The worlds first riddle!

Mark:

HEXAGON

If there are no restrictions, then I'd say 7:
one line through the center that is perpendicular to the plane (like an axle)
three lines through the centers of opposite sides (rotation outside the plane)
three lines through opposite vertices (rotation outside the plane)

The axes of rotational symmetry are the three lines joining opposite vertices, the three lines joining the midpoints of opposite sides, and the line through the centre and perpendicular to the hexagon.


CIRCLE & CHORD
(b) 3/2

By symmetry, ACD is an equilateral triangle. Hence its centroid is the centre O of the circle. Since AO = 1, AE = AO + OE = 3/2.


BEACH RACE
4 - sqrt(2)/4 km



4 - sqrt(2)/4 km
Let y be the solution. Let x=4-y. The time to reach the finish line is:
t=(4-x)/6 + (1+x2)1/2/2.
Set the derivative equal to 0:
dt/dx= -1/6 + 1/2 * 1/2 * 2x * (1+x2)-1/2 = 0.
The solution is x=1/81/2.
y=4-x.


Rap:

Circle Arc

1+sin(30)=3/2



I saw a similar problem in an article on high school mathematics titled "My Dog can do Calculus." The set up was a lakeshore game of fetch by the author and his Labrador.

The shore leg is 4-x km, the sea leg is (x^2+1)^(1/2) km
The time for the race is
t=(4-x)/6+(x^2+1)^(1/2)/2 hr
optimize by setting dt/dx=0
-1/6+2x/(2*2(x^2+1)^(1/2))=0
-2(x^2+1)^(1/2)+3(2x)=0
(x^2+1)^(1/2)=3x
x^2+1=9x^2
8x^2=1
x=(1/8)^(1/2) km
so run down the beach 4-(1/8)^(1/2) before entering the water for a best time of 1 hour 8 min 17 sec (squaring the corners would take 1 hr 10 min)


Next time I will relocate to the mountains in an effort to shake your calculations.



Ok- so now the long lived light bulbs (LLLB) have a lmbda of 1/200 hrs and the short lived light bulbs (SLLB) have a lmbda of 1/100 hrs. and the density function is the probability of a light bulb operating after t hrs of operation (the wikipedia link also explained the definition of memory less, or more properly memoryless)

So now I had an understanding of the problem, and hopefully the solution.

Probability of a single light bulb failing is
Pf(E)=1-exp(-lmbda*t) where lmbda is the MTTF (200 hrs for a LLLB and 1000 hrs for a SLLB)

This problem asks for the expected time for a failure of a single light bulb out of a sample of 5 LLLBs and 5 SLLBs.

Since the probability of failure is one minus the probability of surviving so many hours, then what I want is to figure out what the probability of simultaneous survival of the 10 bulbs, which is (for simplicities sake I'll call the lmbda of LLLBs k0 and the lmbda of SLLBs k1)

Ps(E)=(exp(-k0*t))^5(exp(-k1*t))^5=exp(-5(k0+k1)t)
And
Pf(E)=1-Ps(E)
Moreover the MTTF (lmbda) of the combined set of bulbs is 5(k0+k1)
Putting in k0=1/200 & k1=1/100 then
5(1/200+1/100)=5*3/200=3/40 (hrs^-1)

So I would expect a single light bulb to fail at or about 40/3 hrs. /



The following was written in the certain knowledge that NOONE was going to be able to answer the question. To save my embarrassment, do your self a favor and re-read the answer from Rap.



Warning: This is just a quick explanation which assumes some knowledge of the exponential distribution and integral calculus.
The density function of the shorter lived life bulbs is f(x)=1/100 * e-x/100.

The density function of the shorter lived life bulbs is f(x)=1/200 * e-x/200. The survival is 1 minus the integral of the density function. S(x)=e-x/100, and S(x)=e-x/200.

The survival function for the entire group is (e-x/100)5 * (e-x/200)5.
The density function for the group is the derivitive of 1 minus the survival function = e-3x/40.

The mean for the group is the integral from 0 to infinity of x times the density function for the group = 40/3 hours.


Coffee Can Obscura.

I know they were used in the 16/17th century to project an image to assist artists. But, can they be used to take a photograph?





What is the smallest integer greater than 0 that can be written entirely with zeros and ones and is evenly divisible by 225

1s and 0s

[size=7]225 is 9*25, so this number has to be divisible by both 9 and 25.

I'll start with numbers divisible by 25 that is composed only using 1 and 0. The smallest is 100.

Now fer the part of that number that is divisible by 9 and made only of 1s and 0s.

My granddaddy, a famous builder of moonshine stills in these here hills, teached me this old arithmetic trick called casting 9's. He said it went back to moonshiner called Youkid who was famous for ouzo, a grape based whitedog. Seems fer a number to be divisible by 9, the sum of the digits has to be 9. Since iff'n you add 0 to anyting it's the same so's I'd have to say this part of the number is all 1s. And there had to be 9 of then. So I'd opine this number is 111,111,111. And shonuf, 111,111,111 is evenly divisible by 9.

Putting these to numbers together then the smallest number divisible by 9 and 25 (225) is 111,111,111x100=11,111,111,100.
A check shows that this number is evenly divisible by 225 (49,382,716).

So I'd say this number is 11,111,111,100[/size]

A tin can camera obscura works for a photograph. All you need is a light sensitive plate (I think Kodak still makes it) to focus the image on---this is the pinhole camera that my grandpappy showed me as a pup, and is the basis for the coffee can camera obscura that I built to show my son that solar eclipse.

Rap

Try, You left off my first paragraph (the unoranged one) for the lightbulb problem. That paragraph IMO is the key as to how I was led to the solution

Rap

Rap, thank you for addressing my oversight. You are indeed correct in your assumption, it should inspire others to seek an answer in the unknown.



One of Kerry and Kelly lies on Mondays, Tuesdays and Wednesdays, and tells the truth on the other days of the week. The other lies on Thursdays, Fridays and Saturdays, and tells the truth on the other days of the week. At noon, the two had the following conversation:

Kerry : I lie on Saturdays.

Kelly : I will lie tomorrow.

Kerry : I lie on Sundays.

On which day of the week did this conversation take place

(a) Monday
(b) Wednesday
(c) Thursday
(d) Saturday
(e) Sunday




How many integer pairs (m, n) satisfy the equation m(m+ 1) = 2n

(a) 0
(b) 1
(c) 2
(d) 3
(e) more than 3

Try: "You are indeed correct in your assumption, it should inspire others to seek an answer in the unknown."

Ouch!

[size=8]225
11111111100
225 = 3*3*5*5
Pair the fives up with twos to create tens. A multiple of nine made up of ones and zeros must contain a multiple of nine ones.

LIARS
Kerry lies on M/T/W.
Kelly lies on T/F/S.
The conversation took place on Wednesday.

INTEGER PAIRS
(f) infinitely many[/size]

Lie Detector

[size=7]Neither Kerry or Kelly lie on Sunday, so Kerry is lying and Kelly is telling the truth. If Kelly is telling the truth then tomorrow is Monday or Thursday. Since neither lies on Sunday, Tomorrow is Thursday. Kerry's first lie is unnecessary, but it confirms the conjecture that today is Wednesday(b).[/size]

m(m=1)=2n

[size=7]Quick investigation identifies the following pairs (m,n) as solutions
(0,0)(-1,0)(1,1)(-2,1)(3,6)(-4,6)(4,10)(-5,10)(5,15)(-6,15)(7,28)(-8,28) and so on. Since those listed is more than 3 ordered integer pairs the answer is (e)[/size]

Rap

Mark, of all the people who partake of this site, that statement and yourself are strangers, and likely to remain so. For in the sure knowledge that I offend no one, let me say; your contribution has been without equal. Period!

(As I have been unbelievably nice, perhaps you could see your way clear to give a hint for your: t p o f o f n t s f) It is freekin me out.




No word in the English language is considered a true rhyme for any of the following words: month, silver, purple or orange

True
False




The symbol used to indicate number (American) or pound (English) is known by what term

It is also known as the hashmark.

ellipsis
asterisk
octothorpe
hypen



"In the year of 1492, Columbus sailed the ocean blue" is an example of what term

(can you remember the 8 letter word?)



