The worlds first riddle!

[size=8]3X3 SQUARE
To simplify things, subtract 252 from each number, then divide them all by 83. Now make a standard 3x3 square with magic number 15.[/size]

[size=8]GHOSTS, GHOULS, GOBLINS
There are actually four solutions
0: S
1: B
2: H
3: L
4: N or T
5: G
6: O
7: I or U
8: I or U
9: N or T

Total = 6666660[/size]

Thoh:


1 = 4/4+4-4
2 = 4*4/(4+4)
3 = (4+4+4)/4
4 = (4-4)/4+4
5 = (4+4*4)/4
6 = 4+(4+4)/4
7 = 4+4-(4/4)
8 = 4+4+4-4
9 = 4+4+4/4
10 =(44-4)/4

Very nice


Which is almost the same as:

1 = 44 / 44
2 = (4 x 4) / (4 + 4)
3 = (4 + 4 + 4) / 4
4 = 4 x (4 - 4) + 4
5 = (4 x 4 + 4) / 4
6 = (4 + 4) / 4 + 4
7 = 44 / 4 - 4
8 = 4 + 4 + 4 - 4
9 = 4 / 4 + 4 + 4
10 = (44 - 4) / 4



Mark:

3X3 SQUARE
To simplify things, subtract 252 from each number, then divide them all by 83. Now make a standard 3x3 square with magic number 15.


I am not sure I understand what you mean, no wait, I do understand that I do not understand.


916 335 750
501 667 833
584 999 418



Mark:

GHOSTS, GHOULS, GOBLINS
0: S
1: B
2: H
3: L
4: N or T
5: G
6: O
7: I or U
8: I or U
9: N or T

Total = 6666660


"There are actually four solutions"


I only have the two.

..526090
..526730
5613840 +
------------
6666660



..526090
..526830
5613740+
------------
6666660




This is a real easy question, the answer however may be less so…

How many "Yes or No" questions are needed to guess any seven digit phone number

[size=8]PHONE NUMBERS
23 if the numbers are limited to those that are valid (leading digit is 2-9) according to the North American Numbering Plan.

24 if any 7-digit number is allowed.[/size]

Time: 9.01am, phew, that was lucky

Mark:

PHONE NUMBERS
23 if the numbers are limited to those that are valid (leading digit is 2-9) according to the North American Numbering Plan.

24 if any 7-digit number is allowed.


Use what is known as a "binary search" to ask the questions.
Your first question would be:
Is the number greater than 5000000?

This would divide the 10 million possible phone numbers (from 000-0000 to 999-9999) in half with just one question.

Each succeeding question would divide the remaining numbers in half.
So we are looking for the solution to the equation 2^x = 10000000

You can solve this with logarithms:
x ln 2 = ln 10000000
so x = 23.25

therefore, it would take 24 questions to be absolutely sure that you would be able to guess any 7-digit phone number.

Or, of course, you could just examine
2^24 = 16,777,216 and 2^23 = 8,388,608 to see that 24 questions are needed.




Now, this is not the kind of question you see everyday:

If you multiply two whole numbers together, what is the last digit of the answer most likely to be



Well I did tell ya!

[size=8]LAST DIGIT
0 wins hands down at 27% of the time.
2, 4, 6, and 8 come in at 12% each.
5 is 9%.
1, 3, 7, and 9 come in at 4% each.[/size]

What number differs from its reciprocal by 1

Awesome thread. Will someone tell me what level of mathematics is necessary to be able to solve the riddles that are given out here?
I studied mathematics till high school (before crossing over to the Sciences) and thought may be I should give it a try here!!

Dear Spidergal, how nice to see you on the thread. Thanks to the contributions many have made, there are many thousands of riddles, puzzles and problems of all sorts mixed on this thread. Logic plays a bigger part than pure math knowledge in most questions.

However, if you are like Mark who has more PhD's than my dogs got fleas, then I guess you don't need logic at all. Heck, even I have been known to come up with a correct answer every now and again. Do give it a Try!

