Some interesting brainteasers Starter I used

[ƒ(x) → ∞]

Some interesting brainteasersStarterI used

Some interesting brainteasers

Starter



I used factorials for this one, but it will be interesting to see how you attempt your answer.

Head Scratcher



These questions are designed so that anybody above the age of about 13 can answer them without much mathematical experience.

If this is a success then I will reveal the answer this time next week and come up with a new one [MONDAY 28^TH DECEMBER 2009]

 

[Jimmy Snyder]

I used factorials for this one

Brainteasers factorial?

 

[davee123]

I used factorials for this one

I thought he might mean: 1!

DaveE

 

[ƒ(x)]

Am i missing something? what's the brainteaser?

 

[Jimmy Snyder]

Am i missing something? what's the brainteaser?

That's the brainteaser.

 

[ƒ(x)]

That's the brainteaser.

Oh, I assumed that there was some math problem that i was missing.

 

[D H]

There is something missing. The problem is, what is the problem?

 

[ƒ(x)]

There is something missing. The problem is, what is the problem?

Thats what i said.

 

[ƒ(x)]

Gentlemen, I found the question: https://www.physicsforums.com/showthread.php?t=365013

 

[Jimmy Snyder]

And for the ladies, it's in the same place.

Spoiler

48
8

 

[Borek]

I found the question:

You've missed the idea - it is not about finding questions, but about finding answers.

 

[ƒ(x)]

And for the ladies, it's in the same place.

Spoiler

48
8

How did you do this?

You've missed the idea - it is not about finding questions, but about finding answers.

Considering how I wasn't sure about how to do the questions

 

[Jimmy Snyder]

How did you do this?

For the first one, I counted. I did use a factorial, but it wasn't a big part of the solution.
For the second one, I graphed a few points and the figure appeared.

 

[ƒ(x)]

For the first one, I counted. I did use a factorial, but it wasn't a big part of the solution.
For the second one, I graphed a few points and the figure appeared.

What was your factorial (my teacher for pre-calculus didnt teach)?

For the second, doesn't the function go to infinity?

 

[╔(σ_σ)╝]

The second one is pretty much 4 triangles in the four quadrants with vertices
(-2,0),( 0,2) ,(0,0)

(2,0),( 0,2) ,(0,0)

(0,-2),( 0,0) ,(2,0)

(0,2),( 0,0) ,(2,0)

The area is simply ( 0.5(2)(2)) *4 = 2*4 =8

 

[Borek]

In how many ways can the product of the numbers so that the numbers equal 36?

Wow, I have even no idea what it does mean and Jimmy was able to solve.

Edit: what do we know about colors assigned to dices? What color is four sided dice? Six sided one? Ten sided? Twelve sided? Twenty sided? Not sure what types I had, I can't find my set form RPG times.

 

[ƒ(x)]

The second one is pretty much 4 triangles in the four quadrants with vertices
(-2,0),( 0,2) ,(0,0)

(2,0),( 0,2) ,(0,0)

(0,-2),( 0,0) ,(2,0)

(0,2),( 0,0) ,(2,0)

The area is simply ( 0.5(2)(2)) *4 = 2*4 =8

It is? http://www.wolframalpha.com/input/?i=|y-x|+|y|+=+2

 

[Jimmy Snyder]

Wow, I have even no idea what it does mean and Jimmy was able to solve.

For instance, you could roll a red - 1, orange - 2, yellow - 3, and purple - 6. The product is 36. That's 1, now go find the other 47.

 

[Borek]

red - 12, orange - 1, yellow - 3, purple - 1. Now I need only 46.

 

[ƒ(x)]

For instance, you could roll a red - 1, orange - 2, yellow - 3, and purple - 6. The product is 36. That's 1, now go find the other 47.

I attempted it by finding combinations of 4 whose product is 36 (assuming that they were all 6 sided dice).

I think i got:

1,1,6,6
1,2,3,6
1,3,3,4
2,2,3,3
1,3,3,4

I might have missed one combination. But, since there are 4 identical dice, each one could fill one of the spots. So wouldn't it be 4!*5? Which is 120

 

[Jimmy Snyder]

I attempted it by finding combinations of 4 whose product is 36 (assuming that they were all 6 sided die).

I think i got:

1,1,6,6
1,2,3,6
1,3,3,4
2,2,3,3
1,3,3,4

I might have missed one combination. But, since there are 4 identical dice, each one could fill one of the spots. So wouldn't it be 4!*5? Which is 120

For one thing, you have 1,3,3,4 listed twice. For another, consider the first one 1,1,6,6. In this case I counted yellow-1 purple-1 the same as purple-1 yellow-1 but you counted them as different.

 

[ƒ(x)]

For one thing, you have 1,3,3,4 listed twice. For another, consider the first one 1,1,6,6. In this case I counted yellow-1 purple-1 the same as purple-1 yellow-1 but you counted them as different.

Oh, I did. So its back to 96. Why would you count them the same?

 

[Jimmy Snyder]

Oh, I did. So its back to 96. Why would you count them the same?

Because I can't tell the difference between them just by looking. One yellow die and one purple die look just like one purple die and one yellow one. How do you propose to tell them apart?

 

[ƒ(x)]

Because I can't tell the difference between them just by looking. One yellow die and one purple die look just like one purple die and one yellow one. How do you propose to tell them apart?

Good point. The problem was asking for different outcomes, not total outcomes. 48 it is.

 

[╔(σ_σ)╝]

It is? http://www.wolframalpha.com/input/?i=|y-x|+|y|+=+2

They are equivalent. Graphing exactly is more complicated.If you switch x and y in the original equation you can see my train of thought.

And no the graph does not go to infinity, the graph plotted is not for all values of x and y.

At x=2 or -2 y must be 0 or 2.

If you graph this completely you get some sort of parrallelogram.You get 2 triangles with vertices

(-2,0), (2,2),(2,0)
(2,0), (-2,-2),(2,0)
http://www.wolframalpha.com/input/?i=|y-x|+|y|+=+2,+++-7<+x++<+7,+-2<+y++<+2Again you can see my train of thought by rotating the parralelogram such that the the point (2,2) is on the y axis.

 

[davee123]

Effectively, you can solve by realizing that a function with an absolute value is basically taking TWO functions, one negative, one positive. So, for:

y = |x|

You can graph both:

y = x
y = -x

And that'll give you the basic shape-- you just know that y is always positive, so some areas of the graphs are invalid. With this problem, it's 4 equations:

|y - x| + |y| = 2

A) y - x + y = 2
B) y - x - y = 2
C) -y + x + y = 2
D) -y + x - y = 2

Which simplify to:

A) y = x/2 + 1
B) x = -2
C) x = 2
D) y = x/2 - 1

That's enough to solve the problem, even without invalidating the ranges, which isn't too hard to do. It encloses a parallelogram, whose area is pretty simple to determine.

DaveE

 

[ƒ(x) → ∞]

Spoiler

1,1,6,6 - 6 ways of getting this
2,2,3,3 6 ways of getting this
1,2,3,6 - 24 ways of getting this
1,3,3,4 12 ways of getting this

Therefore the answer is 12+24+6+6 = 48 ways

Spoiler

8