The worlds first riddle!

Liessa, it is the taking part. lol.

Daughters, oh! What a joy.



Mark:
Liessa will finish the race 0.25 meters ahead of Mark.

The girl runs 100 meters in the same time that the boy runs 95 meters; therefore, the rate of the girl is 100 meters per some unit of time and the rate of the boy is 95 meters per the same unit of time.

The Rate-Time-Distance table for the second race:

Runner Rate Time Distance
Boy 95 m/unit 100/95 units 100 m
Girl 100 m/unit 105/100 units 105 m


So the Boy's time is 100/95 or 1.05266316 units of time, and the Girl's time is 105/100 or 1.05 units of time.

Therefore, the Girl won the second race as well! She passed the Boy at 95 meters.





When Mr. Low returned from his fishing trip, his students inquired as to the length of his prize catch. Mr. Low answered,

"The head measured 9 inches."

"The tail was as long as the head and half the body."

"The body was as long as the head and tail."


How long was Mr. Low's prize fish






Suppose you are six feet tall and walk around the Earth's equator.

Which goes further your head or your feet

How much further



Mark, the very cruel wrote;
"Let N = 111...1222...2, where there are 1999 digits of 1 followed by 1999 digits of 2.
Express N as the product of four integers, each of them greater than 1."

How can you think of such a question? The numbers must be enormous. Catch you later.

FISH
6 feet

SIX FEET
head, 12*pi feet

Mark:

FISH
6 feet


1. Let H, B, and T be three variables representing the Head, Body, and Tail of the fish.
2. H = 9".
3. T = H + 1/2 B
4. B = H + T
5. So, B = 9 + 9 + 1/2 B, by substitution.
6. Therefore, B = 36".
7. So, T = 9 + 18 = 27".
8. The Fish = H + B + T or 72" in length.





SIX FEET
head, 12*pi feet


The distance travelled by your feet is
C = 2 pi R, where pi = 3.14 and R is the radius of the Earth.

The distance travelled by your head is
C' = 2 pi (R + 6) or 2 pi R + 12 pi.

Subtracting C' - C, your head goes 12 pi ft farther than your feet (or approximately 37.7 ft)!
The radius of the circular object is not relevant.


"Let N = 111...1222...2, where there are 1999 digits of 1 followed by 1999 digits of 2. Express N as the product of four integers, each of them greater than 1."


After much blood, sweat and tears. Well no blood, I reckon: N = 2 * 3 * 23 *

280008051529790660225544283413848631239935587761672718196457326892109...
500805152979066022544283413848631239935587761674718196457326892109X29...
50080515297906602254436392914653784219...
001610305958132045488566827697262479871175523349436392914653784219X30

Or thereabouts.




The problem is the dice at my local casino have become very worn, and some of the ?'spots' are missing.

Can you help me by counting the total number of ?'spots' that the dice could represent? For clarity (o) = spot, whilst (.) = blank


For example:

o . o
. o .
o . o

Can only be a 5, and never a 6.

I threw:


o . .
. . .
. . .

. o .
. . .
. . .

. . .
. o .
. . .

. . o
. . .
o . .

. . o
. . .
. . o

Total all possible missing combinations of spots for the 5 dice



Try this poetic puzzle involving 5 elves, dressed in 5 different colors, who made 5 different types of toys.


Ole Santa's pack held thirty toys
Made by his elfin crew;
And though none made the same amount,
Each elf made more than two.
The elf named Cher made one more toy
Than the elf who dressed in reds,
But Cher made one less Christmas toy
Than the elf who made the sleds.
Spry Johnny elf made racing cars;
Five toys were made by Jane.
The elf who dressed in yellow suits
Made each and every train.
The elf who always dressed in green
Made one-third as many as Sue.
Cute Marcia elf was dressed in orange,
And one elf dressed in blue.
The elf who made the spinning tops
Made the most toys of them all.
Another perky, smiling elf
Made each and every ball.
Ole Santa's pack held 30 gifts
All tagged for girls and boys.

