Best riddles

love

The destruction of Troy

It caused the destruction of Troy, The worst of tragedies And numerous maladies Yet it is chased, desired and fought for What is it?
Love.
94.80 %
49 votes

clean

Slim and tall

I am slim and tall, many find me desirable and appealing, they touch me and I give a false good feeling, once I shine in splendor, but only once and then no more, for many I am "to die for".
A cigarette.
94.69 %
48 votes

logic

Dad and 3 kids

Once upon a time there was a dad and 3 kids. When the kids were adults, the dad was old and Death came to take the dad. The first son, who became a lawyer, begged Death to let the dad live a few more years. Death agreed. When Death came back, the second son, who became a doctor begged Death to let his father live a few more days. Death agreed. When Death came back the third son, who became a priest begged Death to let the dad live till that candle wick burned out and he pointed to a candle. Death agreed. The third son knew Death wouldn't come back, and he didn't. Why not?
The third son went over and blew out the candle after Death left because the son said "till the candle wick burns out", not "till the candle burns out".
94.69 %
48 votes

clean

The cost of making only the maker knows

The cost of making only the maker knows, Valueless if bought, but sometimes traded. A poor man may give one as easily as a king. When one is broken pain and deceit are assured.
Promise.
94.48 %
46 votes

what am I

Sooner or later everybody needs my help

I dig out tiny caves, and store gold and silver in them. I also build bridges of silver and make crowns of gold. They are the smallest you could imagine. Sooner or later everybody needs my help, yet many people are afraid to let me help them. What am I?
A dentist.
94.48 %
46 votes

logic

Going to St. Ives

As I was going to St. Ives I met a man with seven wives The seven wives had seven sacks The seven sacks had seven cats The seven cats had seven kits Kits, cats, sacks and wives How many were going to St. Ives?
One person is going to St. Ives (the narrator). Because the narrator "met" all of the others mentioned in the poem, this implies that they walked past each other in opposite directions, and thus none of the wives, sacks, cats, or kits was actually headed to St. Ives. If you (like many) think this answer is a bit silly, you can assume that all the people, sacks, and animals mentioned were heading for St. Ives. In this case, we would have 1 narrator + 1 man + 7 wives + 49 sacks + 343 cats + 2401 kits = 2802 total going to St. Ives. However, this isn't the traditional answer.
94.48 %
46 votes

logic

Course of expensive medication

You've been placed on a course of expensive medication in which you are to take one tablet of Sildenafil and one tablet of Citrate daily. You must be careful that you take just one of each because taking more of either can have serious side effects. Taking Sildenafil without taking Citrate, or vice versa, can also be very serious, because they must be taken together in order to be effective. In summary, you must take exactly one of the Sildenafil pills and one of the Citrate pills at one time. Therefore, you open up the Sildenafil bottle, and you tap one Sildenafil pill into your hand. You put that bottle aside and you open the Citrate bottle. You do the same, but by mistake, two Citrates fall into your hand with the Sildenafil pill. Now, here's the problem. You weren't watching your hand as the pills fell into it, so you can't tell the Sildenafil pill apart from the two Citrate pills. The pills look identical. They are both the same size, same weight (10 micrograms), same color (Blue), same shape (perfect square), same everything, and they are not marked differently in any way. What are you going to do? You cannot tell which pill is which, and they cost $300 a piece, so you cannot afford to throw them away and start over again. How do you get your daily dose of exactly one Sildenafil and exactly one Citrate without wasting any of the pills?
Carefully cut each of the three pills in half, and carefully separate them into two piles, with half of each pill in each pile. You do not know which pill is which, but you are 100% sure that each of the two piles now contains two halves of Cirate and half of Sildenafil. Now go back into the Sildenafil bottle, take out a pill, cut it in half, and add one half to each stack. Now you have two stacks, each one containing two halves of Sildenafil and two halves of Citrate. Take one stack of pills today, and save the second stack for tomorrow.
94.48 %
46 votes

logic

Coins on a table

Your friend pulls out a perfectly circular table and a sack of quarters, and proposes a game. "We'll take turns putting a quarter on the table," he says. "Each quarter must lay flat on the table, and cannot sit on top of any other quarters. The last person to successfully put a quarter on the table wins." He gives you the choice to go first or second. What should you do, and what should your strategy be to win?
You should go first, and put a quarter at the exact center of the table. Then, each time your opponent places a quarter down, you should place your next quarter in the symmetric position on the opposite side of the table. This will ensure that you always have a place to set down our quarter, and eventually your oppponent will run out of space.
94.36 %
45 votes

logic

Twelve balls, one different

You have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls. You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls. What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?
Take eight balls, and put four on one side of the scale, and four on the other. If the scale is balanced, that means the odd ball out is in the other 4 balls. Let's call these 4 balls O1, O2, O3, and O4. Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale. If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier. If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter. If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter. If the scale isn't balanced, then the odd ball out is among these 8 balls. Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier"). Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter"). Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls. So now weigh [H1, H2, L1] against [H3, L2, Normal]. -If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter. -So measure [H1, L2] against 2 of the "Normal" balls. -If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier. -If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter. -If the scale is balanced, then H2 is the odd ball out, and is heavier. -If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier. -So measure L1 and H3 against two "normal" balls. -If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter. -Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier. If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter. So weight [H4, L3] against two of the "Normal" balls. If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier. If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter. If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.
94.36 %
45 votes

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