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## Comments

196It's (7 + 1/7) * (7 + 7) = 100! Yay me!

196Fifth riddle:There's a wolf, a chicken and some corn:

They all belong to a farmer who would very much like to transport them across a river using a boat. The problem is the boat is so small he can only bring one "passenger" at a time (either wolf, chicken of corn). There is no limit to the number of crossings.

- If the wolf stays with the chicken, its gonna eat it;
- If the chicken stays with the corn, its gonna eat it.

Is there a way for the farmer to make this work?2,139- Sail back alone.

- Sail the corn to the other side.

- Sail back with the chicken.

- Sail the wolf to the other side.

- Sail back alone.

- Sail the chicken to the other side. Done.

196196Riddle #6You are facing a closed door standing next to 3 light switches:

Behind that door stands a lamp. Exactly one of the 3 light switches controls that lamp. How can you identify that switch without opening the door (and looking at the lamp...) more than once?

53and here is yours answer

here is a smilar riddle just for you JonSnowIsAlive

there is pipe just next to the swichs room, the pipe starts on one wall and ends another wall, no other room has pipes in them, and you can hear water discharge in it but you dont know which way.

is it possible to find out which way this water go without breaking it???

53try to solve the original one. i gurantee you hate mankind immedietly. i disgusted myself for being human alone. why not elf or dwarf? why did i pick human?

6,456196Will try to find a less what-the-hell solution!

536,456If you're interested.

196@FinneousPJ Nice article! In what field are studying exactly? Saw the keyword "medical", then was pleasantly surprised to see an integral...? Math is so cool.

6,456161196I for one frequently model claims frequency in a car insurance context. The common basis of such models is the Poisson distribution which for very small values of its parameter cannot be correctly approximated by the normal distribution.

1615992nd crossing: sister 1 returns to side A.

3rd crossing: sister 1 and brother 1 cross from side A to side B; sister 1 and brother 1 stays on side B.

4th crossing: sister 2 returns to side A.

5th crossing: sister 2 and brother 2 cross from side A to side B; brother 2 stays on side B.

6th crossing: sister 2 returns to side A.

7th crossing: sister 2 and sister 3 cross from side A to side B; sister 2 stays on side B.

8th crossing: sister 3 returns to side A.

9th crossing: sister 3 and brother 3 cross from side A to side B and stay on side B.

alternatively:

7th crossing: sister 3 and brother 3 cross from side A to side B; brother 3 stays on side B.

8th crossing: sister 3 returns to side A.

9th crossing: sister 2 and sister 3 cross from side A to side B and stay on side B.

...but, the picture says that 11 crossings are required at a minimum, so I assume that I've misunderstood one of the rules of the problem.

161I think the problem is that none of the men can be in the presence of any of the women unless her brother is present as well. After your third crossing, S1, S2 and B1 are all on the same side, however briefly. So naturally, B1 will immediately rape S2 before she has a chance to get back to the boat.

A solution avoiding this, I believe, goes thusly:

Crossing

(Side 1 || Side 2)

1: S1,S2

(B1,B2,B3,S3 || S1,S2)

2: S2

(B1,B2,B3,S2,S3 || S1)

3: S2,S3

(B1,B2,B3 || S1,S2,S3)

4: S3

(B1,B2,B3,S3 || S1,S2)

5: B1,B2

(B3,S3 || B1,B2,S1,S2)

6: B2,S2

(B2,S2,B3,S3 || B1,S1)

7: B2,B3

(S2,S3 || B1,B2,B3,S1)

8: S1

(S1,S2,S3 || B1,B2,B3)

9: S1,S2

(S3 || B1,B2,B3,S1,S2)

10: S2

(S2,S3 || B1,B2,B3,S1)

11: S2,S3

( || B1,B2,B3,S1,S2,S3)

599I clearly underestimated the depravity of these men.

53the pipe with water flow. how do we findout which way it flows?

find 100 with 1,7,7,7,7 and elementary school math. which solved correctly by @JonSnowIsAlive

find 100 with 7,7,7,7 and any sicentific calculator function. which hasn't been solved yet so i decide to add some hints here.

and the brothers and their sisters puzzle. which solved correcty by @Morte50.

this puzle has more than one vay to solv it but morte's solution so perfect that i wont try to improve it. i m glad that the decency of those fictional sisters is in safe hands, nevertheless they are still screwed for having such brothers.

thanks for allowing me to contribute in such a quality forum. my (stolen from unknown places) puzzles will continue.

5176,4562,139You have nine identical gold coins. Except of them is counterfeit and weighs a bit less than the others. You also have an old-fashioned balance scale (see below). You can only use the scale two times.

Using the scale, how do you identify which coin is counterfeit?

2,643- Weigh any two of them against each other. If one group is lighter, then the counterfeit coin is in that group. If they are balanced, then the counterfeit coin is in the unweighed group.

