IQ quickie

Which letter is missing? G B L g _ h E Z I

Several, for example 'a'.

"i"

The ordered set " G B L g i h E Z I " is an ESUOS(*).

An ESUOS has the following property:

If someone steals a letter from the extreme positions you still have a copy of the letter inside.

(*) ESUOS: Erratic Silly and Useless Ordered Set

Nope...

s of course!

That's it! :-)

the sum of the numbers of the letters in the alphabet of the last 4 minus the sum of the first 4. 48-28=20, which is s.

Uhhh not even close. Much simpler. :-P

If you read the letters upside down, the solution should reveal itself.

Good one! Think outside the box and if that doesn't work, flip the box.

I thing the real solution is S ;) nice one

I tried turning it but my tablet keeps turning the picture.

You should turn yourself around your tablet ;-)

I will give this puzzle to my students on a pice of paper, instead of on the white board :-)

Find-the-next-element-in-the-sequence-type problems are equivalent to "what number am I thinking of?". It is easy to come up with a rule to generate any number/letter you want for the next element.

See, for example: http://mathforum.org/library/drmath/view/56894.html

I am aware of that, Dustin. Hence the word "quickie" in the title - not meant to be taken too seriously.

No offense meant :)

Unfortunately, the latest political fad in America is "education reform" which is code for "vilify teachers and privatize education". The mechanism that leads to school takeovers and mass layoffs is multiple-choice standardized testing which includes questions like "what's the next number in this sequence."

I take issue with politicians who have no knowledge of math firing math teachers over questions that have no right answer.

If you are interested in pattern matching riddles which have no arbitrary solution, then try the so called "Bongard Problems". http://en.m.wikipedia.org/wiki/Bongard_problem

Why do you say that they don't have arbitrary solutions ?

The definition of wikipedia reads as follows: "All the diagrams from set A have a common factor or attribute, which is lacking in all the diagrams of set B. The problem is to find, or to formulate, convincingly, the common factor."

Clearly, there are arbitrary solutions, such as the property "Being one of the objects in set A". The word "convincingly" is a clear confirmation of that.

Men, this is one of those things that once you've seen you can neve unsee it! 8-)

im stupid?
i turned upside down and don't see anything.

oooops

Am I the only one who "read it upside-down" without turning either myself or it?

now i see the light

It's inspired by the Big Bang Theory episode in which Sheldon asks what the best number is, and Rajesh suggests 5,318,008.

@Carroll - 88 (I knew it already)

@Nathan - I did write it on paper and turned the paper :-)

@Nathan Nope, I can read and write upside down without trouble. Students always try to turn their notes, while I even can correct there notes or add something upside down. Problem is that they are so startled by that fact that they don't listen to what I try to explain :-)

Here is one that only 17% of MENSA members could solve.

If the hour and minute hand on a clock are on top of each other - and the second hand is just beyond the 49 second mark - what time is it?

9h(49+1/11)
My father told me that when he was eight he was asked by his teacher: "at what times of the day are the two hands of the clock on top of each other". He said noon and midnight and then started to compute in his head the time after 1h05, when she said "come on this is easy, there are twelve times when that happen in the morning and same for the afternoon!" He could never convince her she was wrong as he did not have a watch...

It is 4h:21m:49.09s

This means 4h and 20+20/11 minutes.

Crelo must be one of the above average MENSANs

Now try this. If you get it in 30 minutes you are good. If you get it in less that 30 seconds then you are very good.

Take a wood ball and drill a hole through it. The depth of the hole is 6 inches (not the diameter) What is the volume of what is left of this wooden ball. The original diameter of the ball is not given.

That's easy. It's a square pi number :)

For the clock problem the Carroll message helped a lot to figure out the problem.

For the sphere, is obvious from the problem the volume should not depend on the sphere radius. Unfortunately, searching for the volumes formula (sphere and sphere caps, which I do not know by heart) I found a page with the solution.

Still, I cannot imagine this can be done in 30 seconds :-)

Since the diameter of the hole and the diameter of the ball is not given, this means the volume must be the same for any ball for the puzzle to have meaning (proof of this is probably fun, but unnecessary to solve the puzzle). Realizing this, solving the puzzle is easy. I therefore look at the case where the hole has 0 diameter. This corresponds to a ball of diameter 6 inches (radis 3 inches). The volume of a sphere is 4/3*pi*r^3, which cives a solution of 36*pi cube inches. Unfortunately, it took me more than 30 seconds to realize this - I'm guessing around two minutes :-(

Wow, some of these things are still doing the rounds 40 years later :-)

It could be 36*pi cubic inches.

Right around 30 seconds. Using what turned out to be Marius's reasoning.

If it's true that that remaining volume is independent of the sphere's diameter, I think I could prove it, but I need a bigger blackboard.

Let L be half the hole length, r the hole radius, R the sphere radius.

