The worlds first riddle!

[size=8]ORPHANTUS
84[/size]

[size=8]SANTA'S SLEIGH

Row 1: Dasher/buck/magenta, Comet/doe/chartreuse
Row 2: Cupid/buck/white, Dancer/doe/maroon
Row 3: Donner/buck/blue, Vixen/doe/rose
Row 4: Blitzen/doe/ocher, Prancer/buck/cerise[/size]

Francis was absolutely correct.

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


Mark:
SANTA'S SLEIGH

Row 1: Dasher/buck/magenta, Comet/doe/chartreuse
Row 2: Cupid/buck/white, Dancer/doe/maroon
Row 3: Donner/buck/blue, Vixen/doe/rose
Row 4: Blitzen/doe/ocher, Prancer/buck/cerise

Way to go.



ORPHANTUS
84




You could solve the problem using algebra by letting x = the age of Ophantus at his death.

However, there is an easier way to solve the problem. The only number divisible by 6, 12, and 7, and also a number in the human lifespan is 84.
So, he spent 14 years in childhood, seven in youth, and 12 as a bachelor.
He married at 33, and five years afterward, at age 38, a son was born who later died at age 42, when Ophantus was 80.
Four years after that, Ophantus himself died at 84, twice the age to which his son lived.




What is the value of F in the following system of equations
A + B = Z (1)
Z + P = T (2)
T + A = F (3)
B + P + F = 30 (4)
A = 8 (5)

[size=8]F=23
A=8
T=15[/size]

Mark:

F=23
A=8
T=15


I would like to thank Mark for confirming my answer, well done me. Oh, and very well done Mark!


Here is one complete solution:

Substituting equation (1) in equation (2) gives
A + B + P = T
Substituting equation (5) in this last equation gives
8 + B + P = T (6)
Substitute equation (3) in equation (4) to get
B + P + T + A = 30
Substituting equation (5) in this last equation gives
22 - B - P = T (7)

Adding equations (6) and (7) gives the following:
30 = 2T
So T = 15 (8)
Substitute equations (5) and (8) in equation (3) gives
F = 15 + 8 = 23


For an easier way, buy Mark a drink.





I was at an antique auction this morning, and one of the items that came up for sale was a child's sled. It was a hand-made sled, a wooden sled that the auctioneer claimed was made by none other than George Washington himself.
The auctioneer turned the sled over and read the following inscription carved into one of the wooden slats:

"G. Washington, September 10, 1752"

I was thinking of bidding for such a rare item. However, after a little thought I decided it must be a fake.

What do you think, would you have bought it, and why / why not

[size=8]SLED
From ancestry dot com:

In 1582, Pope Gregory XIII (1502–85), who was pope from 1572 to 1585, and his astronomer and mathematician created a new, reformed calendar known as the Gregorian or New Style (N.S.) calendar. It was adopted first in Roman Catholic countries. Protestant countries adopted the calendar during the eighteenth century.

In order to make the calendar adjustment in 1582, ten days were eliminated from October. Thus 4 October 1582 was followed by 15 October 1582.

England and its American colonies did not adopt the reformed Gregorian calendar until 1752. Scotland adopted it earlier, celebrating the New Year on 1 January 1600 and subsequently on January 1st of each year. Interestingly, Alaska did not change from the Julian calendar to the New Style Gregorian calendar until 1867 because, up to that point, it was part of Russia.

In order to make the calendar adjustment, eleven days were dropped from the month of September 1752. An eleven-day adjustment in 1752 was needed because one more day had been lost since the calendar was changed in 1582. The year 1751 began on 25 March and ended on 31 December 1751. The first day of the year was now January 1st and the last day was December 31st—the calendar we use today. Thus, 2 September 1752 was followed by 14 September 1752. In this way, the Julian calendar added one day between 1582 and 1752.[/size]

Mark:

Sled
I think Mark has covered the subject in some depth, and I would only add.