- Take any two of the coins in the unweighed group and weigh them against each other. If one of them is lighter, then that is the counterfeit coin. If they are balanced, then the unweighed coin is the counterfeit.

2,13953you have 12 identical golden coins (or something really valuable) and one of them is fake.

fake one is differ with weight only, but you dont know lighter or heavier.

you have only 3 use with a scale. (i really can't make up a proper story for this rediculous limitation)(as a bonus answer you might try fixing the question, as filling with a background, i don't know maybe scale will be broken if you try using it 4th time)

how could you possibly pinpoint the fake one?

as answer you need the whole possibilities diagram. i mean you must be able to figure out the fake one %100 not %99.

i will try my best for explaning the solution with my rubbish english. just give me a couple of days.

have fun and be warned this puzle is really hard and long!

2,4381) Weigh 2 groups. If scales tip, go to 2), if not go to 3)

2) Remove 2 coins from one side of scale and 1 coin from the other. Swap 1 coin from each side of the scale to the other. Add an unweighed coin to the side that you removed 2 coins from. If the scales tip the same way as in 1), go to 4). If the scale tips the opposite way, go to 5). If the scales do not tip, go to 6)

3) Weigh 2 of the unweighed coins. If the scales tip, go to 7), if not go to 8)

4) Remove all the coins that were added or swapped last step. You will now have 1 coin on one side and 2 coins on the other. Remove the single coin and move one of the move one of the remaining coins to the opposite side (so there's 1 coin on each side). If the side that had 2 coins was high, then whichever side is now high is the fake coin. If the side that had 2 coins was low, whichever side is now low is the fake coin. If the scales are now balanced the 3rd coin is fake.

5) We now know that the fake is one of the 2 coins that were swapped (let's call them A and . Weigh A against any other coin (besides . If the scales tip then A is fake, if not B is fake.

6) We now know that the fake coin is 1 of the 3 removed last step. Take the 2 coins that were removed from the same side (let's call them A and , and weigh them. If A and B were on the low side after 1), then whichever coin is now low is fake. If A and B were on the high side after 1), then whichever coin is now high is fake. If the scales are now balanced, the 3rd coin is fake.

7) We now know 1 of the 2 coins just weighed (let's call them A and is fake. Weigh A against any other coin (besides . If the scales tip, A is fake. If not, B is fake.

8) We now know 1 of the 2 remaining un-weighed coins (let's call them A and is fake. Weigh A against any other coin (besides . If the scales tip, A is fake. If not, B is fake.

53i think you got it. but i think i couldn't recognize the answer if i didn't know the answer for sure, i must say.

it may because of my bad english or too complicated demostration.

first weight 4to4 and leave the 4

there is two possibilities, sides stay same or different.

if they same this means we know the 8 of them genuine and you have 2 wheighting left to analyse rest4,

you need to leave 1 posible fake here, and weight 3 posible fake with 3 genuine, this time we have 3 possibilities. either one of 3 possible fake is light, heavy or genuine, last weighting should clear all possibilities with prowing the fake one is light or heavy.

now if in first weighting sides may different and a scale could be heavier then the rest. this means 4 of them possibly heavy and 4 of them possibly light and 4 of them genuine.

this time we need to label the coins as what are they, may be we have a pen or something. and mark on them as H,L,G. then you have to put one side H,H,L and other one same H,H,L and leaving 2 possible light and genuine ones.

again 2 possibilities either sides same or different.

same sides means all 6 of them genuine and you use 3rd weighting with two L to check which one is lighter.

and if the sides different this means either H's are heavy or the L is light, again the rest is genuine.

3rd scale must be weighting remaining possible heavy ones to check which one is heavier.

as a bonus answer for backround story for this puzzle is: you are near graduation of jewelcrafting school. and this puzzle is one of the tests. and someone allready solved this with 4 weighting. you are desperatly trying to impress your teachers. ok lame background i could only come up whit this lame story in this time... i realy wanna hear your imagination about this.

i hope its understandable enough for first time encounterers. i know its a wall of writing but the solution is rewarding imho. have fun

1,1502,415A bard equips a long sword that should deal 1d8 damage; the bard has proficiency in all weapons. The bard whacks a goblin with the long sword; the goblin takes 472 slashing damage and is obliterated. The bard hits several more goblins with the long sword; each one takes exactly 472 damage.

The bard then equips a bastard sword that should deal 2d4 damage. The bard whacks a goblin with this sword and the goblin takes 724 damage; its guts fly all the way to the moon. Several other goblins are slashed by this bastard sword, and each of them takes exactly 724 damage.

The bard then equips a two-handed sword that should deal 1d10 damage. The bard whacks the last goblin in the encounter. How much damage does this goblin take? (Note: It is a specific amount of damage, not a range)