The worlds first riddle!

middle of the night? that was around 11:00am my time.


Where in H-E-double hockey sticks are you?

I get
5 calves
1 lamb
94 pigs
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Farmer Mark hits pay dirt again.
What are the odds that he could have guessed the correct answer?

Here are a few ways he didn’t use to guess the answer:
10c + 3s + p/2 =100
20c + 6s + p = 200
but c + s + p = 100

Subtract
19c + 5s = 100
c = 5, s = 1 and p = 94
Cost 50 + 3 + 47 = 100
5 cows @ $10 ($50) , 1 sheep @ $3 ($3) and 94 pigs @ 50c ($47)

Restating, you are to buy three items, say x, y, and z (all integers) and the sum of these items is 100. You have $100 to spend on the three items with their individual costs being $10.00, $3.00, or $.50, respectively. You are to buy at least one of each item.

With all of this information, you can derive the following relationships:
(1)--x + y + z = 100 (1)
(2)--10x + 3y + .5z = 100 (2)

By Observation
3---Multiplying (1) by .5 yields .5x + .5y + .5z = 50 (3)
4---Subtracting (3) from (2) yields 9.5x + 2.5y = 50 (4)
5---Dividing (4) through by 2.5 yields 3.8x + y = 20 (5)
6---By observation, the only value of x that will make 3.8x an integer is 5
7---Therefore 3.8(5) + y = 20 making y = 1
8---Thus, z = 94.
Check: 5 + 1 + 94 = 100 and 10(5) + 3(1) + .5(94) = 50 + 3 + 47 = 100. Okay.

By Observation
Another way of viewing it is to say that the average cost of each item is $1.00. Each x's cost differs from the average by +$9.00, each y's cost by + $2.00, and each z's cost by - $.50. Therefore, for each x he must buy 18 z's and for each y he must buy 4 z's. Thus, since 5(1 + 18) + (1 + 4) = 100, he must buy 5 x's, 1 y, and 94 z's.

Analytical method
(3)---Multiplying (2) by 2 yields 19x + 5y = 100 (3)
(4)---Dividing (3) through by 5 yields 3x + 4x/5 + y = 20 (4)
(5)---Rearranging, 4x/5 = 20 - 3x - y
(6)---By definition, 4x/5 must be an integer but we want a unit coefficient for x.
(7)---Multilply 4x/5 by 4 and divide by 5 again yielding 16x/5 = 3x + x/5.
(8)---x/5 must also be an integer so let x/5 = k making x = 5k
(9)---Substituting back into (3) we get 95k + 5y = 100 or y = 20 - 19k
(10)---In order to have at least one of each, clearly k can only be 1 making
(11)---k 0 1 2
x 0 5 10
y 1 1 neg neg = negative
z 99 94 neg
(12)---Therefore the only valid solution is x = 5, y = 1, and z = 94.



Missy, will y’all hush up; iffin’ the bayou boyz get to hear I woz up before midday they would hogtie me to a Liriodendron tulipifera. Ima in a right state; books everywhere!

Are you volunteering to call on over for a little light dusting, because I need to measure your waist? Oh, BTW, can you bring a hula hoop?

Now, imagine if you will, a vertical Missy (no, I didn’t say horizontal – that would be to easy!!! Hhaha) and she has a circular waist, and is temporarily at rest procrastinating.

Around her salubrious waist rotates a hula hoop of twice its diameter.

Can you show that after just one revolution of the hoop, the point originally in contact with Missy has traveled a distance equal to the perimeter of a square circumscribing her sultry waist?

Dear friends and reprobates, thank you for your words of concern for Mark’s health after he failed to answer the last conundrum in under seven minutes.

May I assure you he is fit and well and probably on vacation. The answer is so deceptively simple he would have had no difficulty with it.


Since motion is relative (although Missy is not related to me) consider the hoop as fixed and the poor girl whirling around.

The original point of contact on Missy traverses the diameter of the hoop twice, and this is the required distance.

I provide (at no cost to the reader) a pictorial cartouche (except it is circular) so it is no cartouche.




As Missy whirls provocatively, the original point of contact C on her succulent waist traverses the diameter BD, since:
Arc AB = R0 = (R/2) (20) = arc AC.



As today is Friday, does this make sense …
if ¼ of 20 is 6, then what is 1/5 ?

Happy Easter dudes.