George Washington was born on February 11, 1732, by the Julian calendar which was then in use. In 1752, when the Gregorian calendar was adopted and 11 days were dropped, Washington shifted his birth date in order to keep the "real" day. That's why we now celebrate his birthday on the 22nd.




Mark goes into a store and says to the owner: "Give me as much money as I have with me and I will spend $10."

It is done and he repeats the operation at a second and a third store, after which he has no money left.

How much did he start with

[size=8]$8.75[/size]

Mark knows how I love to see an answer before the ink has dried, so…



You are presented with two urns and 10 red marbles and 10 white marbles. You may place the marbles in the urns as you please. Then a friend will enter the room, select an urn at random, and draw one marble from it.

How should you place the marbles in order to maximize the probability that your friend draws a white marble

What is the probability that your friend will draw a white marble

[size=8]MARBLES
1W, 0R
9W, 10R

P(W) = .5 + .5 * 9/19 = 0.7368[/size]

Mark:

$8.75


Let x = how much the man started with.
Then in the first store:
x + x - 10 is how much he had left.
In the second store:
(2x-10) + (2x-10) - 10 is how much is left
This simplifies to 4x - 30
In the third store:
(4x-30) + (4x-30) - 10 is how much is left
This simplifies to 8x - 70

So, 8x - 70 = 0 (no money left)
8x = 70
x = 70/8 or $8.75




MARBLES
1W, 0R
9W, 10R

P(W) = .5 + .5 * 9/19 = 0.7368




You should put one white marble in one urn and then put all the other marbles in the other urn.

The probability that your friend picks a white marble is
(1/2)(1) + (1/2)(9/19) = 1/2 + 9/38 = 28/38 or 14/19 or 73.68%.

It is interesting that the problem did not say that ALL the marbles had to be used, so you could just put white marbles in the two urns. Therefore, the probability would be 1 or 100%.





My local toy store sells items according to its own pricing system. A doll costs $10, a toy train costs $20, a cradle costs $20, a jumping jack costs $30, and a Tickle Me Elmo costs $50.

Using the same system, how much does a drum cost

Is it a "toy train" or a toy "train?"

Good point - Go with the latter.

[size=8]Then I'll say $10 per vowel.[/size]

Mark:

Toy store
Then I'll say $10 per vowel.

If you are gonna say that, then I'll say; well done.




The numbers in the sequence 2, 7, 12, 17, 22, 27, 32, 37, ... increase by 5s.

The numbers in the sequence 3, 10, 17, 24, 31, 38, 45, 52, ... increase by 7s.

What are the first 10 numbers which appear in both sequences


CLUE: The first is 17




What comes after once, twice, thrice

[size=8]17 + 35x where x ranges from 0 to 9.[/size]

Mark:

17 + 35x where x ranges from 0 to 9.



Answer: 17, 52, 87, 122, 157, 192, 227, 262, 297, 332


Since 5 and 7 have no common factors, the only values the sequences have in common are 17, plus multiples of 5x7.



Given the set of letters below, you have to spell the word MATHEMATICS; by starting at the top of the rhombus and moving down through the letters.

At each level, you must move to a letter which is below and to the immediate left or below and to the immediate right of the letter you are at.

In how many ways can you spell out the word mathematics by moving from the top letter and ending at the bottom letter

[size=8]C(10,5) = 252[/size]

Mark:

C(10,5) = 252




You could use "Pascal's Triangle" to help solve it. I have replaced the letters of mathematics by numbers representing how many different ways you could reach that particular letter, and can confirm the answer given by Mark.








Einstein may have said, "This problem will not be solved in my lifetime". If there was a million dollar prize, could you figure it out?



Place two Aces, 2 Kings, 2 Queens, and 2 Jacks in the spaces below so that:

• Every Ace borders* a King.
• Every King borders a Queen.
• Every Queen borders a Jack.
• No two of the same cards border each other.
• No Ace borders a Queen.

*Border means horizontal or vertical.

two solutions:

__K
QJQ
_AKJ
__A

__K
QJQ
_AKA
__J