Jiggles got it right this time.. Well done :) Your turn.
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Jiggles got it right this time.. Well done :) Your turn.
ok,
this is sort of similar to yours,
a woman shoots her husband.
then she holds him under water for over 5 minutes.
finally, she hangs him.
but 5 minutes later they both go out together and enjoy a wonderful dinner together.
how can this be?
yes your right bigboab,
here is the full answer-
The woman was a photographer. She shot a picture of her husband, developed it, and hung it up to dry.
Given a list of integer variables A. B .C. D. E. F.
each contain an integer.
You are required to reverse the contents of the variables so that A = F. B = E. C = D. D = C. E = B. F = A.
You are not allowed to use any other variables.
i.e. If you start of with
A = 1. B = 2. C = 3. D = 4. E = 5. F = 6.
You should end up with
A = 6. B = 5. C = 4. D = 3. E = 2. F = 1.
Computer programmers should find this easy.:)
Good one! :smilie4:
Obviously, normally I'd use another variable, but without that it makes it interesting.
I think this should work:
Spoiler: Show
Set A = A + F
Set F = A - F
Set A = A - F
Set B = B + E
Set E = B - E
Set B = B - E
Set C = C + D
Set D = C - D
Set C = C - D
Not very elegant, but if you work through the example..
Starting with A=1,F=6
A = A + F : ( A = 1 + 6 = 7 )
F = A - F : ( F = 7 - 6 = 1 )
A = A - F : ( A = 7 - 1 = 6 )
etc.
:smartass:
Two numbers m and n are chosen such that 1 < m <= n < 100.
Mr. S. is told their sum.
Mr. P. is told their product.
The following dialogue takes place:-
Mr. P.: "I don't know the numbers."
Mr. S.: "I know that you don't know the numbers. I don't know them either."
Mr. P.: "Now I know them."
Mr. S.: "Now I know them, too."
What are the two numbers?
Is this the answer?
Spoiler: ShowM = 2
N = 2
here's a riddle-
If a man carried my burden, he would break his back.
I am not rich, but leave silver in my track.
What am I?
lol, its maths, im crap at maths, and its proably to do with 4 or 2 or something.
lol at the spoiler, your correct, your too good :(.
well, im out of riddles, so yours still stands.
, but i checked it up, but i still dont get it, its about logicial thinking n stuff, heres a link- http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/logic_sum_productSpoiler: Showits 4 and 13
i still dont understand, how does it work then?
Afraid not :([/quote]
Help!:(
Spoiler: ShowI need a wee bit of help. Is it the split of one of the primary numbers between 11 to 97?
I must admit I have no clue how on earth this sum/problem works.. :blink: Barbs - Any chance of a simple and full explanation?
barbs??
im dying to find out what this riddle is about.
:lol:
I heard this ages ago, and I managed to solve it by a process of trial and error, it's to do with number theory and such like.
There are plenty of write-ups about it on the internet... :whistling
that was a quick response, how fast is your pc????
I will attempt an explanation next week, I'm off sick at the moment, and don't feel up to it :(
next week?
how complex is this riddle?
ok, thats some complex riddle, i read the explanation of it like 5 times and still dont understand it.
why did you post such a riddle? huh? why? why? my brain hurts.
i have a big headache, so aspirin wont do :P
i told barbs to explain it, i'll have to wait till next week :blink:
moving along.....
How are a jeweller and a jailer alike?
I found it on the interwebs as I had no chance of working out for myself after 3 full minutes of struggle.
Spoiler: Show
Two Mathematicians Problem
Date: 05/18/98 at 04:46:56
From: Yusuf Kursat TUNCEL
Subject: 2 Mathematicians Problem
Dear Dr. Math,
We are given that X and Y are two integers, greater than 1, are not
equal, and their sum is less than 100. A and B are two talented
mathematicians; A is given the sum, and B is given the product of
these numbers:
B says "I cannot find these numbers."
A says "I was sure that you could not find them."
B says "Then, I found these numbers."
A says "If you could find them, then I also found them."
What are these numbers?
Here is my approach:
The reason that B cannot find the numbers from their products is that
the product has more than 2 prime factors that are not the same (like
2 * 3 * 5, which results in products like 10 * 3, 2 * 15, and 5 * 6)
So, for A to be sure that B could not find the numbers, A must have a
sum that does not contain a two prime solution pair such as
2 + 13 = 15. A cannot have this because he wouldn't be sure if the
product is given 2 * 13 = 26 or not.
I have written all of the prime numbers from 2 to 100, and added them
with each other, I have eliminated the the results from the set of
numbers 2 - 100 (also 3, 4, 5, 6, 7, 8, 9, and 10 can not be sum and
are eliminated automatically) so that I have a set, which contains
probable sum of the numbers:
{11, 17, 23, 27, 29, 33, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67,
71, 75, 77, 79, 83, 87, 89, 93, 95, 97, 98}
Any of the numbers in this set cannot be constructed just by using two
prime numbers. But I couldn't go far. Would you help me?
Date: 05/18/98 at 17:29:46
From: Doctor Rob
Subject: Re: 2 Mathematicians Problem
Your approach is the right one, but you must be careful.
First of all:
2 + 2 = 4 <= X + Y <= 99 and
2 * 2 = 4 <= X * Y <= 2450 = 49 * 50
B says, "I can not find these numbers."
Then X * Y cannot be prime, since it is the product of two numbers
greater than 1. It also cannot be the square of a prime number,
because X and Y are not equal. If X * Y had exactly two proper
divisors, then B would know the two numbers. This eliminates the
product of two distinct primes, and the cube of any prime.
