The worlds first riddle!

Try had three illegitimate sons from three different women, all of whom were quite good at math and logic (Try chose his women well). To get a share of Try's inheritance, each had to correctly determine a positive integer which he had chosen. He told them that the number had four different non-zero decimal digits in ascending order.

He prepared three sealed envelopes, each of which contained a number. The first contained the product of the four digits, the second contained the sum of the squares of the four digits, the third contained the sum of the product of the first two digits and the product of the last two digits, and the envelopes were clearly marked as such. He showed the three envelopes to the three sons and had them each take one at random.

The sons were stationed at three different computers so that they couldn't communicate with one another (but were linked to Try's computer). After one hour they could submit a number or decline. Anyone who submitted a wrong answer would be eliminated and get nothing. If one or more submitted the correct answer they would each receive a share of the inheritance, and the contest would end with the others getting nothing. If no one submitted the correct answer they would be instructed to work on the problem for another hour. The process would repeat as often as necessary.

At the end of the first hour, no one had submitted an answer.
At the end of the second hour, no one had submitted an answer.
At the end of the third hour, no one had submitted an answer.
At the end of the fourth hour, all three of them submitted the correct answer!

Can you determine the number before Try kicks the bucket

Thoh:

7:28PM

4 x 182 = 728



Mark:


$20.19



$10 - 1
$5 - 1
$1 - 4
.50 - 1
.25 - 1
.10 - 4
.05 - 0
.01 - 4

$20.19


Mark wrote, "Can you determine the number before Try kicks the bucket?"

Well you had better hurry up, I am already feeling a little pail.





A traveler at the Dallas Airport had lots of time on her hands between flights, so she decided to conduct an experiment on one of the moving walkways. She found she could walk the length of the walkway, moving in its forward direction, in 1 minute.

Walking at the same rate against the forward direction of the walkway, it took her 3 minutes to cover the same distance.

How long would it take her to cover one length of the walkway if the walkway were to stop

[size=8]TRAVELER
90 seconds[/size]

Try chose his women well - Puzzle!

That ain't no puzzle - that be a fact! Oh, I make the number [size=7]3479[/size] and that's a fact!


Mr. Q retired and invited his math students and his computer students to his retirement party.

Mr. Q shook hands with all his guests.
Among his guests, his math students shook hands with each other, and his computer students shook hands with each other; but the math students and computer students simply greeted one another with a nod of the head.
One hundred twenty-three handshakes were exchanged at the party.
No math student was also a computer student.

How many people were at the party (including old Mr. Q) if at least three math students and at least three computer students attended

22
123=9C2 + 12C2 + 9 + 12 12+9+1=22

sequence: 3 3 5 4 4 3 5 5 4 3 ?

sequence: 3 3 5 4 4 3 5 5 4 3 [size=7]6 7 8[/size]

yes no yes...but the ? is right

6 6 8 (Try must have mistyped)

Mark:


TRAVELER
90 seconds


In 1 minute, the traveler can walk 1 length in the forward direction but only one-third of a length in the backward direction.
Factoring out the effects of the walkway's speed, we find that in 1 minute she can walk (1 + 1/3) / 2 = 2/3 of a length in one minute.
This means the woman can walk one length of the stationary walkway in 3/2 x 60 = 90 seconds.



Thoh:

22
123=9C2 + 12C2 + 9 + 12 12+9+1=22




For those who may find Thoh's answer far too simple, may be amused by this explanation:



The answer is 22 people, including Mr. Q, 9 members of 1 class, and 12 members of the other group.

Mr. Q shook hands with all 21 students.
The nine math students shook hands with the other 8 math students for
a total of 36 handshakes (9 x 8 / 2).
The 12 computer students shook hands with the other 11 computer students for
a total of 66 handshakes (12 x 11 / 2).

To solve this problem,
let c = number of computer students who came to the party.
Let m = the number of math students who attended.
The number of handshakes which take place among a group of n people can be determined by the formula for the combinations of n things taken r at a time, where r = 2.
The formula nCr = (n!) / (r!((n - r)!))

