The worlds first riddle!

If the 9 dots are represented by:
123
456
789

then:
Line 1= 159
Line 2= 963#
Line 3= #24#
Line 4= #789

#= going outside the box. I hope that makes sense even though (thoh) lmao2, someone had to help me in how to write it. :wink:

yup TTH, that's correct
you have to think outside the box for that one

Mark:

HYMNS
Here's how many are required of each digit:
0 - 8
1 - 9
2 - 9
3 - 9
4 - 9
5 - 9
6 - 9
7 - 8
8 - 8
9 - 8
Total = 86

However, if 6 can do double duty as 9 we have:
6 - 12
9 - 0
Total = 81


Way to go Dude!


TTH: "I will place my bet on Mark"

Good choice.



?'Everyone can row except the cat.

THOH: "im sorry...but LMFAO"


Paws for laughter!



I see THoH and TTH have been playing around, nuff said I think. :wink:





Three Forums -- Riddles, Trivia and History -- competed in a track meet. Each team entered one person, and one only, in each event. Stormy, a teecher at Trivia High, sat in the bleachers to cheer her friend, the A2kay's shot-put champion. When Stormy returned home later in the day, her father asked how her Forum had done.

"We won the shot-put all right," she said, "but Riddles High won the track meet. They had a final score of 22. We finished with 9. So did History High."

"How were the events scored?" her father asked. "I don't remember exactly ," She replied, "but there was a certain number of points for the winner of each event, a smaller number for second place and a still smaller number for third place. The numbers were the same for all events."

(By "number" She of course meant a positive integer.)

"How many events were there altogether?" "Gosh, I don't know, Dad. All I watched was the shot-put."

"Was there a high jump?" asked Try. (Don't ask what he was doing there) Stormy nodded.

"Who won it?"
Stormy didn't know.

Incredible as it may seem, this last question can be answered with only the information given. Which Forum won the high jump AND how many points were awarded for first place in each event


Even more incredible, you can actually reconstruct the entire track meet from the comments above! You can determine the number of events and the place in which each Forum finished in each event. You can also determine the score for each event and the total scores for each Forum! *



* When I say, ?'you' I of cause mean me. Coz I iz smarter than a dead squirrel.


BTW: In reply to the correspondent who wanted to know if I had any information about ?'Royalty', All I can say is; Once a King always a King, but once a Knight never seems enough!

[size=8]TRACK MEET
Well, I had typed up the whole explanation (factors of 40, etc.), then hit a wrong key and lost it all. Since I don't feel like retyping it, I'll just give the final results:

Event Riddles Trivia History
----------------------------------
SP . . . 2nd/2 . 1st/5 . 3rd/1
HJ . .. . 1st/5 . 3rd/1 . 2nd/2
#3 . . . 1st/5 . 3rd/1 . 2nd/2
#4 . . . 1st/5 . 3rd/1 . 2nd/2
#5 . . . 1st/5 . 3rd/1 . 2nd/2
----------------------------------
. . . . . . . . 22 . . . . 9 . . . . . 9[/size]

Mark:

TRACK MEET
Well, I had typed up the whole explanation (factors of 40, etc.), then hit a wrong key and lost it all.

(I know that feeling, there should be a ?'put it back like it was meant to be' button.)


Since I don't feel like retyping it, I'll just give the final results:

Event Riddles Trivia History
----------------------------------
SP . . . 2nd/2 . 1st/5 . 3rd/1
HJ . .. . 1st/5 . 3rd/1 . 2nd/2
#3 . . . 1st/5 . 3rd/1 . 2nd/2
#4 . . . 1st/5 . 3rd/1 . 2nd/2
#5 . . . 1st/5 . 3rd/1 . 2nd/2
----------------------------------
. . . . . . . . 22 . . . . 9 . . . . . 9





Riddles won the high jump event in the track meet involving the three forums.

Five points were awarded for each first place Three different positive integers provide points for first, second and third place in each event. The integer for first place must be at least 3.

We know there are at least two events in the track meet, and that Trivia (which won the shot-put) had a final score of 9, so the integer for first place cannot be more than 8. Can it be 8? No, because then only two events could take place and there is no way that Riddles could build up a total of 22 points. It cannot be 7 because this permits no more than three events, and three are still not sufficient to enable Riddles to reach a score of 22.