From the clues that you've been given, guess who made what toys

DICE
225

"Let N = 111...1222...2"
The 2 and the 3 are correct. Try it for a smaller case, say 1111122222. What if you divide this by 100002?

ELVES
Cher - 6 trains - yellow
Jane - 5 balls - red
Johnny - 3 cars - green
Marcia - 7 sleds - orange
Sue - 9 tops - blue

DICE
225


No, no, way too high.

BIG clue;

. o .
. . .
. . . can only be a 6





Mark:

ELVES
Cher - 6 trains - yellow
Jane - 5 balls - red
Johnny - 3 cars - green
Marcia - 7 sleds - orange
Sue - 9 tops - blue


Solution:

Each elf made at least three toys (line 4). One elf made three times as many as another (13-14), and three elves made consecutive amounts (5-8). If you eliminate duplicate amounts (3), two possible number combinations total 30: 12/6/5/4/3 or 9/7/6/5/3.


With either solution, Sue made the most (13-14). And we know that she made tops 17-18). She was not dressed in green (13-14), orange (15), red (5-6), or yellow (11-12), so she must be the elf in blue.

The two elves who made the least must be dressed in green (13-14) and red (5-6). Therefore the elf who made the least can not be Marcia (15). It also can not be Cher (5-8), or Jane (10), so Johnny must have made the least. And he made cars (9).

The elf who made the second most was not Jane (10)or Cher (5-8), so it must have been Marcia.And Marcia dressed in orange (15).

Since the two elves who made the least were in green and red, the elf in yellow must have made the third-smallest amount of toys. And that elf made trains (11-12).

Since the elf making sleds could not have made the second-smallest amount of toys (7-8), Marcia must have made the sleds. Cher is therefore the elf in yellow, and the person who made the second-smallest amount of toys is in red (5-8).

By elimination, Johnny is in green, and Jane is in red and made the balls. Since the elf in green made three toys, Sue must have made nine toys (13-14), and our amounts of toys must be: 9/7/6/5/3.

Sue (blue) made 9 tops.
Marcia (orange) made 7 sleds.
Cher (yellow) made 6 trains.
Jane (red) made 5 balls.
Johnny (green) made 3 cars.


"Let N = 111...1222...2"




N is even so it is a multiple of 2, but the last two digits are not divisible by 4, so N is not a multiple of 4. The sum of its digits is 1999 ⋅ 1 + 1999 ⋅ 2 = 1999 ⋅ 3, so it is a multiple of 3 but not 9. 5 is not a factor because N does not end with 0 or 5.
Potential third factors can be tested by a technique that I devised based on remainder repetition intervals. I will use the division of 7 into a short series of ones and twos as an example:

1 5 8 7 3 0 3 1 7 4 6 7 ) 1/1
14 16 15 12 10 22 21 25 23 24 20
As you can see, it divides evenly into 6 ones, and divides evenly into 6 twos. Because the remainder at the end of each section is zero, it can loop around as many times as desired. In addition, at equal distances from the change from 1 to 2 are two remainders of 2, so the cycle can be interrupted at that point. Therefore, 7 is a factor of any number like N with 6x ones and 6x twos, or a number with 6x+5 ones and 6x+5 twos, with x as an integer greater than or equal to 0. I write this cycle structure as 6x and 6x+5. Since 1999 does not equal 6x or 6x+5 for any integer x, 1999 does not fit seven's cycle, and 7 is not a factor.
Now, this method of testing primes until a factor is found can be used. The results starting after 7
are:
Prime Cycle structure
11 2x
13 6x
17 16x, 16x+2
19 18x, 18x+8
23 22x, 22x+19

1999 = 22 ⋅ 90 19 + =22 ⋅ x 19 + , so the third factor of N is 23. The results of division cycle with the remainder cycles, so N/23 can be expressed as follows:
(Notation: X indicates that the previous series of numbers should be repeated the indicated number of times, ... indicates that the next number should be concatenated to the previous one)
N/23 = 0048309178743961352657X90...
0048309178743961352661837.
00966178357487922705314X.
Each section alone is not a multiple of 3, but the three sections together are. In addition, the middle section (put together) is also a multiple of 3. So
N
-------------=
323
⋅
001610305958132045088566827697262479871175523349436392914653784219X30...
0016130595813204508872785829307568438... 003220611916264090177133655394524959742351046698872785829307568438X30
Only the first section is odd, but the remainder can be contained in the first digit of the next repetition.