Number of people Number of Handshakes
Mr. Q (1) c + m
m math students (m!) / ((m - 2)! x 2!) = (1/2)m(m - 1)
c computer students (c!) / ((c - 2)! x 2!) = (1/2)c(c - 1)


Totalling the three columns above, the number of handshakes =
c + m + (1/2)c^2 - (1/2)c + (1/2)m^2 - (1/2)m.
Set this equal to 123 and simplify to get the following:

c^2 + c + m^2 + m = 246.

I wrote a computer program in C++ to solve this problem.
Listed below is the nested loop which checks this condition:



When I ran the program, I discovered two solutions to the problem:
2 math students and 15 computer students and
9 math students and 12 computer students.
So I added the additional clue about having at least 3 students from each class at the party.







Three men, travelling with their wives, came to a river which they wanted to cross. The one available boat would accomodate only two people.
Since the husbands were very jealous, no woman could be with a man unless her own husband was present.

Under these severe handicaps, how can they get across the river using the one boat

Perhaps it would be a good idea to refer to the three couples as:
A, a, B, b, C, and c, where the capital letters refer to husbands and lower case letters refer to their wives (or vice versa!).




Thank you for the typo lifeline. Word has defaulted to type size 5 and reading is impossible.

We have the same here but with a goat, a cabbage and a wolf, you see...

Mrs. b and Mrs. c cross, and Mrs. b returns with the boat.
Mrs. b and Mrs. a cross, and Mrs. a returns with the boat.
Mr. B and Mr. C cross, and Mr. B and Mrs. b return with the boat.
Mr. B and Mr. A cross, and Mrs. c returns with the boat.
Mrs. b and Mrs. c cross, and Mr. A returns with the boat to pick up his wife....

Francis, it would appear that your frequent trips to market to sell your cabbages with your goat have given you an edge.


A visitor to the Zoo asked the zookeeper how many vultures and how many cows were in the zoo.

The zookeeper, a young math wizard called Thoh replied;
"There are 45 heads and 150 feet."

How many vultures and cows are at the zoo

Coming from the market with a cabbage for the goat with a wolf as companion, Try, paying a visit to the zoo...

Francis I forgot your trips to the market passed the zoo, so you would know the number of animals. However, the real question is:

Place 16 chess pieces (or 16 checkers) on an 8 x 8 chess board so that no three are in a line (horizontally, vertically, or diagonally).

Can it be done, and if so how

Assuming diagonally means 45 degree angle, here's a solution:

Francis:

Boat:
We have the same here but with a goat, a cabbage and a wolf, you see...

Mrs. b and Mrs. c cross, and Mrs. b returns with the boat.
Mrs. b and Mrs. a cross, and Mrs. a returns with the boat.
Mr. B and Mr. C cross, and Mr. B and Mrs. b return with the boat.
Mr. B and Mr. A cross, and Mrs. c returns with the boat.
Mrs. b and Mrs. c cross, and Mr. A returns with the boat to pick up his wife....


Indeed we did, and it fooled no one.

Not content with that, he continued:

"There are 45 heads and 150 feet."
How many vultures and cows are at the zoo?
v+c=45
2v+4c=150
v=45-c
2(45-c)+4c=150
90-2c+4c=150
90+2c=150
2c=150-90
2c=60
c=30
v=45-30
v=15

30 cows
15 vultures

Now, although all his numbers are written in French, the answer is clear.

For the non-French speaking amongst you:

Let C = the number of cows
and V = the number of vultures.
From the total number of feet, we know that
V(2 feet/bird) + C(4 feet/cow) = 150 or
2V + 4C = 150

The total number of creatures in the zoo is
V + C = 45, so
V = 45 - C.

Now we can substitute this equation into the feet equation to obtain:
2(45 - C) + 4C = 150
90 - 2C + 4C = 150
2C = 60
C = 30 Cows and V = 45 - 30 = 15 Vultures



Mark:

Chess problem

That is the first answer I have. Here is another

----XX --
----XX--
XX------
XX------
------XX
------XX
--XX----
--XX----





The Candy Factory manufactures of Mac Candy, want to be able to fill any demand for candy that ranges between 1 and 1000 pieces.
The production manager, a retired math teacher from the local high school, determined that they only needed 10 different size bags of candy on hand to be able to fill any order!

