The worlds first riddle!

[size=8]PIRATES
Abe: 75
Bob: 150
Cal: 300[/size]

Mark wrote, "Can you remind us of the question?"

No can do, but it was about three tanks that were filling up with rainwater and had to be emptied. The rain stopped, started, doubled etc. Do not tell me you have forgotten all about it, while I have been working night and day to make sense of it!


BTW. As I may be away tomorrow, may I congratulate you as you start your third year in the forum. The number of questions you have answered will never be equalled. A truly magnificent record.



In the sixth, seventh, eighth, and ninth basketball games of the season, Mame scored 23, 14, 11, and 20 points, respectively.

Her points-per-game average was higher after nine games than it was after the first five games.

If her average after 10 games was greater than 18, what is the least number of points she could have scored in the 10th game

Thanks. It's been fun.

[size=8]MAME
29 points

She averaged 17 points in games 6-9. Therefore, she could have scored at most 84 points in games 1-5 for a total of 152 in games 1-9. After ten games, she needed a total of at least 181 points. So, she scored at least 29.[/size]

The problem:

Three identical, empty open-top containers (C1, C2, C3) are outside.

- At 0 minutes, a steady rain begins and starts filling the containers at R liters per minute.

- Seven minutes later, Try starts emptying C1 and empties it at rate of T liters per minute.

- As soon as C1 is empty (instantaneously), Try starts emptying C2 and empties it in 42 minutes.

- Then Try starts emptying C3. At that point, the rate of the rainfall doubles to 2R liters per minute, and Try doubles his speed to 2T liters per minute. Try empties C3, but before he does, the rain stops.

- Try goes back to C1, then to C2, emptying them at the original rate of T liters per minute in 44 minutes.

At what minute does Try finish emptying C3?

Try's answer: 132 minutes

I'll get back to you. I have to solve it again.

Mark:

Basketball
29 points

She averaged 17 points in games 6-9. Therefore, she could have scored at most 84 points in games 1-5 for a total of 152 in games 1-9. After ten games, she needed a total of at least 181 points. So, she scored at least 29.
(Please note 45 words)

(My version 182 words. What did I tell ya'll!)

The least number of points Mame could have scored was 29.
For an average of 10 points to exceed 18 points, there must be at least 181 points. Thus 181 points is our target...

Mame scored a total of 68 points in games 6 through 9, which is a 17 point average.
In order to score least in game 10, Mame must score as much as possible in the first 5 games, but without violating the stated requirement that her average after 9 games was higher than after 5 games.

This means she must average just a little less than the 17 point average that she had in the last 4 games (games 6 through 9).
So, assign 17 points for all but 1 of the first 5 games, and then 16 for the other game. So, she totalled 84 points in the first 5 games. The total for the first 9 games is then 68 + 84 = 152.

To get the least number of points in game #10, she needs
181 (target) - 152 (number so far) = 29 points


PIRATES
Abe: 75
Bob: 150
Cal: 300



Let x = number of coins that Abe brought.
Let y = number of coins that Bob brought.
Let z = number of coins that Cal brought.

After Abe steals coins from Bob and Cal,
Abe has x + y/2 + z/2 which equals 300 coins,
Bob has y/2 coins, and
Cal has z/2 coins.
After Bob steals coins from Abe and Cal,
Abe has 150 coins,
Bob has y/2 + z/4 + 150 coins, and
Cal has z/4 coins.
After Cal steals coins from Abe and Bob,
Abe has 75 coins,
Bob has y/4 + z/8 + 75 coins, and
Cal has z/4 + 75 + y/4 + z/8 + 75 coins.
Therefore, Abe must have brought 75 coins, so x =75.
Since x + y/2 + z/2 = 300 and since x = 75, it follows that y + z = 450.

Since the number of coins that Bob brought equals the number of coins that he wound up with, y = y/4 + z/8 + 75
Multiplying by 8, we get 8y = 2y + z + 600 or 6y - z = 600.

Solving these two equations simultaneously,
y + z = 450 and
6y - z = 600, we get y = 150 and z = 300.

(If there is an easier way, that was the one Mark chose)


Mark wrote, "Thanks. It's been fun."

Sheesh, I nearly swallowed my teeth. ?'Fun' you say! I thought you were doing this as some sort of community service, to escape five years in the slammer.

I raise a glass to your ability. I will see if I can keep it going till Christmas.

So, what do you make of this?


