But there's an annoying resistor in the "middle" after reducing it to 5 resistors in a whetstone bridge, so i still had to apply 1 volt, use symmetry, sum voltages and currents, solve, and divide the applied voltage by the resultant current to get 1.31 ohms instead of the 0.773 solution in the math world link.

Question 10: I built the circuit with 1K ohm resistors and it came out to 1.30k...so my 1.31 appears correct. To clarify my previous post: to analyze it, break off the lattice where it starts repeating indefinitely and solve that part the standard way for 3, 6, 9, and 12 resistors to get 3, 2.75, 2.733, 2.7321. Those are maximums. Now short out the last resistor to get the minimums: 2, 2.67, 2.727, 2.7317. For this second minimums case, on your calculator you're punching 2 inverse + 1 inverse + 2 inverse + 1 inverse....infinity. In the first case it's the same, but start with 3. Now this resistor goes in parrarel with a 1 ohm resistor and then in series with another to get 1.7320 for an equivalent R in two opposing sides of a square, 1 ohm resistors on the 2 other sides, another one across the diagonal (the one that's a real pain), then the resistance is measure across the other diagonal.

Question 15: somewhere i heard 5, but if the team members have equal skill in the same area, then the answer is 1 because comminication within a brain is much more efficient than the communication channels between brains.

Question 12: i've always thought bell curve stuff was pretty wild. It quantifies that which we do not know and we were able to derive it from the binomial (1's and 0's) even before communication theory. Maybe i should say the boolean algebra of NAND gates since any non-quantum thinking machine can be created from just NAND gates.

Q 11: work on the computer

Q 5: It's ease of use for the masses. Get Google to realize this and then fix it.

Q 19: a senseless question

Q 21: Peace.

Q 17: TTS

oops, i did the resistors for a ladder, not a lattice....correct answer is 4/pi - 1/2= 0.77324

Q.1 wwwdot = google dotcom
wwwdot = go to lego code
lego = latin to build or to gather
lego ergo sum = I gather therefore I am.

Just curious but could this be one way to "solve this cryptic equation" since lego is entwined in the google culture.

Q 3

In the strict sense 312211 but in the playful sense the next line could begin "You are in amaze..." from Q4

Q 8
can a icosahedron be hollow? and if so would that double the faces?

There was a riddle "I have two coins that add up to 15 cents one is not a nickel, what are they? answer...the one that is not a nickel is a dime the "other one is a nickel"

so the three colors...the question does not state it is the same three colors that you are picking from each time.

There is no definite answer to #12 it is an opinion. With that said, in my opinion simplicity is beauty 2+2=4 looks rather attractive to me.

want to take the aptitude test to enter google.How can go about it?

No self-respecting person would not want to work for a company that is pretentious enough to publish such a test.

As for the basic colors, red, green and blue can form any colors when ADDING the wavelengths (as when you point two light beams of different colors to a white surface).
Cyan, magenta and yellow can create all colors when subtracting wavelengths (if you mix two colors of paint the result will absorb the wavelengths color 1 absorbed plus the wavelengths color 2 absorbed).

That is why computer screens use RGB (they EMMIT light) and printers use CMY (they absorb light).

I believe the most beautiful math equation, if I had to choose one, would be that a^n+b^n=c^n is impossible when a, b c and n are natural and n>2.

For question #3 the counting of the previous line above will essentially give us our next line which is 312211 or three ones two twos one one, BUT...if you add your totals on the side a sequence forms.. 1 =1
1 1 =2 (2+1)=3(next line)
2 2 =3 (3+2)=5(next line)
1 2 1 1 =5 (5+3)=8(next line)
1 1 1 2 2 1 =8 (8+5)=13?
but 312211 dosen't equal 13 it equals 10 and thows off this pattern.

For question #3 the counting of the previous line above will essentially give us our next line which is 312211 or three ones two twos one one, BUT...if you add your totals on the side a sequence forms..
1 =1
1 1 =2 (2+1)=3(next line)
2 1 =3 (3+2)=5(next line)
1 2 1 1 =5 (5+3)=8(next line)
1 1 1 2 2 1 =8 (8+5)=13?
but 312211 dosen't equal 13 it equals 10 and thows off this pattern.

re: #10.
Folks, the answer to this must be at _least_ .5 ohm. Each node on the grid connects to everything else by 4 1-ohm resistors. Four 1-ohm resistors in parallel is .25 Ohm (IIRC). That's .25 Ohm at each node in the knight's move. So, imagine one node, 4 resistors leading out connected to a metal plate connected to 4 resistors leading into the 2nd node. In other words, even ignoring the rest of the grid, you absolutely cannot avoid going through two groups of 4 resistors (which amounts to .5 ohm). So, give up this zero (or infinite) resistance silliness.