It is easy to see that r^2+L^2=R^2

Now look at a cross section (perpendicular to the hole) that you'll have to integrate on 2L :
S=piR^2 - pi r^2 = pi(R^2-(R^2-L^2)) = piL^2 independent of R...

I let the integral as an exercise to the reader.

I found a blackboard, but I got stuck on the integral of the cube of the sine of theta.

Take R as the radius of the sphere, and 2*d (=6 inches) as the depth of the hole [d=3 inches is half the depth of the hole]. The volume of the entire sphere is 4*pi/3 * R**3, and it consists of two spherical caps, the hole (which is a cylinder of height 2*d and radius r) and the remainder that we're looking for.

The second formula for the volume of a spherical cap (When all else fails, and usually a lot sooner than that, try Wikipedia.) is

pi/3 * h**2 * (3*R - h)

where h is the height of the spherical cap and R is the radius of the sphere. The height of the spherical cap is the sphere radius minus half the depth of the hole:

h = R - d

Substituting for h in the cap volume, we get

pi/3 * (R - d)**2 * (3*R - R + d) =

pi/3 * (R**2 - 2*d*R + d**2) * (2*R + d) =

pi/3 * (2*R**3 - 4*d*R**2 + 2*R*d**2 + d*R**2 - 2*R*d**2 + d**3) =

pi/3 * (2*R**3 - 3*d*R**2 + d**3)

The volume of the cylinder of height 2*d and radius r, is

2*pi * r**2 * d

The radius of the cylinder, r, is determined from

d**2 + r**2 = R**2

r**2 = R**2 - d**2

making the volume of the cylinder

2*pi * (R**2 - d**2) * d =

2*pi*d*R**2 - 2*pi*d**3

At long last, the volume of sphere minus the volume of the two spherical caps minus the volume of the cylinder is now

4*pi/3 * R**3 - 2*pi/3 * (2*R**3 - 3*d*R**2 + d**3) - 2*pi*d*R**2 + 2*pi*d**3 =

factoring out pi/3,

pi/3 * (4*R**3 - 2*(2*R**3 - 3*d*R**2 + d**3) - 6*d*R**2 + 6*d**3) =

pi/3 * (4*R**3 - 4*R**3 + 6*d*R**2 - 2*d**3 - 6*d*R**2 + 6*d**3) =

pi/3 * (4*d**3) =

4*pi/3 * d**3

That is, the volume of the remainder is independent of the volume of the sphere, and when d=3 inches, the volume of the remainder is 4*pi/3 * 3*3*3 = 36 pi cubic inches.

It took a lot longer than 30 seconds to come up with the proof. I too, looked up the volume of a cap on Wikipedia and could thus prove that the answer is independent of the diameter of the ball.

But what only takes 30 seconds is that since no diameter is given then I can assume any ball diameter.

And as Marley correctly said, just assume the hole has a diameter of zero and all you need to calculate is a volume of a ball with diameter of 6 (radius of 3)and that should only take about 30 seconds.

Some of you guys are pretty damn smart. Now how many of you smart guys are successful with girls. HA!

I don't get the relationship you do between drilling this hole, ball diameter, less than 30 seconds and success with girls, can you elaborate?

More seriously, speaking of smart people, have you seen that a Chinese mathematician called Yitang Zhank may have proved a weaker form of the twin primes conjecture?
The twin primes conjecture states that there are infinitely many primes p, p+2.
What he seems to have proved is that there are infinitely many primes with small gap, p,p+g, with g less than 70 millions.
This is another hint that primes behave mostly like random numbers.
See here for more details.

That is fantastic result! Um, but I don't agree that "primes behave like random numbers" - if they did, then the "probability" that a random integer x is prime should be asymptotically equal to the product over the primes p less than sqrt(x) of (1 - 1 / p), while it was proved by Mertens that it is asymptotically off by a factor 2 e^(-gamma) where gamma = 0.5772156649...

Yes this was an over simplification, Primes are in P and "of course" random numbers are not.

The reference at the end of the pre-cited paper : "Terry Tao on the dichotomy between structure and randomness" states things more precisely.

"Marley correctly said, just assume the hole has a diameter of zero..."

Marius was first to publish.

Without the proof, you don't really know that the correct answer isn't, e.g., "It depends on how big the piece of wood was initially, but the minimum left over is 36 pi cubic inches." (which would be the correct answer if the question were about drilling an "oblate ellipsoid" rather than "sphere")

To answer Carroll's question:

While girls should be impressed by math abilities they are usually not. They seem to be more impressed by bank account size and whether or not you are captain of the football team.

Fortunately, my girl is an ex-research scientist and now a college professor and pretty damn cute so I am very lucky. But she may be more impressed by my mountain biking ability than my math or Twixt skills. Who knows :)

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I don’t get the relationship you do between drilling this hole, ball diameter, less than 30 seconds and success with girls, can you elaborate?

The 56 pages paper from Yitang Zhang has been posted by a mathoverflow fellow on his dropbox. Enjoy!