The worlds first riddle!

[size=7]STEER
~10300 calories[/size]

[size=7]Heavy Rain:
nary a hive
I hear navy
hire a navy

Conversation:
voices rant on[/size]

Mark:

HANDSHAKES
15


You might suppose that each person has to shake hands with 5 other people, and since there are 6 people, there would be 6x5=30 handshakes, but then you would have counted each handshake twice. So, just divide 30 by 2 to get 15 handshakes. Another approach: The first person shakes hands with 5 people, the next person, 4 people, and so on. So, 5+4+3+2+1=15 handshakes.



ROBERT
He is 6 and she is 2.


Right now he's three times as old. However, in two years, Robert will be 8 and Susan will be 4, so he'll only be twice as old.


SCORES
90


You would have to score a 90. 50 + 60 + 70 + 80 +x = 260 + x
(260 + x)/5 = 52 + (x/5) = 70
70- 52 = 18
x/5 = 18
x = 90

~OR~
The difference between 50 and 70 plus the difference between 60 and 70 equals 30. The difference between 70 and 80 is 10. To make the differences equal, 30, you need a difference of 20. 70 + 20 = 90.



STEER
~10300 calories

Weight is proportional to linear dimension (length or girth of the steer) cubed.
Surface area is proportional to linear dimension squared.
Therefore, 13,500 x [ ( 420/630)1/3 ] 2 = 10,300 calories


The Morse Code: here come dots


Heavy Rain:
nary a hive
I hear navy
hire a navy

Conversation:
voices rant on



Mother-in-law:



Today's questionmark ?



A racing driver drove around a 6-mile track at 140 miles per hour for 3 miles, 168 miles per hour for 1.5 miles, and 210 miles per hour for 1.5 miles.

What was his average speed for the entire 6 miles



Ten years from now Raju will be twice as old as Juhi was when Mena was nine times as old as Raju.

Eight years ago, Mena was half as old as Juhi will be when Juhi is one year older than Raju will be at the time when Mena will be five times as old as Raju will be two years from now.

When Raju was one year old, Mena was three years older than Raju will be when Juhi is three times as old as Mena was six years before the time when Juhi was half as old as Raju will be when Mena will be ten years older than Mena was when Juhi was one-third as old as Raju will be when Mena will be three times as old as she was when Juhi was born.

How old are they now

[size=7]Mother-in-law: Hitler woman[/size]

Mark:
Mother-in-law: Hitler woman


A B C D
E F G H
J K L M

This cute diagram represents a small sheet of 12 postage stamps, as they are usually sold, all perforated at the edges and all of the same value. (The letters are only there to identify the separate stamps).

You need four of the stamps in order to mail a letter, but would like all four to be properly joined together at their edges (not at their corners).

For example: ABCD, EFGH, JKLM, FGHL would all do, but NOT EFLM.

In how many different ways can you get such a group of four
(Only the number of ways is required, not a listing)

[size=7]STAMPS
I get 113.
8 4x1
9 2x2
48 L-shaped
24 T-shaped
24 Z-shaped[/size]

Mark:

STAMPS
I get 113.
8 4x1
9 2x2
48 L-shaped
24 T-shaped
24 Z-shaped

A most interesting total. I could only manage 65 by brute force. The only question I would raise is; Are all the shapes unique? DCBA in any combination is the same as ABCD.




Mr Dippy Doppolous was exactly 5 years older than his wife and they had two children, who were twins and went to the local school.

One evening, one of the twins (who had just been given a calculator) announced that if you multiplied together the parents' ages, and then added together the twins' ages, and divided one result by another, the answer was 57 exactly.

How old is Mr Dippy Doppolous

[size=7]MR DIPPY
38

Otherwise, the local school is a college, or he had unlawful sex with a minor.[/size]

This could be shocking news.



This brief message has been written in cipher. No letter represents itself.
Each letter in the original message has been substituted by some other letter.

LYCW ZQHNX FYZZ XYCN RYOBLW XBYX

Can you work out what the original message might have been


We are taking on water!

[size=7]RACING
160mph[/size]

STAMPS AGAIN
Oops! I solved it for a 4x4 block. Here's the answer for a 3x4 block:
3 4x1
6 2x2
28 L-shaped
14 T-shaped
14 Z-shaped
65 total
You were right.

Mark:
MR DIPPY
38

Cant argue with that logic.

RACING
160mph

I thought this one had passed unnoticed.


STAMPS AGAIN
Oops! 65 total

You were right.

Well, fancy that.


I have just noticed:

Carter fenced in a rectangular plot of land and then divided it into two halves by putting another fence along the main diagonal. This diagonal fence measured 41 metres exactly.

Dale also fenced in a rectangular plot of land in a similar way (including the diagonal fence) but his rectangle was one-third greater in area than Carter's plot and used less fencing.

In both plots, all the dimensions (lengths, widths and diagonals) were a whole number of metres.