Isn't there a glaring problem with the resistor question....You cannot have an infinite "rectangle" on side is infinitely long, no other side can be longer, thus there appears to be no answer since the question is invalid. I'm not a math guru, please email me if this logic is wrong.

re:#10
Their can be no answer, because the question is invalid. You cannot have an infinite "rectangle". A rectangle has one side longer than another, but if one side is infinitely long, how can the other side be longer? If this logic is flawed feel free to let me know...

"Red, green, and blue are not the colors that create all others, Vijay. On can eisily see when they buy printer cartridges(or take a high school art class) that the true primary colors are cyan, magenta, and yellow. Red it magenta darkened with black, green is yellow and blue, and blue is cyan darkened with black."

You're confusing the subjects. RGB nor CMYK are "primary colors" (whatever that means). In strictly physical terms, any monochrome colour has an equal right to be called "primary" or whatever. In broader terms, "color" is a mere subjective reflection of the visible light, a concept that exists in our minds only.

When reffering to RGB or CMYK as to "primary colors" we speak about the two basic methods of producing colored output: Additive (RGB, when we need to add certain colors to the otherwise black background to achieve the desired color - CRT is an excellent example) and subtractive (CMYK, when we need to take away certain colors to the otherwise white background to achieve the desired color - color printing on a white paper is an excellent example).

In CMYK you get clear red by mixing clear magenta and clear yellow. Magenta darkened with black would have to be dark magenta. :-)

Note that you can achieve every possible color through additive method, but not so when using subtractive methods. There are certain colors which no color-printer on this planet could produce, and every color monitor/TV on this planet will produce them (easily).

Hello there. I went to Google seeking employment, took this test, and answered all the questions correctly. I eventually got tired of working in the mailroom, and now teach creative writing at Harvard. Just kidding!

I wrote perl scripts that solve #1 and #17.

They're on my personal website at:
www.ontologistics.net/GLAT/

Answer to Problem 1:
W == 7
D == 5
O == 8
T == 9
G == 1
L == 0
((E == 6) AND (M == 3)) OR ((E == 3) AND (M == 6))

188103 + 589486 == 777589
W = 7 D = 5 O = 8 T = 9 G = 1 L = 0 E = 3 C = 4 M = 6
188106 + 589483 == 777589
W = 7 D = 5 O = 8 T = 9 G = 1 L = 0 E = 6 C = 4 M = 3

Answer to problem
199981

I wrote perl scripts which solve #1 and #17.

The answers, scripts, and justifications are on my personal web page at: http://www.ontologistics.net/GLAT/

Problem 1: (both answers)
W = 7 D = 5 O = 8 T = 9 G = 1 L = 0 E = 3 C = 4 M = 6
------------ OR -----------------
W = 7 D = 5 O = 8 T = 9 G = 1 L = 0 E = 6 C = 4 M = 3

Problem 17:
199981

#1
it's either no answer or multiple answer, depends on allowence of repeat number...
since:
L=0
T-E=M
10+D-G=C
W-1-O=T
W-O=O
W-G=D

and also there's another set of possibility.. either of these sets of possibility can have at least one answer if same value is allowed for different letter... stupid question!

problem 1

Yes, of course there are lots of answers to problem #1 if you let some of the variables equal each other.

If you require that all variables be unique, then there are only two answers.

These two answers happen to coincide with the statement about M and E being interchangable.

The question is a good one. It's just intentionally ambiguous.

Have to agree with pshb20 that the most beautiful math equation is Euler's equation expressed as e^(pi * i) = -1. It contains all the two most famous transcendental numbers e and pi, the two binary digits, and the imaginary number (-1)^0.5.
When my second year math prof finished the proof of e^(i*theta) = cos (theta) + i*sin (theta), and then substituted pi for theta, the class applauded.

I think you all are overthinking number 17. Note that it says "What is the next largest n such that f(n)=n?" rather than "What is the next n." Bearing in mind that people at Google also have to be creative thinkers, and given that "next largest" generally means the largest one smaller than the current one (ie "I have three kids. Tom is the largest, Joe is the next largest."), I believe the answer they are looking for would be zero. It's right (provided they consider zero a whole number :cP), and it takes a lot less time to find.

Of course, given the previously mentioned typos/grammar errors, maybe they just wrote the problem incorrectly.