Difficult clean riddles

logicclean

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.
73.10 %
93 votes
cleanclever

There are five vowels in English language (a, e, i, o, u, and sometimes y). Easy version: Can you tell us a word contains of these 5 vowels? Hard: Can you tell us a word contains of all these 6 vowels with y? Very difficult: Can you tell us a word contains of all these vowels with y in their alphabetical order?
Unquestionably In their alphabetical order: Facetiously Abstemiously Adventitiously
73.01 %
62 votes
cleanfunnylogiclove

A doctor and a bus driver are both in love with the same woman, an attractive girl named Sarah. The bus driver had to go on a long bustrip that would last a week. Before he left, he gave Sarah seven apples. Why?
An apple a day keeps the doctor away!
72.99 %
258 votes
logicsimpleclean

Even though the odds are always in favor of the gambling house, why does the establishment insist on a house limit on stakes?
Every casino in the world would go bankrupt without a house limit on stakes. Without it, gamblers would keep doubling their stakes until they won. No matter how bad a losing streak they were on, they would eventually win. For more information, search: Martingale
72.84 %
66 votes
cleanpoemssimple

Die without me, Never thank me. Walk right through me, Never feel me. Always watching, Never speaking. Always lurking, Never seen.
Air.
72.81 %
216 votes
logiccleansimple

A boy goes and buys a fishing pole that is 6' 3" long. As he goes to get on the bus, the driver stops him. The driver tells him that he can't take anything longer than 6' onto the bus. The boy goes back into town, purchases one more thing, and the driver allows the boy on the bus. What did the boy buy, and what did he do with it?
The boy bought 6' long box. He put the fishing pole in diagonally and the entire package was only 6'!
72.80 %
79 votes
logicmathcleanclever

2+3=8, 3+7=27, 4+5=32, 5+8=60, 6+7=72, 7+8=? Solve it?
98 2+3=2*[3+(2-1)]=8 3+7=3*[7+(3-1)]=27 4+5=4*[5+(4-1)]=32 5+8=5*[8+(5-1)]=60 6+7=6*[7+(6-1)]=72 therefore 7+8=7*[8+(7-1)]=98 x+y=x[y+(x-1)]=x^2+xy-x
72.54 %
52 votes