[size=8]RECIPROCAL
phi - the golden ratio[/size]

Mark:

LAST DIGIT
0 wins hands down at 27% of the time.
2, 4, 6, and 8 come in at 12% each.
5 is 9%.
1, 3, 7, and 9 come in at 4% each.

I can add nothing to that answer.



RECIPROCAL
phi - the golden ratio


The answer is 1.61803...
Other possibilities are -1.61803..., or .61803..., or - .61803...
These are the decimal approximations for
(1 + sqrt(5))/2 and (1 - sqrt(5))/2 or
(-1 + sqrt(5))/2 and (-1 + sqrt(5))/2

This number is called what Mark said.

To solve for this number,
let n = the number and 1/n = its reciprocal.
Then n - 1/n = 1 (or 1/n - n = 1)
Multiplying by n, n^2 - 1 = n
So n^2 - n - 1 = 0
Using the quadratic formula,
n = (1 + sqrt(5)) / 2
and n = (1 - sqrt(5)) / 2





Two A2K astronomers were looking through a telescope and observed an alien mathematics class.

On the chalkboard, the following equations were written:

13 + 15 = 31
10 x 10 = 100
6 x 3 = 24

One astronomer said to the other, "How many fingers do they have?"

What is the answer to that question

[size=8]FINGERS
7

18 = 2x + 4
x = 7[/size]

I don't believe it Five (5) minutes to answer, I will have to speak to the management to slow you down. They suggested:


There is a cubical water tank that is one mile long, one mile wide, and one mile deep. It is completely filled with water.

A hole is punched in it at the bottom, and water begins to flow at the rate of one gallon per second.

How long does the tank take to empty



(Before you do any calculations, have a guess. I bet you will be miles out. Well not miles exactly)

[size=8]guess: years

calculations:
5280^3 * 7.48 = 1,101,040,680,960 gallons
1,101,040,680,960 / (60 * 60 * 24 * 365.242199) = a bit over 34,890 years[/size]

Mark:

FINGERS
7

18 = 2x + 4
x = 7



Or, to put it another way:

First, observe that seven numerals are used in the equations:
0, 1, 2, 3, 4, 5, and 6.

So began with base seven and notice that all three equations were correct in that number system.
If the three equations had not worked in base 7, you could have tried base 8, then base 9, ...






guess: years (My first thoughts were two years )


calculations:
5280^3 * 7.48 = 1,101,040,680,960 gallons
1,101,040,680,960 / (60 * 60 * 24 * 365.242199) = a bit over 34,890 years



Since 1 cubic foot of water is equivalent to 7.48 gallons of water. One mile = 5,280 feet, so one cubic mile = 5280 x 5280 x 5280 cubic feet which equals
147,197,952,000 cubic feet.

Multiply by 7.48 to obtain the number of gallons of water in the tank: 1,101,040,680,960 gallons!

Draining at one gallon per second, this represents the number of seconds it would take to drain the tank!
Divide by 60 to convert to minutes;
Divide by 60 again to convert to hours;
Divide by 24 to convert to days;
And finally divide by 365.25 to convert to years!

The number of years is approximately 34,889.87. This is a little less than Mark, but then he is in another time zone.





Notorious female bandit Spider Gal was counting the 7 bags of gold coins she had just acquired from holding up the Santa Fe Stagecoach. One bag had 10 coins, another 20 coins, and so on in increments of 10 coins up to 70.

Her bandit brother gringo Mark rode up on a horse and demanded the 3 bags with the most coins.

She labeled the bags A to G and informed Mark that he could have the three biggest bags if he could work out which they were from these simple facts:

Bags B, D, E, and G contain a total of 150 coins.
Bags A, C, D, and E had 100 coins.
Bags A, B, D, and F had 170 coins.


Which three bags should Mark take

[size=8]BAGS OF GOLD
B, F, and G

Only the second clue is necessary. 100 can be gotten from four bags in only one way: 10 + 20 + 30 + 40. That means any bag not mentioned in that clue is one of the three largest bags.[/size]

Mark:

BAGS OF GOLD
B, F, and G


(Mark makes a very valid point)

"Only the second clue is necessary. 100 can be gotten from four bags in only one way: 10 + 20 + 30 + 40. That means any bag not mentioned in that clue is one of the three largest bags."