John: so Jack, how many kids you got now?
Jack: less than 5
John: well...how many?
Jack: figure it out; the product of their ages equals twice the sum.
John: that's nice; but not enough info.
Jack: my wife gave birth a year ago.
John: still not enough.
Jack: the age of my oldest is the same as the number on this door.
John: hmmm...that helps...still not enough.
Jack: the two in the middle...
John: stop there! I know how many kids you have, and also their ages...

Can you figure it out

4 kids: 5, 2, 2, 1

has to be 4 kids.... less than five, and 2 middle kids

youngest is 1, and 2 middle are B and B, and oldest is A years old

AB^2 = 2(A + 2B + 1)

A = (4B+2)/(B^2-2)
then i used a graphing calculator to find integer A when B is an integer and got my answer

RAIN
Try: 132 minutes

How did he do it?


The first column shows the elapsed time with variable (unknown) quantities.
Columns 3-5 show the amount of water in the three containers at the various times.
Column 7 shows the equation that can be derived from the information at that point in time.
From the first two equations, we get x = 14 minutes, T/R = 3/2.
Knowing T/R, from the fourth equation, we get y = 6 minutes.
Knowing x, y, and T/R, from the third equation, we get z = 19 minutes.
The second column shows the elapsed time computed by plugging in the values for x, y, and z.

Thoh:

4 kids: 5, 2, 2, 1

has to be 4 kids.... less than five, and 2 middle kids

youngest is 1, and 2 middle are B and B, and oldest is A years old

AB^2 = 2(A + 2B + 1)

A = (4B+2)/(B^2-2)
then i used a graphing calculator to find integer A when B is an integer and got my answer


Nice and logical.



For those with only pen and paper:
Age possibilities:
3 6 *
4 4 *
1 3 8 **
1 4 5
2 2 4 *
1 1 4 6 **
1 2 2 5
1 2 3 3 **
2 2 2 2 *

* eliminated due to youngest(s) being 1 (wife gave birth a year ago).
** eliminated: if number on the door was equal to oldest, then John would know.
So, it's 1-4-5 or 1-2-2-5; 1-2-2-5 is it, due to "two in the middle"...


RAIN
132

Wow! That was a lucky guess of mine . I wonder how many ways there are to reach the same conclusion. I think I have it as:


R = Rain speed
T = Trys speed
A = hours to empty C1
B = hours of rain in C3
C = hours to empty C3 (Note: AR means A*R)
Try enters C1 at hour 8

Gallons already in C1 : 7R
Rain while Try empties: AR
Trys emptying of C1 : AT
so: AT = 7R + AR ; T / R = (A + 7) / A (1)

Try empties C2 in 42 hours
Gallons already in C2 : R(A + 49)
Rain while Try empties: 42R
Trys emptying of C2 : 42T
so: 42T = R(A + 49) ; T / R = (A + 49) / 42 (2)

(1) (2) : (A + 7) / A = (A + 49) / 42
A^2 + 7A - 294 = 0
(A + 21)(A - 14) = 0 ; A = 14 (3)

(1) (3) : 2T = 3R

Try empties C3; both speeds double
Gallons already in C3 : R(A + 49) = 63R
Rain while Try empties : 2RB
Trys emptying of C3 : 3RC
so: 3RC = 2RB + 63R ; C = (2B + 63) / 3 (4)

Try empties C1 and C2 in 44 hours, normal speed
Gallons now in C1 and C2: R(42 + 4B)
Trys emptying of C1/C2: 44(3R / 2)
so: R(42 + 4B) = 44(3R / 2) ; B = 6 (5)

(4) (5) C = 25

Hour at which Try enters C3 : 7 + 14 [A] + 42…………………= 63
Hour at which rain stopped : 7 + 14 + 42 + 6 ………….= 69
Hour at which Try is finished: 7 + 14 + 42 + 25 [C] + 44 = 132



It is starting to rain on Marks parade:

-open-top empty container C is outside
-at 0 hours, a steady heavy rain starts
-rain enters C at rate of R gallons per hour (gph)
-Mark jumps in C as soon as the rain starts
-Mark takes water out at 1/6 R gph for 9 hours
-at this point, rain slows to 1/3 R gph, and Mark triples his speed
-13 hours later, the rain stops
-at this point, Mark reduces his current speed by 8 gph,
and empties C in 10 hours

OK; if Mark handled a total of 778 gallons in this 32 hour period, what is the size of C in gallons, and the rain speed at start

[size=8]RAIN 2
492 gallons
R=66[/size]

Whilst we agree on the R=66. There appears to be a difference of 99 in the gallons, this is interesting. Time will tell.

I got the same R and C as Try....maybe the mistake is in assuming Markr is volumeless

I see the error of my ways.
[size=8]393[/size]gallons

Thoh wrote, "I got the same R and C as Try"

This has never happened before; I have added you to my Christmas card list.