Slightly more involved arguments eliminate 6, 4 and 3 as the integer for first place. Only 5 remains as a possibility. p If 5 is the value for first place, there must be at least five events in the meet. (Fewer events are not sufficient to give Riddles a total of 22, and more than five would raise Trivia's total to more than 9.)

Trivia scored 5 for the shot-put, so its four other scores must be 1. Riddles can now reach 22 in only two ways: 4, 5, 5, 5, 3 or 2, 5, 5, 5, 5. The first is eliminated because it gives History a score of 17, and we know that this score is 9. The remaining possibility gives History a correct final tally, so we have the unique reconstruction of the scoring shown by Mark.

Riddles won all events except the shot-put, consequently it must have won the high jump.

This problem can be solved without any calculation whatever. The necessary clue is in the last paragraph. The solution to the integer equations must indicate without ambiguity which forum won the high jump. This can only be done if one forum won all the events, not counting the shot-put; otherwise the problem could not be solved with the information given, even after calculating the scoring and number of events.

Since the forum that won the shot-put was not the over-all winner, it is obvious that the over-all winner won the remaining events. Hence without calculation it can be said that Riddles won the high jump.

Way to go Riddles!!!



To celebrate the extraordinary dexterity of the Riddles victory…


You're given a hundred dollars and told to spend it all purchasing exactly a hundred animals at the pet store.

Dogs cost $15. Cats cost a dollar, and mice are 25 cents each.
You have to purchase at least one of each animal.

The question is, how many of each animal do you have to purchase to equal one hundred animals purchased at exactly one hundred dollars

[size=7]a+b+c=100
5a+b+1/4c=100
so
20a+4b+c=400
subtracting
4a+4b+4c=400
you get
16a-3c=0
since 16 and 5 are relatively prime
a=3 and c=16
and
b=100-3-16=81
So there are 3 dogs, 81 cats and 16 mice
Check
3*5+1*81+16/4=15+81+4=100.[/size]

Rap

Pet Shop

3 dogs.............$45
41 cats............$41
56 mice...........$14
_________________
100 animals...$100

Speaking of cats...

there's a bunch of cats...
each cat sees three cats...
each cat sits on a cat's tail...
What's the least amount of cats possible?

CATS
4

CATZ

Two and a mirror.

Rap

CATS:

If a cat can see its self; then 3.




Rap^:

a+b+c=100
5a+b+1/4c=100
so
20a+4b+c=400
subtracting
4a+4b+4c=400
you get
16a-3c=0
since 16 and 5 are relatively prime
a=3 and c=16
and
b=100-3-16=81
So there are 3 dogs, 81 cats and 16 mice
Check
3*5+1*81+16/4=15+81+4=100.

What!
3Dogs=$45
81 Cats=$81
16 Mice=$4
Total = $130

You dun only got $100! :wink:


Stormy:

Pet Shop

3 dogs.............$45
41 cats............$41
56 mice...........$14
_________________
100 animals...$100


Which all goes to show; There is only one way to skin a cat!





Let D stand for the number of dogs, C for the number of cats, and M stand for the number of mice.

Then you can write two equations:
D + C + M = 100 and
15 D + 1 C + 0.25 M = 100.00.

It wasn't going to be five dogs because five dogs would eat up $75 and then you'd only have $25 with which to buy another 95 animals, and it wasn't going to work.

And you know it wasn't zero dogs because that was one of the conditions of the problem.

So, use trial and error with dogs = 4, then 3, 2, and 1 to see which works. The answer turns out to be 3 dogs, 41 cats and 56 mice.




The following Javascript solves the problem:


for (Dogs=1; Dogs<=6; Dogs++) {
for (Cats=1; Cats<=99; Cats++) {
for (Mice=1; Mice<=399; Mice++) {
Cost=1500*Dogs+100*Cats+25*Mice;
if ((Dogs*1+Cats*1+Mice*1 == 100) && (Cost == 10000))
{alert(Dogs+' '+Cats+' '+Mice);
}}}}





A certain A2K member has 5 daughters: Emily, Jane, Betsy, Abigail, and Nancy.

Their fortunes are as follows:

The first two and the last two have $19,000;
The first four, $19,200;
The last four, $20,000;
The first and the last three, $20,500;
The first three and the last, $21,300.

They all live in CA so they are pretty rich.


What is the fortune of each

Now Try...you know that math is not my strong suit. Where are my wacky word games my dear?

Oops $15 per dog, not $5!