I think the sands of time are running out:

The hands of a clock are together at 12 noon.
At what time will they next be together again


Sure you can use a little math, but you can use your own watch to find an answer.

(In years, months, weeks, days, hours, minutes and seconds) Thank you!

[size=8]CLOCK HANDS
two hands or three?

two: 11 times in 12 hours, so 12/11 of an hour. 1:05:27.27
three: probably midnight - I haven't done the math[/size]

Mark:

CLOCK HANDS
two hands or three?

two: 11 times in 12 hours, so 12/11 of an hour. 1:05:27 .27



circle on the face of the clock is divided into 60 equal parts. The minute hand travels around the circle at a speed of 60 spaces per hour. The hour hand follows it at a speed of 5 spaces per hour. So the minute hand gets ahead at a speed of 55 spaces per hour. The two hands will be together again when the minute hand is a full lap, or 60 spaces, ahead. To get that far ahead will take 60/55 hours, which is 1 1/11 hours, or 1 hour 5 5/11 minutes. So the hands will be together at 5 5/11 minutes past 1 o'clock.

Another solution is to use a RATE-TIME-DISTANCE table.
Let y = the spaces travelled by the hour hand.
Then y + 60 = the spaces travelled by the minute hand.
Let x = hours travelled by both the hour and the minute hands.

………………Rate …. Time Distance
Minute Hand 60 sp/hr x hr y + 60 sp
Hour Hand …..5 sp/hr x hr y sp


This yields 2 equations:
60 x = y + 60
5 x = y

Solving simultaneously, x = 60/55 hours or
approximately 1:05:27 (hr:min:sec).



Three: probably midnight - I haven't done the math

(Anyone out there know the answer)





When V. Stormy went to Teacher Training College, she was assigned a password to use on the computer system. The password consisted of a series of five different digits. She was told to memorize it and not write it down.

She decided to invent a reminder, but all she could come up with was:

(1) The sum of the first two digits was exactly twice the sum of the last four, and

(2) The sum of the three digits in the middle was twice the sum of the last two.

On the first day of classes, poor Stormy forgot her password. From her reminder, can you reconstruct the password for her

Mark:

CLOCK HANDS
two hands or three?

two: 11 times in 12 hours, so 12/11 of an hour. 1:05:27 .27



circle on the face of the clock is divided into 60 equal parts. The minute hand travels around the circle at a speed of 60 spaces per hour. The hour hand follows it at a speed of 5 spaces per hour. So the minute hand gets ahead at a speed of 55 spaces per hour. The two hands will be together again when the minute hand is a full lap, or 60 spaces, ahead. To get that far ahead will take 60/55 hours, which is 1 1/11 hours, or 1 hour 5 5/11 minutes. So the hands will be together at 5 5/11 minutes past 1 o'clock.

Another solution is to use a RATE-TIME-DISTANCE table.
Let y = the spaces travelled by the hour hand.
Then y + 60 = the spaces travelled by the minute hand.
Let x = hours travelled by both the hour and the minute hands.

………………Rate …. Time Distance
Minute Hand 60 sp/hr x hr y + 60 sp
Hour Hand …..5 sp/hr x hr y sp


This yields 2 equations:
60 x = y + 60
5 x = y

Solving simultaneously, x = 60/55 hours or
approximately 1:05:27 (hr:min:sec).



Three: probably midnight - I haven't done the math

(Anyone out there know the answer)





When V. Stormy went to Teacher Training College, she was assigned a password to use on the computer system. The password consisted of a series of five different digits. She was told to memorize it and not write it down.

She decided to invent a reminder, but all she could come up with was:

(1) The sum of the first two digits was exactly twice the sum of the last four, and

(2) The sum of the three digits in the middle was twice the sum of the last two.

On the first day of classes, poor Stormy forgot her password. From her reminder, can you reconstruct the password for her


I am sure she will make it worth your while!!!