If all the bags are to hold a different amount of candy, and if you want to be able to fill any order of candy from 1 to 1000 pieces (without breaking any of the bags).

How many pieces of candy should be put into each of the ten bags

[size=8]CANDY
2^n for n=0 to 9 until he runs out of candy.[/size]

Mark:

CANDY
2^n for n=0 to 9 until he runs out of candy.


Mark gives us an answer in French, which when I ran it through a cheese grater translator it came out as:

The ten bags should hold the following amounts of candy:


Bag 1: 1
Bag 2: 2
Bag 3: 4
Bag 4: 8
Bag 5: 16
Bag 6: 32
Bag 7: 64
Bag 8: 128
Bag 9: 256
Bag 10: any amount from 489 to 512

For example, to fill an order of 63 pieces of candy, you would give the customer bags 1 through 6!
To fill an order of 300 pieces, you would give bags 3, 4, 6, and 9.
This is based on the binary (base 2) number system.





Twas the night before the holidays, and at the South Pole
The last-minute planning was taking its toll.
As Santa was hastily making a scheme
For the placement of deer in his sleigh-pulling team,
The good Mrs. Claus was crocheting bright bows
To be worn by these reindeer (four bucks and four does).

The ribbons were colored in eight festive hues:
One ocher, one rose, one cerise, one chartreuse,
One maroon, one magenta, one white, and one blue.
(These ribbons helped Santa keep track of who's who.)
The deer pulled the toy-laden sleigh in four rows,
Arranged so no row held two bucks or does.

The order of pullers was changed year by year,
For Santa was thoroughly fair with his deer.
He summoned the elves and instructed them thus:
"Let's hitch up the reindeer with minimum fuss.
The bow on the buck behind Dasher is white,
While Blitzen, a doe, sees cerise to her right.

The blue bow is nearer my sleigh than is Dancer,
But nearer the front of my team than is Prancer.
The doe in chartreuse gets a front-of-team honor,
But not on the same side as Cupid or Donner.
Now Comet stands two spots ahead of the rose.
And three deer of four on the right side are does.

The cerise bow is worn two in back of maroon,
One of which is beside the bright ocher festoon.
Oh-Cupid's in front of a buck, by the way.
Well, that's how they line up for pulling my sleigh.
I trust that you elves, being clever, now know
Each reindeer's position and color of bow."

In no time each colorful ribbon was tied
And the team was hitched up for the transglobal ride.

Can you ascertain where each member fits in
Who's Comet? Who's Cupid" Where's Donner? And Blitzen
Who's Dasher? Who's Dancer" Where's Vixen? And Prancer
With logical thought, you'll determine the answer
And write down the color and place for each deer
Happy Holidays to all, and to all much good cheer!



Row / Position / Name / buck:Doe / Ribbon Color

Fairy Nuff, that was a most difficult riddle. Perhaps ya'll will find this a might easier - but don't bet on it!





In 1987, Mark (no relation I am sure) made the following statements:

I lived 1/8 of my life before going to school.
Then I went to school for 2/5 of my life.
Immediately after graduation, I married my wife Carol.
Then I taught at High School for 1/8 of my life.
I have spent the rest of my life teaching at High School in CA.
I have been married for 19 years.
After 3/5 of my life, my daughter Amy was born.
After 3/4 of my life, my daughter Sarah was born.
My son Michael was born 2/3 of the way between my two daughters' births.

How old was he
How old was he when he got married
How old was he when Amy was born
How old was he when Michael was born
How old was he when Sarah was born

How old was he? - 40
...when he got married? - 21
...when Amy was born? - 24
...when Michael was born? - 28
...when Sarah was born? - 30

Francis, those numbers look familiar, I am surprised it took you a whole eight minutes


I have gotta leave a little something for Mark to ponder.

Ophantus, the Greek mathematician known as the "Father of Algebra", supposedly had the following inscribed on his tombstone:

"Ophantus passed 1/6th of his life in childhood, 1/12th in youth, and 1/7th more as a bachelor.
Five years after his marriage, there was born a son who died four years before his father, at half his father's age."

How old was Ophantus when he died