In any event, we all agree that:


R = Rain original speed
Easy to see that Mark's 778 gallons are this way:
9(R / 6) + 13(R / 2) + 10(R / 2 - 8) = 778
Therefore: R = 66

Total rained gallons = 9(66) + 13(22) = 880

1- C overflowed during start 9 hour period (102 gallons)
2- 13hour period begins with a full C
3- Mark emptied C plus 286 gallons (the 13 hour rainfall), in last 2 sessions, or over 23 hours:
C + 13(R / 3) = 13(R / 2) + 10(R / 2 - 8)
Since R=66, then C = 393 gallons.





For the purposes of this puzzle, we now have a new weight:
Similar to the pound/ounce or Kilo/whatever. We now have the JIZZ/KIXX.

(I took the opportunity to give the lonely j/k/x/z some work!).
Mark orders a bag of Ontario potatoes, asking for j JIZZES and k KIXXES.
The potato man goofs and prepares a bag weighing k JIZZES and j KIXXES.
The total KIXXES Mark ordered are less than 500.

YOU: good; see ya later
ME : the clerk's error caused Mark's order to be reduced by P%, P being an integral in the range 1 to 99.
YOU: gotta go
ME : ah c'mon, how many KIXXES in a JIZZ?
YOU: oh boy (leave, come back with printout); hmmm...83 possibilities; gimme a clue
ME: P is equal to this number here
YOU: that sure helps, but need another clue
ME : Mark's order was over 100 KIXXES ...and now you know...

How many KIXXES in a JIZZ

Don't feel bad than no one in the whole wide world could answer the above question, because the same applies to this one.




Count, but spell out the values: ONE, TWO, THREE, ...
In other words, what is the first number that contains the letter "C"


What numbers have ?'J' ?'K' or ?'Z' in it

Note - No Zillion or Roman numerals

haven't started on the jizzes, kixxes problem yet

[size=8]1,000,000,000,000,000,000,000,000,000 (octillion)

zero[/size]

[size=8]KIXXES IN A JIZZ
I found 243 different valid combinations (84 different total KIXXES values). I'm not sure why the problem says 83.

A 35% reduction is the only case where knowing that the order was greater than 100 KIXXES reduces the possibilities to one.
8 KIXXES/JIZZ yields a total order of 60 KIXXES.
43 KIXXES/JIZZ yields a total order of 480 KIXXES.

There are 43 KIXXES/JIZZ.[/size]

Now, here's one for you:

Same problem but different last statement.
Replace
"Mark's order was over 100 KIXXES."
with
"Mark's order was a multiple of five KIXXES."

I hope you kept your data from the original problem.

Mark:

Number with a ?'C'.

1,000,000,000,000,000,000,000,000,000 (octillion)

numbers have ?'J' ?'K' or ?'Z' in it.
zero

KIXXES IN A JIZZ
I found 243 different valid combinations (84 different total KIXXES values). I'm not sure why the problem says 83.

A 35% reduction is the only case where knowing that the order was greater than 100 KIXXES reduces the possibilities to one.
8 KIXXES/JIZZ yields a total order of 60 KIXXES.
43 KIXXES/JIZZ yields a total order of 480 KIXXES.

There are 43 KIXXES/JIZZ.



[code]


TOTAL………..KIXXES IN……………………………..KIZZES IF……….% REDUCTION
KIXXES……..A JIZZ……KIXXES..JIZZES…REVERSED……….TO TOTAL KIZZES
60………………8……………………7………….4…………….39………………….35 ( 60*.65=39 )
480…………..43…………………11………….7……………312………………….35 (480*.65=312)


[CODE]



Among the 83 possibilities <500>100 total KIXXES clue eliminates the smaller one.
So there are 43 KIXXES in a JIZZ.
These are the ONLY possibilities that end up with you KNOWING after the clues.
All others are either 1 choice (then you'd know as soon as shown the number) or 3 choices or more with more than 1 in
the >100 range.









After Tim had set the table for four people -- his parents, his brother Tom, and himself -- they immediately sat down to eat.

Your mission is to state the exact time that they started dinner (the value of ATE)

Each letter must represent the same digit, and no leading zeros, if you please.


SET
SET
SET
SET
-----+
ATE

7:28PM

4 x 182 = 728

Thoh, you make a convincing case. I will go check my watch. In the meantime:






Using only pennies, nickels, dimes, quarters, half dollars, $1 bills, $5 bills, and $10 bills, what is the most money that you can have WITHOUT being able to change a $20 bill

[size=8]$20.19[/size]