The worlds first riddle!

Treasure Chest
[size=7]I'll take the heavier one & that will be the one filled with the smaller coins. The larger coins will be one of two cases. They will be a larger diameter or they will be thicker.

Larger diameter coins will tile the chest bottom in a larger cell and leave a larger void volume than the smaller diameter coins in the fixed bottom of the chest.

Thicker coins will have a greater layering restriction than thinner ones.

So at best the treasure chest of larger coins will be as heavy as the treasure chest of smaller ones?-so the smaller coins is the better bet[/size]

Favorite number
[size=7]#*9*12345679=#*111,111,111[/size]

Magic 3
[size=7](2x+12)/4-x/2=(2x-2x+12)/4=12/4=3
so with the holy hand grenade thou shall count to three. Thou shall not count to two for that shall be one too little, nor shall thou count to four for that is one too many. Three is the number thou shall count, no more, no less.[/size]

Rap c∫;?/

Rap:

3 identical numbers
I realize this involves a multiplication, but I like it bettr'n the traditional 11+1 answer-----3*3+3=12

Less than 18
This is a sight gag
How about 171->17.1 & 2->.2 & 17.1+.2=17.3<18
or I could modify 171 with a / so 171->17 1/2<18
or put the 2 over the top so it becomes 2/171<18 or 17/12<18 or 1/712<18 and so on

Half of 11
Roman numeral sight gag
11=XI, clice XI in half horizontally and throw away the bottom half so XI->VI which is 6.

Five Digit #
ABCDE
no digit identical
2 are perfect squares (1,4,9)
@ are prime (2,3,5,7)
One is neither (6,8)
B-D-6
E=C/2
A=E+3

E has to be the starting point.
E<5 so E is a perfect Square or is prime
try E=4 (perfect sq)
then C=8 (neither prime or sq)
A=4+3=7 (prime)
now need a remaining perfect sq and prime s.t. B=D-6
if D=9 (sq) then B=3 prime
so the number is
73894 two prime (7,3), two sq (9,4), and one neither (8), the first is three more than the last, the third is twice the last, and the second is 6 less than the fourth---but is it unique?

E=3 C=6, then A=6 won't work
E=2 then C=4, A=5 and D=9 and B=5, 2 sq, 3 prime won't work
E=1, then C=2, A=4 and D=9, B=5, 3SQ, 2 prime won't work

So 73894 is the unique answer.




Mark:

THREE IDENTICAL NUMBERS
1 + 11

HALF OF ELEVEN
11 = XI
cut it in half horizontally and keep the top half
6 = VI

FIVE-DIGIT NUMBER
73894 if probably what you're looking for
counting 0 as a square, 50462 is another solution




TREASURE CHESTS
I'll take the heavier one.
I'll take the one with small coins.

(The same value in each box! Solve with a quantity/value table.)
Example: Gold coin Quantity/ Value /Total
Small: 60/ $10/ $600
Large: 30/ $20/ $600


FAVORITE NUMBER
9*12345679 = 111,111,111

(You will never see a more concise explanation than that)

PICK A NUMBER
(2N+12)/4 - N/2 = 3

(On the other hand he could have said):

Solution solve by algebra:
Let X be the number.
Double it => 2X.
Add 12 => 2X + 12.
Divide by 4 => (2X + 12)/4.
Subtract ½ of the original number => (2X + 12)/4 - X/2.
When this algebraic expression is simplified, you get 2X/4 + 12/4 - X/2. This is then reduced to X/2 + 3 - X/2, which is just 3.
So, regardless of the selected number, the arithmetic operations introduce a 3 and removes the originally selected number!


Rap:

Treasure Chest
I'll take the heavier one & that will be the one filled with the smaller coins. The larger coins will be one of two cases. They will be a larger diameter or they will be thicker.

Larger diameter coins will tile the chest bottom in a larger cell and leave a larger void volume than the smaller diameter coins in the fixed bottom of the chest.

Thicker coins will have a greater layering restriction than thinner ones.

So at best the treasure chest of larger coins will be as heavy as the treasure chest of smaller ones?-so the smaller coins is the better bet

"The larger coin has twice the amount of gold, is twice the size, and is twice the value of the smaller coin. " The coins were ?'square' Chinese coins.

Favorite number
#*9*12345679=#*111,111,111

Magic 3
(2x+12)/4-x/2=(2x-2x+12)/4=12/4=3
so with the holy hand grenade thou shall count to three. Thou shall not count to two for that shall be one too little, nor shall thou count to four for that is one too many. Three is the number thou shall count, no more, no less.


(Can you change something in that problem and make the result a different number)


Rap is a poet, and boy, he sure do know it.

An engineer was crossing a road one day when a frog called out to him and said, "If you kiss me, I'll turn into a Material Girl".

He bent over, picked up the frog and put it in his pocket.
The frog spoke up again and said, "If you kiss me and turn me back into a Material Girl, I will stay with you for one week." The engineer took the frog out of his pocket, smiled at it and returned it to the pocket.

The frog then cried out, "If you kiss me and turn me back into a girl, I'll stay with you and do ANYTHING you want." Again, the engineer took the frog out, smiled at it and put it back into his pocket.

Finally, the frog asked, "What is the matter? I've told you I'm a Material Girl, that I'll stay with you for a week and do anything you want. Why won't you kiss me?"

The engineer said, "Look I'm an engineer. I don't have time for a girlfriend, but a talking frog, now that's ."



Incredible INTEGERS

Form a 3 x 3 square with integers 1 - 9, where any row, column or diagonal adds to 15

Form a 4 x 4 square with integers 1 - 16, where any row, column or diagonal adds to 34


A circle is only a square with the sharp bits cut off.

Six boys pick a captain by forming a circle then eliminating every n'th boy. The 2nd boy in the counting order can choose n. If he wants to be captain what's the smallest n he should pick


12 black and one white mouse are in a circle. Where should a cat start so that if he eats every 13th mouse the white mouse will be last

[size=7]3X3
2 7 6
9 5 1
4 3 8

4X4
16 03 02 13
05 10 11 08
09 06 07 12
04 15 14 01

SIX BOYS
10

MICE
Put the white mouse in the eighth position. So, start seven mice before the white mouse.[/size]

Magic Squares
[size=7]3by3
8,1,6
3,5,7
4,9,2
4by4
16,3,2,13
5,10,11,8
9,6,7,12
4,15,14,1

I used an algorithm I learned in a Math History Class--BTW I believe this is the magic square in Durer's 'Melancholia'[/size]

The Cabin boy
[size=7]n=3
1st 2,3,4 4 out
2nd 5,6,1 1 out
3rd 2,3,5 5 out
4th 6,2,3 3 out
5th 6,2,6 6 out
2 sole survivor[/size]

Cat-n-'white' mouse
[size=7]put the white mouse in the eighth position and it will be the last.[/size]

Rap

Magic 3
(Can you change something in that problem and make the result a different number)

Solution:
Yes, you can add 16 instead of 12, and the result is always 4.
Or, you can add 20 instead of 12, and the result is always 5, etc.

Mark:

3X3
2 7 6
9 5 1
4 3 8

4X4
16 03 02 13
05 10 11 08
09 06 07 12
04 15 14 01

SIX BOYS
10

MICE
Put the white mouse in the eighth position. So, start seven mice before the white mouse.




Rap:

Magic Squares
3by3
8,1,6
3,5,7
4,9,2
4by4
16,3,2,13
5,10,11,8
9,6,7,12
4,15,14,1

I used an algorithm I learned in a Math History Class--BTW I believe this is the magic square in Durer's 'Melancholia' (Made me sad) :wink:



The Cabin boy
n=3
1st 2,3,4 4 out
2nd 5,6,1 1 out
3rd 2,3,5 5 out
4th 6,2,3 3 out
5th 6,2,6 6 out
2 sole survivor


I have 10, but I like your thinking.


Cat-n-'white' mouse
put the white mouse in the eighth position and it will be the last.




?'If the white mouse is 1st in the counting order, the cat should start at the 7th mouse'

So say I, but then I guess that is what you are all saying.





Interesting point is, we all had a different lay out for the same answer.

4 3 8
9 5 1
2 7 6

1 15 14 4
12 6 7 9
8 10 11 5
13 3 2 16



20 passengers are in a sinking ship. 10 are mathematicians. They all stand in a ring. Every 7th climbs into the lifeboat which can only hold 10 people. Where should the mathematicians stand in the ring


30 passengers are in a sinking ship. They all stand in a circle. Every 9th passenger goes overboard. The lifeboat holds 15. Where are the 15 lucky positions in the circle that the engineers take


Einstein never answered these questions.


Alan, Bill and Chris dug up 9 nuggets. Their weights were 154, 16, 19, 101, 10, 17, 13, 46 and 22 kgs. They took 3 each. Alan's weighed twice as much as Bill's. How heavy were Chris's nuggets


The product of 3 brothers' ages is 175. Two are twins. How old is the other one


A man has 2 bankcards, each with a 4 digit number. The 1st number is 4 times the 2nd. The 1st number is the reverse of the 2nd. What is the first number

[size=7]20 PASSENGERS
Assuming we're trying to save the mathematicians,
positions 1, 4, 5, 7, 8, 9, 14, 15, 16, 17.

30 PASSENGERS
positions 1, 2, 3, 4, 10, 11, 13, 14, 15, 17, 20, 21, 25, 28, 29

Einstein wasn't asked these questions.

NUGGETS
That sounds like a personal question!
272kgs (Alan's were 84, Bill's were 42)

3 BROTHERS
7 (or 175)

BANKCARDS
8712[/size]

Mathematicians Lifeboat order
[size=7]1,4,5,7,8,9,14,15,16,17[/size]

Engineers Lifeboat order
[size=7]1,2,3,4,10,11,13,14,15,17,20,21,25,28,29[/size]

Nuggets
[size=7]Still Pondering[/size]

Brother's ages
[size=7]175=7*5*5 the twins are 5 and the other brother is 7[/size[/color]]

Bank Cards
[size=7]4(2178)=8712[/size]

Rap c∫;?/

Mark:

20 PASSENGERS
Assuming we're trying to save the mathematicians,

?'To the engineer, the glass is half full. To the engineer, the glass is half-empty. To the mathematician, the glass is twice as big as it needs to be.'

positions 1, 4, 5, 7, 8, 9, 14, 15, 16, 17.

30 PASSENGERS
positions 1, 2, 3, 4, 10, 11, 13, 14, 15, 17, 20, 21, 25, 28, 29


"Einstein wasn't asked these questions."

Quid pro quo; Einstein never answered these questions. Coz he was never asked.

NUGGETS
That sounds like a personal question! (I am trying to pile the gold up in neat piles)

272kgs (Alan's were 84, Bill's were 42)

3 BROTHERS
7 (or 175)

BANKCARDS
8712


Rap:

Mathematicians Lifeboat order
1,4,5,7,8,9,14,15,16,17

Engineers Lifeboat order
1,2,3,4,10,11,13,14,15,17,20,21,25,28,29

Nuggets
Still Pondering (Checkout BIG Mac's chicken)

Brother's ages
175=7*5*5 the twins are 5 and the other brother is 7

Bank Cards
(2178)=8712


Are these questions getting easier?


Tom has 7 sandwiches, Jan has 5, Simon has none. They share them out equally. Simon leaves, paying for his sandwiches by leaving 12 biscuits. What's the fairest way for Tom and Jan to share out the biscuits


A cyclist buys a cycle for 15 pounds paying with a 25 pound cheque. The seller changes the cheque next door and gives the cyclist 10 pounds change. The cheque bounces so the seller paid his neighbor back. The cycle cost the seller 11 pounds. How much did the seller lose


A greengrocer was selling apples at a penny each, bananas at 2 for a penny and pears at 3 for a penny. A father spent 7p and got the same amount of each type of fruit for each of his 3 children. What did each child get


A woman bought something costing 34c. She only had 3 coins: $1, 2c and 3c. The shopkeeper had only 2 coins: 25c and 50c. Fortunately another customer had 2 10c coins, a 5c coin, 2 2c coin and a 1c coin. How did they sort things out


Mr and Mrs A are 120 km apart. A bee is on Mr A's nose. The couple cycle towards each other, Mr A at 25km/h and Mrs A and 15km/h. The bee dashes from Mr A's nose to Mrs A's nose and back again and so on at 60km/h. How far does the bee travel before the cyclist's crash


Editors note: No cyclists were hurt in the formulation of this problem.

[size=7]BISCUITS
Tom gets 9, Jan gets 3

BOUNCED CHECK
$21

FRUIT
1 apple, 2 bananas, and 1 pear

COINS
This is what they end up with:
original woman: her purchase, 50c, 10c, 10c, 1c
shopkeeper: $1, 3c, 2c, 2c, 2c
other customer: 25c, 5c

BEE
180km

Was the bee hurt?[/size]

Sandwich problem
[size=7]Simon gives Tom 9 Biscuits, Jan three[/size].

Bicycle Thief
[size=7]The seller is out 25 pounds for the rubber cheque and 11 pounds for the bicycle (36 pounds)[/size]

Friut Dividend
[size=7]to each child
2 apples 1 pear
1 apple 2 banannas 1 pear
1 apple 4 pears
4 banannas 1 pear
2 banannas 4 pears
7 pears[/size]

Cange and a haircut 2 bits.
[size=7]Store keeper gives other customer a quarter for 2 dimes, the 2 two cent pieces and a penny. The customer gives the storekeeper the dollar and the 2 and 3 cent piece The storekeeper gives the customer the the half buck, two dimes and a penny.[/size]

Cycle Crash
[size=7]The flying A's are approaching each other at (25+15=40)kph. They're initially about 120 km apart. 40kph covers 120 km in about 3 hrs. The fly flies at 60 kph for 3 hours. 3 times 60kph=180 km. The bee travels 180 km before the crash[/size].

Rap

Mark:

BISCUITS
Tom gets 9, Jan gets 3

BOUNCED CHECK
$21

FRUIT
1 apple, 2 bananas, and 1 pear

COINS
This is what they end up with:
original woman: her purchase, 50c, 10c, 10c, 1c
shopkeeper: $1, 3c, 2c, 2c, 2c
other customer: 25c, 5c

BEE
180km

Was the bee hurt?
(You cannot bee serious! You really take the biscuit) apple,pear etc


Rap:

Sandwich problem
Simon gives Tom 9 Biscuits, Jan three.

Bicycle Thief
The seller is out 25 pounds for the rubber cheque and 11 pounds for the bicycle (36 pounds)

?'Let's just call it 21 pounds, and don't go opening a store'. :wink:

Friut Dividend
to each child
2 apples 1 pear
1 apple 2 banannas 1 pear ***
1 apple 4 pears
4 banannas 1 pear
2 banannas 4 pears
7 pears

Change and a haircut 2 bits.
Store keeper gives other customer a quarter for 2 dimes, the 2 two cent pieces and a penny. The customer gives the storekeeper the dollar and the 2 and 3 cent piece The storekeeper gives the customer the half buck, two dimes and a penny.

?'I have lost part of my answer which went, "They pool the money. The woman takes 74, the shopkeeper takes 109 and the customer 28…."
I think they all work'.


Cycle Crash
The flying A's are approaching each other at (25+15=40)kph. They're initially about 120 km apart. 40kph covers 120 km in about 3 hrs. The fly flies at 60 kph for 3 hours. 3 times 60kph=180 km. The bee travels 180 km before the crash.


Another day, another batch of ?'string' theory questions, take your pick


Pick a number. If it's even, divide by 2. If it's odd multiply by 3 and add 1. Continue this until you reach "1". Eg 3-10-5-16-8-4-2-1. Which integer less than 100 produces the longest chain


Pick a number. Multiply the digits together. Continue until you get a single digit. What is the only 2 digit number which would require more than 3 multiplication


Starting with 1, place each integer in one of 2 groups so that neither contains a 3 term Arithmetic Progression. How far can you go

[size=7]LONGEST CHAIN
97 reaches 1 after 118 steps

MORE THAN 3
77

NO ARITHMETIC PROGRESSION
I got to 8:
1 4 5 8
2 3 6 7[/size]

An engineer, a mathematician and a riddler have two identical marbles each that will break if they drop them from a certain height, from one of the floors, of a building with a 100 floors. It could be that the marbles will break if they drop them from the 1st floor, but it could also be that they will break from the 100th floor (or any floor in between). If one breaks you have only one left so be careful what strategy you choose in dropping them from specific floors. Put the mop down Whim it is not 100.

It is your job to find out from what floor the marbles will break.

Don't tell the others but what is the fewest number of drops needed to cover every possibility

[size=7]MARBLES (glass balls)
14[/size]

Collatz conjecture
[size=7]The 3n+1 problem---neat. I've googled this before. The integer 97 thakes 116 steps to get to the 4-2-1 loop. So 97 takes 119 steps[/size].

Two digit products
[size=7]Tried several combinations and quickly converged on 77 (unlike Collatz Conjecture) 77=>49=>36=>18=>8 4 steps[/size]

Two Groups
[size=7]I'm not sure I understand the problem but I go
1,4,5,8,9,....
&
2,3,6,7,10,11....
since the repeated regression of the first group is +3.+1,+3,+1 ad infinitum and the second is +1,+3,+1,+3 i'd hypothesis that the repeat cannot go on farther than 3 terms so n<8. Or it could be 7.[/size]
Rap

Marble drop
[size=7]Using a modified halfie sort I say I can name that floor in 8 notes Bob.
Drop the first marble from floor 50, then 75/25 depending on the outcome. Then splitting the difference each time working down to the specific floor until the difference differes by one floor (drop#7). The last drop would focus on that particular floor.[/size]

Rap

Marbles additional
You could very well fine the floor in 7 marbles as 100<2^7=128

Rap

Mark:

First post 8.07
This post 8.30…23 minutes, including typing! Are you human?


LONGEST CHAIN
97 reaches 1 after 118 steps (prehaps 119)

MORE THAN 3
77

NO ARITHMETIC PROGRESSION
I got to 8:
1 4 5 8
2 3 6 7

MARBLES (glass balls)
14

Clearly not.


Rap:

Collatz conjecture
The 3n+1 problem---neat. I've googled this before. The integer 97 thakes 116 steps to get to the 4-2-1 loop. So 97 takes 119 steps.

Two digit products
Tried several combinations and quickly converged on 77 (unlike Collatz Conjecture) 77=>49=>36=>18=>8 4 steps

Two Groups
I'm not sure I understand the problem but I go
1,4,5,8,9,....
&
2,3,6,7,10,11....
since the repeated regression of the first group is +3.+1,+3,+1 ad infinitum and the second is +1,+3,+1,+3 i'd hypothesis that the repeat cannot go on farther than 3 terms so n<8. Or it could be 7.