Talking of Santa's little helper:


Santa needs some help from you this year - he'll supply the laughs (HO HO) if you do the wrapping. In each of the words and phrases below, Santa's filled in HO twice, but the rest of the letters are missing.

Can you reconstruct each word or phrase from the clues and blanks given?



1. Hershey addict: _ H O _ _ H O _ _ _


2. Listing of famous and influential people:

_ H O ' _ _ H O

3. Word pronounced like another, as "heir" or "air"

H O _ _ _ H O _ _


4. Garage where stolen cars are dismantled for parts:

_ H O _ _ H O _


5. Be Patient!:

H O _ _ _ _ _ _ H O _ _ _ _ !

DICE
first: 2, 3, 4, 5, or 6 (5 possibilities)
second: 6 (1 possibility)
third: 1, 3, or 5 (3 possibilities)
fourth: 2, 3, 4, 5, or 6 (5 possibilities)
fifth: 4, 5, or 6 (3 possibilities)

5*1*3*5*3 = 225

Have I misinterpreted the problem?


"Let N = 111...1222...2"
Divide 111...1222...2 by 1000...02 to get 111...1
Divide 1000...02 by 6 to get 1666...67
The factors are 2, 3, 111...1, and 1666...67
111...1 consists of 1999 ones
1666...67 consists of 1997 sixes

I'm not claiming that this is the only solution, but it is probably the simplest

HO HO

chocoholic
who's who
hole whole
chop shop
hold your horses

Mark:

DICE
first: 2, 3, 4, 5, or 6 (5 possibilities)
second: 6 (1 possibility)
third: 1, 3, or 5 (3 possibilities)
fourth: 2, 3, 4, 5, or 6 (5 possibilities)
fifth: 4, 5, or 6 (3 possibilities)

5*1*3*5*3 = 225

Have I misinterpreted the problem?


Absolutely not. I didn't think past totals. 2,3,4,5,6,=20 etc. Total 70.
Had I thought it through to the obvious conclusion, 225 was the best answer. A classic case of the ?'inventor' not seeing the wood for the trees.


Mark:

"Let N = 111...1222...2"
Divide 111...1222...2 by 1000...02 to get 111...1
Divide 1000...02 by 6 to get 1666...67
The factors are 2, 3, 111...1, and 1666...67
111...1 consists of 1999 ones
1666...67 consists of 1997 sixes

"I'm not claiming that this is the only solution, but it is probably the simplest."


Simply brilliant.


HO HO

chocoholic
who's who
hole whole
chop shop
hold your horses




6. Chemical element used in making matches:

_ H O _ _ H O _ _ _

7. Big Cheese: H O _ _ H O


8. The furthest extent one can go:

_ H O _ _ H O _


9. Game in which players try for ringers:

H O _ _ _ _ H O _ _

10. Talented Oxford Attendee:

_ H O _ _ _ _ _ H O _ _ _


11. One-handed basketball throw: H O _ _ _ H O _


12. Sanctimonious:

H O _ _ _ _ - _ _ _ _ - _ H O _



Three local sportscasters, Dan, Ben, and Ray, each predicted the winners of the four local football games. The only team not picked to win was Clarke County.

Dan picked: Handley, Warren County, James Wood, and Strasburg.

Ray picked: Loudoun Valley, James Wood, Sherando, and Handley.

Ben picked: Warren County, Musselman, Handley, and Loudoun Valley.


Which teams are playing each other




Starting tomorrow; a short series of puzzles with answers. All you have to do is choose the right one.