Then its
a+b+c=100
&
15a+b+1/4c=100
so
56a-3c=0
& 56 & 3 are relatively prime
so c=56 & d=3 & a=41
or
56 mice
3 dogs
& 41 cats
$45+$41+$14=$100

BTW my answer was the only one for $5 dogs.

Rap

Emily, 5,000
Jane, 4,500
Betsy, 6,000
Abigail 3,700
Nancy 5,800

What's the penalty for quigamy in CA?

[size=7]Set it up as a matrix and you can reduce to a=5, b=4.5, c=6, d=3.7 & e=5.8

All I kin K in a way

Emily=$5K
Jane=$4.5K
Betsy=$6K
Abigail=$3.7K
Nancy=$5.8K[/size]

Rap

$25,000, but they all look like Try; so it wouldn't be worth it no matter what their fortunes amounted to!

of course
just arrange the cats in four corners of a room and have each cat sit on its own tail

Sweet Shari, how very nice to see your back, and your front too!
May I wish you everything you would wish for yourself, for your upcoming birthday.

I will be working that day, so I am very sorry to say I will not be able to attend your party.

Have a great time!



Rap, "BTW my answer was the only one for $5 dogs."


Nice move; writing your own questions to fit the answer!



Rap2:

Set it up as a matrix and you can reduce to a=5, b=4.5, c=6, d=3.7 & e=5.8

All I kin K in a way

Emily=$5K
Jane=$4.5K
Betsy=$6K
Abigail=$3.7K
Nancy=$5.8K





Imur:

Emily, 5,000
Jane, 4,500
Betsy, 6,000
Abigail 3,700
Nancy 5,800


Good to see you dude, where have you been? It's the Irish elections, is that what kept you?


To solve this problem, set up five equations with five variables and solve simultaneously.
The five equations are:


E + J + A + N = 19,000
E + J + B + A = 19,200
J + B + A + N = 20,000
E + B + A + N = 20,500
E + J + B + N = 21,300



"What's the penalty for quigamy in CA?"

I don't know, but it sure would be fun to find out!



Mark, "$25,000, but they all look like Try; so it wouldn't be worth it no matter what their fortunes amounted to!"


Oh! So they are all blondanddelicious are they?




Thoh: CATS - "…each cat sit on its own tail"


Very clever, I would never have thought of that.





Anyhoo, the five chicks mentioned above were supposed to fill in the missing numbers, they have all gone round to Mark's pool. In their absence can anybody assist

Determine the missing entries A, B, C, D, and E so the sum of each row, column, and diagonal is the same.

The matrix is
13 4.25 10.5
6.75 9.25 11.75
8 14.25 5.5
all sum to 27.75

Rap

Rap:

The matrix is
13 4.25 10.5
6.75 9.25 11.75
8 14.25 5.5

All sum to 27.75


Yo bro, way tago.



Since the 3 numbers in each row must be the same, equate the middle row and the last column to find C:
6.75 + C + D = 10.5 + D + 5.5

Therefore C = 16 - 6.75 = 9.25
Then the sum must be 8 + 9.25 + 10.5 = 27.75.

A = 13
B = 4.25
E = 14.25 and
D = 11.75






In the snack bar at A2kay towers there five different products: Italian Ices, Cream Ices, Gelati, Misto, and Frozen Custard.

A visitor states that there are so many different choices that they have a difficult time making up their minds.

(Ya, I know the standard of visitor is not up to much, but this is the Wall Street HQ.)

Can you determine how many different choices there are using the information below

There are 26 flavors of Italian Ice.
There are 5 flavors of Cream Ice.
Gelati is made by layering either vanilla or chocolate Frozen Custard with the customer's favorite flavor of Italian Ice or Cream Ice.
Misto Shakes are made by blending either vanilla or chocolate frozen custard with the customer's choice of Italian Ice or Cream Ice.
There are two flavors of Frozen Custard.





Due to the festival of Lesbos, I am of to the Virgin Isles to help out. I leave you with a little something for the W/E




Out west, there are long stretches where Interstate 80 runs parallel to the Union Pacific railroad track. Last summer, I set my car on cruise control at 64 m.p.h. It took me 34 seconds to pass a 900 foot long train.

(i.e., it took 34 seconds from the time my car caught up to the end of the train until it pulled even with the front of the train).

If both the train and car moved at constant rates of speed, how fast was the train going

How far did the car travel before it overtook the train