How much more fencing did Carter use than Dale

[size=7]FENCES
Carter: 9x40 with diagonal of 41 - total = 139
Dale: 16x30 with diagonal of 34 - total = 126

Carter used 13 more meters than Dale.[/size]

Mark: (A master fencer)

FENCES
Carter: 9x40 with diagonal of 41 - total = 139
Dale: 16x30 with diagonal of 34 - total = 126

Carter used 13 more meters than Dale.

Clever, very clever.

Dale's Pythagorean triplet was 9, 40, 41
Carter's was 16, 30, 34


TT and DP were going to cut the grass on their square lawn. They agreed that each should cut one-half of the area.

TT went first and cut a border 2 metres wide all the way around.

After doing a few sums, DP agreed that exactly one-half had been cut and happily cut the rest.

What was the area of the lawn


A group of people were gathered together in our barn, and it was possible to count them using the following descriptions -

1 grandmother
1 brother
1 grandfather
2 sisters
2 fathers
2 sons
2 mothers
2 daughters
4 children
1 father-in-law
3 grandchildren
1 mother-in-law
1 daughter-in-law

This gives a total of 23 people. (Take my word for it) OK count them yourself!

However, there was less than that. (If you think it should read; there ?'were' less than that) Get bent!

What is the LEAST number of people that could have been present in the barn

[size=7]LAWN
186.51 square meters

BARN
I didn't know you have a barn.
4 - two males and two females since it doesn't say that the people all have to have the relationships with each other. In fact, they can all be unrelated.

For instance, as I sit here alone trying to unbend myself, I can be described in at least eight different ways:
father, son, brother, cousin, grandson, uncle, nephew, great nephew

You're probably thinking of an unprintable ninth about now, and you're probably right.[/size]

Mark:

LAWN
186.51 square meters

(To two DP)


BARN
I didn't know you have a barn.

(Sorry, that should read; yarn) :wink:

4 - two males and two females since it doesn't say that the people all have to have the relationships with each other. In fact, they can all be unrelated.

I think I could have worded that question a little ?'tighter'

I was thinking: 7 people.
Call them Arthur, Belinda, Charles, Delia, Edward, Fiona, Georgette.
Arthur is the father of Charles.
Belinda is the mother of Delia.
Charles and Delia are married and
Edward, Emma, Georgette are their children.


For instance, as I sit here alone trying to unbend myself, I can be described in at least eight different ways:
father, son, brother, cousin, grandson, uncle, nephew, great nephew

(Point, very sharply taken.)

You're probably thinking of an unprintable ninth about now, and you're probably right.

(Even though I have a one-track mind, I would never think along those lines.)



Starting in any cell of the 3x3 grid moving up, down or across, but NOT diagonally, it is possible to make a tour that visits all 9 cells of the grid.

A tour may start in any one of the 9 cells.

Given that no cell may be visited more than once, (so the tour cannot cross over itself) how many different tours are possible

A B C
D E F
G H J

[size=7]9-CELL TOUR

I get 40, but ignoring rotations, reflections and direction of travel, I get only three unique paths:
one that starts and ends in adjacent corners
one that starts and ends in opposite corners
one that starts/ends in a corner and ends/starts in the center[/size]

Use the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and use each digit once and once only.

Divide the ten digits into two groups of five, and then arrange the digits in each group of five to make a multiplication sum.

It could be done like this
0, 1, 5, 6, 9 used to make 106 x 59 (which is 6254)
2, 3, 4, 7, 8 used to make 437 x 82 (which is 35834)
and the two multiplication sums have different answers.


Can you find a way that makes the answers to both multiplication sums the same for:

a) Smallest

b) Largest

Surely you mean product instead of sum.

Can each group be divided into more than two subgroups? For instance, is
10*6*59
3*47*8*2
valid (although not correct)?

Assuming the answer is no,
[size=7]min: 1*6970 = 2*3485 = 6970
max: 64*915 = 80*732 = 58560[/size]

Mark:

9-CELL TOUR

I get 40, but ignoring rotations, reflections and direction of travel, I get only three unique paths:
one that starts and ends in adjacent corners
one that starts and ends in opposite corners
one that starts/ends in a corner and ends/starts in the center

True, very true.

"Assuming the answer is no"

My dear Sir, your assumptions are, as eminently correct as the perspicacity of your answers.

min: 1*6970 = 2*3485 = 6970
max: 64*915 = 80*732 = 58560



I recently found just the things I wanted for Christmas at Beth's yard sale.
The way the prices were displayed was interesting.

FIVE for 10c
TWELVE for 18c
FIFTEEN for 20c
NINETEEN for 24c


What did ELEVEN cost me



A deltahedron is a 3-dimensional shape all of whose faces are equilateral triangles.

The simplest of these is a regular tetrahedron, which is made of 4 equilateral triangles.

The deltahedron I have has 9 vertices.

How many faces does my deltahedron have

[size=7]YARD SALE
19c (1c per stroke)

DELTAHEDRON
14 faces and 21 edges[/size]