The worlds first riddle!

Treasure Chest


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∫;?/

[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

[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∫;?/

[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

[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

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