No vacation - just didn't have an answer.

1/4 of 20:
Since 6 is 1/4 of 20 in base 12, I'll go with 4 4/5 (four and four fifths)
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I hear that Mark; I told Missy it would be much fairer if she was to post a picture of her lissom midriff so one could get a much better handle on the problem!

It will therefore not come as a surprise if I state that the thought of Missy’s waist is to blame for a typographical malfunction in which: ‘of 10’ was misplaced from the end of the last question.

If ¼ of 20 is 6, then what is 1/5 (of 10)?

However as Mark has already stated the computation must be in the duodecimal scale, which has the base twelve.

Hence (if I had written it complete) 1/5 of 10 is 22/5, since 1012 = 1210.
.......................................................Hidden ^



I was fixin’ up supper last night frying some ham and eggs, when I made the most amazing discovery…

7(F R Y H A M ) = 6(H A M F R Y)

Can you ID the digits?



Being a dirt farmer I nose (I said I was a farmer, not an English professor!) that you get ham from a lamb. So when my neighbor died he left his two sons a herd of cattle which they promptly sold.

The number of dollars received per head was the same as the number of heads. With the proceeds of the sale the sons bought sheep at $10 each and one lamb for less than $10.

The sheep and lamb were divided between the two brothers so that each received the same number of animals.

So I ask you; how much should the son who received only sheep pay to the son who received the lamb, in order that the division should be equitable?

Note: The answer should be self-evident to anyone who had a Ph.D. in Quantum coherence or works in the A2K cafeteria.

What’s for supper tonight Missy?

Pork Roast and carrots, onions and peas...a little rice and some crusty rolls. And sweet tea of course.

That sure sounds mighty fine eatin’; iffin it weren’t for all this here rain we is having I would rightly mosey on over and seek sweet charity.

Perhaps you could save a roll in the barn for me?

It is rainy and cold. Hate it.

Roll in the barn - just for you.

FRYHAM:
6994*FRY = 5993*HAM
Divide both sides by 13:
538*FRY = 461*HAM
FRY = 461
HAM = 538
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lamb:
The number of dollars is equal to the number of cattle squared. For an equal distribution of animals, the number of sheep must be odd, so the number of dollars must be an odd multiple of 10 plus the price of the lamb. Let the number of cattle be 10x + y (y < 10).
(10x + y)^2 = 100x + 20xy + y^2
Since 100x and 20xy are even multiples of 10, y^2 must provide an odd digit in the tens place. Only 4 and 6 do this, and they both yield a 6 in the units place.
Therefore, a lamb costs $6.
So, the son with sheep only must pay (10-6)/2 dollars to the son who got the sheep.
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There is more than one way to fry a ham:

Let FRY = x and HAM = y, then
7(1000x + y = 6(1000y + x)
6994x = 5993y
538x = 461y
Since the numerical coefficients are relatively prime, it follows that;
x = FRY = 461 and y = HAM = 538.

Which is just what Mark said.



There are times when an answer is framed so well it supersedes the question itself; I submit this answer of Mark’s is one of those:

A sheepish lamb:

The number of dollars is equal to the number of cattle squared. For an equal distribution of animals, the number of sheep must be odd, so the number of dollars must be an odd multiple of 10 plus the price of the lamb. Let the number of cattle be 10x + y (y < 10).
(10x + y)^2 = 100x + 20xy + y^2

Since 100x and 20xy are even multiples of 10, y^2 must provide an odd digit in the tens place. Only 4 and 6 do this, and they both yield a 6 in the units place.

Therefore, a lamb costs $6.
So, the son with sheep only must pay (10-6)/2 dollars to the son who got the sheep.


Which is a beautiful way of saying:

If x is the number of cattle, y the number of sheep, z the price of the lamb, we have x2 = 10y + z with y odd and z<10.

But the penultimate digit of a square is odd if and only if the last digit is 6.
Thus z = 6, and the luckier son should hand over $2.

So let us give thanks for such a magnificent breath taking, heart stopping performance-
No, not you Mark; I refer to the roll in the hay with Missy!
Thank you for your succour and for giving me the opportunity to sow some wild oats – Hallelujah!


Needless to say that after such gambolling I just wanted to lay back and have a smoke, when suddenly Missy said, “Hey fatso, put down the defibrillator and if you wish for dessert; I wanna know:”

If five cards are drawn at random from a pack of cards which have been numbered consecutively from 1 to 97 and thoroughly shuffled.

What is the probability that the numbers on the cards as drawn are in increasing order of magnitude?



My guess is about the same probability as The New York Yankees winning twice as many pennants as the next closest team.

What do y’all think?

5 cards:
Interestingly, it doesn't matter how many cards are in the deck. If you draw N cards, the probability is 1/N!.

The number of ways to draw N cards from a pack of M cards (order matters) is M!/(M-N)!.
The number of ways to draw N cards from a pack of M cards in increasing order is the number of ways to draw N cards from a pack of M cards since each combination corresponds to a unique increasing set. This number is C(M,N) which is M!/(N! * (M-N)!).

Therefore, the probability is 1/N! and M (the size of the deck) is irrelevant.
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Normally it is Mark who is punctilious and succinct, so I leave it to the erudite reader to decide:

Interestingly, it doesn't matter how many cards are in the deck. If you draw N cards, the probability is 1/N!.

The number of ways to draw N cards from a pack of M cards (order matters) is M!/(M-N)!.

The number of ways to draw N cards from a pack of M cards in increasing order is the number of ways to draw N cards from a pack of M cards since each combination corresponds to a unique increasing set. This number is C(M,N) which is M!/(N! * (M-N)!).

Therefore, the probability is 1/N! and M (the size of the deck) is irrelevant.


Mark is indeed correct in his assertion that we need be concerned only with the order of the cards drawn.

Therefore the permutations of any five numbers are 5! so the probability is 1/120.



Have you ever been a sippin’ a glass of Egon Muller-Scharzhof Scharzhofberger Riesling Trockenbeerenauslese and suddenly had the overwhelming desire to know:

What square is the product of four consecutive odd integers?

No, neither have I; but the question still stands iffin’ y’ll want to get into Harvard.

square:
9 = -1 * -3 * 1 * 3
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At $6000/bottle, no.

9 = -1 * -3 * 1 * 3

Another pearl of wisdom from Mark the jeweller.

If n (n+2 (n+4) (n+6) = m2, then (n 2 +6n+4)2 = m 2 +16.
But only 0 and 9 are squares of the form a2 -16, and since m2 is odd, the square sought must be 9 = (-3) (-1) (1) (3).

$6.000 a bottle you say; yuck! I would use that as a mouthwash.


Ships that pass in the night… I had arranged a clandestine meeting with Missy down near Mobile Bay. She was on the East bank and I on the West. Two ferry boats ply back and forth across this river with constant speeds, turning at the banks without loss of time (don’t ask).

We embarked at 10pm. And both boats left the opposite shores at the same instant, with my heart pounding they met for the first time 700 feet from one shore, horror upon horror, they continued on their ways to the banks, returned and meet for the second time 400 feet from the opposite shore.

Anxious to splice the mainbrace; what the hell is the width of this here river?

Aintry? This river don't go to Aintry!

river:
518.16 meters
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Mark has spoken in some foreign devils language and speaking in tongues has come up with: 518.16 meters!

Ha, ha; now let’s see what the answer is in good o’le American distances…



Boat A left one shore, traveled 700 feet and met B. Together they had travelled the width of the river. A continued across the river to the opposite shore and back 400 feet, where it met B again.

Together they had traveled a total of three times the width of the river.

As their speeds were constant, A traveled three times 700 feet, or 2100 feet.
The width of the river was therefore 400 feet less than the distance A travelled; that is 1700 feet.

And what did Mark say; 518.16 meters! You have just gotta laugh… Wait, the jokes on me, I have just been told by a French interpreter that 518.16 meters = 1700 feet!

Dang! These foreign devils is smart!


I have decided to start the A2K Riddles club; an exclusive fraternity named after Eta Sigma Chi. If there had been five more in the group, the initial expense to each member would have been $100 less.

What was the initial cost per person?

Ps. cash only – leave in reception in plain brown envelope - thank you.

cost:
each member paid $100 + $20 per member
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Damn. It has happened again:
Words disappeared from the post!

Second try –

I have decided to start the A2K Riddles club; an exclusive fraternity (with an initial 10 members) named after Eta Sigma Chi. If there had been five more in the group, the initial expense to each member would have been $100 less.

What was the initial cost per person?

Post script: Excellent answer Mark, seeing how little you had to work with.