HO HO
phosphorus
honcho
whole hog
horseshoes
rhodes scholar
hookshot
holier than thou

FOOTBALL
Handley vs. Clarke County
Warren County vs. Sherando
James Wood vs. Musselman
Strasburg vs. Loudoun Valley

JELLYBEANS
I don't believe this was ever answered.

Let's play a game. Start with n jellybeans placed around the circumference of a circle. To play, one of us removes from the circle either one jellybean or two adjacent jellybeans. We alternate turns. The person removing the last jellybean from the circle wins (and gets to eat all the jellybeans).

Do you want to go first, or should I?

What is your winning strategy?

I'll go second (and win). :wink:
Depending on whether n is odd or even and how many you take, I'm going to take one or two from the opposite side of the circle such that the circle is now broken into two arcs with an equal number of jellybeans (I can always do this). Then, I mirror your moves on the opposite arc.

You should go first. The person who goes second should always win.

Mark:

HO HO
phosphorus
honcho
whole hog
horseshoes
rhodes scholar
hookshot
holier than thou

FOOTBALL
Handley vs. Clarke County
Warren County vs. Sherando
James Wood vs. Musselman
Strasburg vs. Loudoun Valley



First, it is important to realize that none of the four teams listed by each sportswriter could be playing each other.
Begin with Dan's predictions: Handley, Warren, James Wood, and Strasburg cannot play each other.
From Ben, we know that Musselman and Loudoun Valley must be James Wood's and Strasburg's opponents.


Mark:

JELLYBEANS

"I'll go second (and win)."


Welcome Vengoropatubus. May I thank you for your timely update for your thread, ?'CRAZY logic problem' I very much doubt anyone could have come up with the answer.

"The person who goes second should always win. "




The second player (almost) always has a winning strategy. The only exceptions are when you start with only 1 or 2 jellybeans, in which case the first player obviously wins.

Once the first player moves, the second player is faced with (essentially) a row of jellybeans. Regardless of how many were originally taken, the second player can now take the row of jellybeans and remove 1 or 2 jellybeans so as to leave two rows of jellybeans with an equal number of candies. Now whatever the first player does to one row, the second player mimics on the second row and is guaranteed to take the last candies.



Questions with answers! I must be


If you were writing a computer program with the intention of breaking the world record for calculating the most digits of pi, why would it be foolish for you to use the well-known formula pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...

• It is more fun to use the Monte Carlo method

• It gives incorrect digits after the 875th decimal has been reached

• It converges terribly slowly

• It would be too hard to implement this formula in any programming language

• Computers aren't good at doing such repetitive tasks as calculating series iterations





No, only the first question has the answer. :wink:

A man has two children. He opens college savings accounts for each of them, starting each with seven dollars and fifty cents.
(From what he remembers of tuition, he thinks that this will be helpful.)

Then the rigorous savings plan begins.
During the first month he deposits some positive amount (dollars and cents) into the son's account.

During the second month he deposits some positive amount into the daughter's account.

During the ensuing months he continues to alternate between the accounts, but from the third month onward the amount of each deposit is equal to the sum of the previous two months' deposits.

(So the son's deposit during the third month is the sum of the son's desposit in the first month plus the daughter's deposit in the second month.)

The accounts bear no interest.
(This is another clue into the man's naiveté.)

At the end of two years and four months, the daughter is a millionaire (exactly).


How much does the son have in his account (exactly)
What colleges will he be able to afford

PI/4
It converges extremely slowly.

Mark:
PI/4
It converges extremely slowly.

Dang me, how could you know that? Well, assuming you were not about in 1897, try this little known fact:


In 1897 a government body unanimously passed a bill setting pi equal to 16/(sqrt 3), which approximately equals 9.2376. Where did this occur

• California

• Indiana

• Tennessee

• Ohio

• Canada




Train A and train B are crossing the country, from coast to coast, over 3,000 miles of railroad track.

Train A is going from east to west at 80 miles per hour, against a 12 mph headwind.

Train B is going from west to east at 90 miles per hour against a 24 mph headwind.