Which word has definitions which are antonyms of each other

lay
cleave
down
cut



What term refers to replacing a disagreeable or possibly offensive term or phrase with a more agreeable one

allegory
euphemism
hyperbole
idiom

month, silver, purple or orange - True


And that's an euphemism...

[size=8]true
octothorpe
cleave
euphemism[/size]

Hint in other topic.

Unrhymable words--True
#-crosshatch
1492-Mnemonic
antonyms---cleave and cut are synomyms
lay and down hace many meaning, but non of them are antonyms---so I don't know.
less offensive-euphemism

Rap

From my wifes OED

Oh! Self antonym! My forehead has an increasing tilt
Cleave to separate(1) or cling(3)
# crosshatch---octothorpe (ent--made from eight thorns)

Rap

Sorry for being late. It would appear that while copying your answers, I tried to download the whole thread/site! In short, the file and system jammed and I lost it.
It took a little time to come up with something new.


The prime factorization of 225 is 5*5*3*3. Therefore, the answer will be both a factor of 25 and of 9.
All factors of 25 end in either 00, 25, 50, or 75. The only one of these composed of 0's and 1's is obviously 00, so the answer must end in 00. The hard part is finding a series of 0's and 1's preceeding the 00 that will make the entire number divisible by 9.

If you didn't already know the following trick then this problem would be very hard. If you did know it then the problem was was likely very easy. The trick is that if the sum of digits of a number is divisible by 9 then the number itself is also divisible by 9. Note that this is true for 3 also. For example the number 17685 is divisible by 9 because 1+7+6+8+5=27, and 27 is divisible by 9.

To prove this let's consider any five digit number, abcde. This number can be expressed as follows.
a*10000 + b*1000 + c*100 + d*10 + e =
a*(9999+1) + b*(999+1) + c*(99+1) + d*(9+1) + e*1 =
a*9999 + b*999 + c*99 + d*9 + a + b + c + d + e =
9*(a*1111 + b*111 + c*11 + d*1) + a + b + c + d + e
9*(a*1111 + b*111 + c*11 + d*1) must be divisible by 9 because it is a factor of 9. So if a+b+c+d+e is also a factor of 9 then the entire number must be a factor of 9. Note also that the remainder of abcde/9 is the same as the remainder of (a+b+c+d+e)/9.

The smallest number consisting of all 1's and divisible by 9 is thus 111,111,111. Adding the two zeros at the end results in the answer to the problem: 11,111,111,100.


True*

Octothorpe*

"In the year of 1492, Columbus sailed the ocean blue" is an example of what term?
Answer: Mnemonic

cleave*

euphemism*



LIE

Kerry is clearly lying, and is the one who tells the truth on Saturday. Hence the conversation takes place Monday, Tuesday or Wednesday, and Kelly's statement is only true on the last day.


M &N

Either both m and m + 1 are powers of 2, or both are negatives of powers of 2. The two solutions are (m, n) = (1,1) and (m, n) = (-2,1).





Row, row, row your boat!

On one side of a river are three humans, one big monkey, two small monkeys, and one boat.

Each of the humans and the big monkey are strong enough to row the boat. The boat can fit one or two bodies (regardless of size).

If at any time at either side of the river the monkeys outnumber the humans the monkeys will eat the humans.

How do you get everyone on the other side of the river alive



Rap, please thank your wife for her participation.

Perhaps I should have made it clear that, there is nothing to stop the rower getting out of the boat.

HS | <B | HHS = eaten!

Oh!

too late

Rap

sorry about that - unfortunately, it can't be hidden and neatly formatted at the same time.

Mark:

Let:
H=human
B=big monkey
S=small monkey

He is of cause correct.

Interesting to note it only took 25 minutes from start to finish for him to rethink.

Rap, I may be outta the loop, but what does, "too late " refer to?




You are blindfolded before a table. On the table are a very large number of pennies. You are told 128 of the pennies are heads up and the rest are tails up. How can you create two subgroups of pennies, each with the same number of heads facing up




There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage you to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four

[size=8]PENNIES
Separate 128 pennies from the large group and turn them all over[/size]