Marble drop
Using a modified halfie sort I say I can name that floor in 8 notes Bob.
Drop the first marble from floor 50, then 75/25 depending on the outcome. Then splitting the difference each time working down to the specific floor until the difference differes by one floor (drop#7). The last drop would focus on that particular floor.


Time out buddy, you ain't got but two marbles and you dun used them both up.

Say they smashed on 50 and 25! Now what - go ask Mark, or perhaps Harry. :wink:

"You could very well fine (please note the ?'d' is one row down) the floor in 7 marbles as 100<2^7=128"

Have you been on the ?'wackey baccy'?

If not, pass it over.


I hafta go along with Marks ?'lucky' guess: 14.

Throw the first marble from level 14, and - as long as it continuous to survive - increase the floor level one less than the last increase:

14, 27, 39, 50, 50, 69, 77, 84, 90, 95, 99

and then use the second marble to work up one floor at a time from the one but last floor tested (i.e. the last floor for which the first marble survived).

Maximum number of drops needed: 14


Rap wrote elsewhere, "RAP is an engineer. I'm in charge, I get to hold the clipboard."

That is sooo cool. I once got to look at the clipboard, but it was a long way off.


Are you the driver?

Nelson Mandela is sitting at home watching TV, and drinking a beer when he hears a knock at the door. When he opens it, he is confronted by a little Japanese man, clutching a clipboard and yelling:

"You Sign! You sign!" Behind him is an enormous truck full of car exhausts.

Nelson is standing there in complete amazement, when the Japanese man starts to yell louder: "You sign for these! You sign!". Nelson says to him, "Look, you've obviously got the wrong man", and shuts the door in his face.

The next day he hears a knock at the door again. When he opens it, the little Japanese is back with a huge truck of brake pads. He thrusts his clipboard under Nelson's nose, yelling: "You sign! You sign!"

Mr. Mandela is getting a bit hacked off by now, so he pushes the little Japanese man back, shouting: "Look, go away! You've got the wrong man! I don't want them!" Then he slams the door in his face again.

The following day, Nelson is resting, and late in the afternoon, he hears a knock on the door again. On opening it, there is the same little Japanese guy thrusting a clipboard under his nose, shouting:

"You sign for these parts!! You sign!" Behind him are TWO very large trucks full of car parts. This time Nelson loses his temper completely, he picks up the little man by his shirt front and yells at him:

"Look, I don't want these! Do you understand? You must have the wrong name!

Who do you want to give these to?"

The little Japanese man looks at him very puzzled, consults his clipboard and says…


…(Get your Japanese accent ready).....






"You not Nissan Main Dealer?"



The following two questions use the following result -

"Given integers a and b the biggest number that can't be expressed in the form ia + jb is ab - a - b."

Apples are packed in boxes of 8 and 15. What is the biggest number of apples that would require loose apples


A country only has 5p and 7p coins. What is the highest price that you couldn't give exact money for



If D = the day (1-366) in year Y, then the day of the week can be calculated using: d = D+Y+(Y-1)/4 - (Y-1)/100 + (Y-1)/400 mod7.

Where d=1 would mean Sunday, etc. Can the first day of each century (e.g. 1st Jan 2001, 1st Jan 1901) be any day


Pick 3 digits (not zero) and make 6 2-digit numbers from them. Add up all these numbers, add up all the original digits and divide the first total by the second. I make it 22. How about you

Now you're providing the answers before you pose the questions?!?

[size=7]CALENDAR
[1+Y+INT((Y-1)/4)+INT((Y-1)/100+INT((Y-1)/400))] mod 7
can take on any value from 0 to 6 for Y=100n+1
So, if the formula is correct, then the answer is yes.

3 DIGITS
AB + BA + AC + CA + BC + CB = 22(A+B+C)
since each letter appears in the tens and units place twice.[/size]

Mark, "Now you're providing the answers before you pose the questions?!?"

I am trying to cut you out of the loop. No, seriously I just can't stop giving…

How many presents did I give "my true love" during the 12 days of Christmas