A says, "I was sure that you could not find them."
Yes, X + Y cannot be the sum of two distinct primes, or the sum
of a prime and its square. This forces X + Y to be one of the
following numbers:
11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67,
71, 77, 79, 83, 87, 89, 93, 95, 97
Since all the remaining sums are odd, one of the two numbers must be
odd and the other even. X * Y must then be even.
B says, "Then, I found these numbers."
B can throw away any factorization of X * Y in which the sum of the
two factors is not one of these numbers. There must be exactly one
such possibility left for him to know the numbers. This eliminates
X * Y = 12, for example, because even though it has factorizations
2 * 6 and 3 * 4, the sums of the two factors, 8 and 7 respectively,
are not in the set of possible sums, so there are no possibilities
left. It also throws out X * Y = 120, because 120 = 5 * 24 = 15 * 8,
and 5 + 24 = 29 and 15 + 8 = 23, which are both on the list, so there
are two possibilities left. This limits the possibilities for the
product to:
18 = 9 * 2, 9 + 2 = 11
24 = 8 * 3, 8 + 3 = 11
28 = 7 * 4, 7 + 4 = 11
50 = 25 * 2, 25 + 2 = 27
52 = 13 * 4, 13 + 4 = 17
54 = 27 * 2, 27 + 2 = 29
76 = 19 * 4, 19 + 4 = 23
92 = 23 * 4, 23 + 4 = 27
96 = 32 * 3, 32 + 3 = 35
98 = 49 * 2, 49 + 2 = 51
100 = 25 * 4, 25 + 4 = 29
112 = 16 * 7, 16 + 7 = 23
124 = 31 * 4, 31 + 4 = 35
140 = 35 * 2, 35 + 2 = 37
144 = 48 * 3, 48 + 3 = 51
148 = 37 * 4, 37 + 4 = 41
152 = 19 * 8, 19 + 8 = 27
160 = 32 * 5, 32 + 5 = 37
172 = 43 * 4, 43 + 4 = 47
176 = 16 * 11, 16 + 11 = 27
188 = 47 * 4, 47 + 4 = 51
192 67
198 29
208 29
212 57
216 35
220 59
222 77
228 79
230 51
232 37
234 35
238 41
244 65
246 47
250 35
... ...
A says, "If you could find them, then I also found them."
This means that the sum in the last preceding list must occur only
once. That eliminates X + Y = 11, 27, 29, 23, 35, 51, 37, 41, 47, ...,
leaving only a single sum that occurs only once in the above table.
This tells you X and Y and how A and B figured them out.
Nice explanation.
I'd got as far as the list of possible sums. In fact that list is incomplete since in our puzzle both numbers could be between 2 and 99 so the sum could be as high as 198, but the reasoning is the same and it turns out that the extra possible sums do not change anything.
no shit,
i was told this joke a day before i posted it, it took me ages to work out, even then i had to google it.
I see someone posted an explanation about the previous puzzle. :rolleyes:
Thank fuck! :happy:
that had to be the worst riddle ever, it was maths.
We don't seem to have done the "U2 riddle" yet, so let's do it now...
==================================================================
U2 has a concert that starts in 17 minutes and they must all cross a bridge to get there. All four men begin on the same side of the bridge. You must help them across to the other side in under or at 17 minutes. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each band member walks at a different speed. A pair must walk together at the rate of the slower man’s pace:
= Bono: 1 minute to cross
= Edge: 2 minutes to cross
= Adam: 5 minutes to cross
= Larry: 10 minutes to cross
For example: if Bono and Larry walk across first, 10 minutes have elapsed when the get to the other side of the bridge. Larry then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission. There is no trick behind this. It is simple movement of resources in the appropriate order. There are two known answers to this problem. This is based on a question Microsoft used to give to all prospective employees. Note: Microsoft expects you to answer this question in under 5 minutes!
Good Luck!
Bono and Edge cross - 2 mins
Bono goes back - 1 mins
Adam and Larry cross - 10 mins
Edge goes back with the light - 2 mins
Bono and Edge cross the bridge - 2 mins
im pretty sure its right
No, that's totally wrong :lol:
Bono and Adam would take 5 minutes, and bono and edge would take 2 minutes.
They have to walk the speed of the slowest person. :whistling
EDIT: Now you've got it right. Get it from Google, did ya? :rolleyes:
no,
the trick is
thats the key, put the slowest people together to take advantage of the speedy dudes.Quote:
" They have to walk the speed of the slowest person"
Ok, so can you do the classic farmer, fox, chicken, corn problem (without looking it up)
A farmer has to get a sack of corn, a chicken, and a fox across a river. The farmer is only able to bring one of the above items along with him at a time. The only problem is if he leaves the fox alone with the chicken, the fox will eat the chicken, and if he leaves the chicken along the corn sack, then the chicken will eat the corn sack. How does the farmer get all 3 items across safely?
ok,
he first gets the chicken and puts it on the other side
goes back and gets the fox and puts it on the other side then takes the chicken
takes the chicken puts it on the starting side, then gets the corn
puts the corn with the fox, then goes back and gets the chicken.
that was easy, and i didnt check it up.
heres a riddle-
two shops are facing each other on the same road and they sell the same goods. but they are not competing. how can this be ?
Feed the corn to the chicken, eat the chicken, shoot the fox and carry its dead body across.
That's how a real farmer would have done it at one time.
These days, it's shoot the fox, feed its dead body to the chicken, convert the corn into biofuel and drive across with the chicken.