Which train will be closer to the west coast when they meet

TRAINS
If "meet" means when the fronts of the trains reach the same point, then the only thing that matters is direction. The west-to-east train will be closer to the west coast until the front of the east-to-west train passes the back of the west-to-east train.

SAVINGS
I forgot about this. I'll work on it today.

SAVINGS
a) $618,036.85
b) any college he wants to attend

1897 BILL
Indiana (googled it)

Mark:
1897 BILL
Indiana (googled it)

No need for confessions. I Googled to find it.


In the USA the value of p gave rise to heated political debate. In the State of Indiana in 1897 the House of Representatives unanimously passed a Bill introducing a new mathematical truth:

Be it enacted by the General Assembly of the State of Indiana: It has been found that a circular area is to the square on a line equal to the quadrant of the circumference, as the area of an equilateral rectangle is to the square of one side. (Section I, House Bill No. 246, 1897)

The Senate of Indiana showed a little more sense and postponed indefinitely the adoption of the Act.



TRAINS
If "meet" means when the fronts of the trains reach the same point, then the only thing that matters is direction. The west-to-east train will be closer to the west coast until the front of the east-to-west train passes the back of the west-to-east train.


I doubt anyone could be more exact than that. I would happily have acknowledged:

When the trains meet, they will be at exactly the same point. Therefore, they will each be the same distance from the west coast.


SAVINGS
a) $618,036.85
b) any college he wants to attend


Lets not quibble over a few cents. I thought it would take a week to come up with an answer. $618,042.45.


Each penny deposited to the son's account in the first month eventually causes a total of $1,964.17 in his account and $3,178.11 in the daughter's account.

Each penny deposited to the daughter's account in the first month eventually causes a total of $1,964.17 in her account and $1,213.94 in the son's account.

If the initial deposits are x cents and y cents, the daughter's wealth tells us that 3178.11x + 1964.17y + 7.50 = 1000000.00,
whose only solution (in positive integers) is x=73, y=391.

The son's wealth is then
1213.94*391 + 1964.17*73 + 7.50 = 618,042.45.

The monthly deposits are

3.91, 0.73, 4.64, 5.37, 10.01, 15.38, 25.39, 40.77, 66.16, 106.93, 173.09, 280.02, 453.11, 733.13, 1186.24, 1919.37, 3105.61, 5024.98, 8130.59, 13155.57, 21286.16, 34441.73, 55727.89, 90169.62, 145897.51, 236067.13, 381964.64, 618031.77.




What professor was dismissed from his position in 1934 for teaching in an "un-German" style after saying (correctly) that pi/2 is the value of x between 1 and 2 for which cos x vanishes

• Edmund Landau

• J. Robert Oppenheimer

• Scott Adams

• Elias Bröms

• Kip Thorne




Whim invited us to play golf in the Tyrolean Alps.
The air is so thin up there that air resistance is negligible.
And the course is laid out a bit strangely...

The first hole is 150 yards, due horizontal.
(The cup is on the same vertical level as the tee.)
We hit the ball at top speed, at the correct angle to maximize distance, and find that it precisely reaches the cup. Hole in one.

The second hole is also 150 yards, but due vertical: the cup is directly below the tee. This time we require no force at all: we can merely drop the ball into the cup.

The third hole is a bit trickier: 150 yards, at a 45 degree angle below the horizon. At what angle should we hit the ball, and at what speed (as a percentage of the speed required on the first hole), if we wish to minimize the speed required

Solve this and I will see you in the 19th.

Golf: Tap the ball as lightly as possible, it will roll down the incline into the cup. (Or are you looking for a ballistic solution?)

DrewDad, known as ?'the Shark' on the pro circuit writes, "Tap the ball as lightly as possible, it will roll down the incline into the cup."

As this is a 4 ball better ball comp. I think you might have won it for us…if you can answer:

"At what angle should we hit the ball, and at what speed (as a percentage of the speed required on the first hole), if we wish to minimize the speed required"

That is the ?'bummer'. Cos if you Sin you will be well on your way. :wink: