Riddles, brain puzzles and mathematical problems

[BAndrew]

I have solved it in the past (I think 2 years ago). It's too easy and boring. No thought required, you just have to be patient. I bet Einstein never wrote it. They just gave it this name to popularize the riddle. I don't like it at all. I won't post the solution so others can solve it.

Not to mention the vagueness some statements have. For example, in statement NO4 The green house is left of the white house, but from which perspective?

[Red]

I have solved it in the past (I think 2 years ago). It's too easy and boring. No thought required, you just have to be patient. I bet Einstein never wrote it. They just gave it this name to popularize the riddle. I don't like it at all. I won't post the solution so others can solve it.

Indeed, i managed to get to that group of 2% too. But i've still seen many struggle with this riddle.

[Bridge]

The perspective is always facing the front side of the houses if it's not specified - I didn't personally find any of the statements to be vague.

Spoiler below!

The German.

[Red]

Correct. Congratulations you're smarter than 98% of earths people. According to this.

[BAndrew]

We have a scale with two disks like this:



and we have 27 identical coins, but 1 of them has less weight than the rest.
How many weighings are needed to find this coin? Specify also how this will be done. Give the answer with the minimum possible weightnings.

[Froge]

13 coins on the left, 13 on the right. If both balance, then remaining coin not weighed is the special one. If not balanced, take side that weighs less and divide it 6-6. If still no balance, the process is repeated with 3-3, and finally 1-1.

4 weighings necessary.

[BAndrew]

13 coins on the left, 13 on the right. If both balance, then remaining coin not weighed is the special one. If not balanced, take side that weighs less and divide it 6-6. If still no balance, the process is repeated with 3-3, and finally 1-1.

4 weighings necessary.

Great try, but there is a way to do it in 3 weightnings. Maybe I didn't make it clear, but you have to find the "best" possible answer or equivalently the minimum number of weightnings.

[Oscar House]

Make 3 piles of 9 coins each. Compare two of them and you'll know which of the piles contains the lighter coin. Divide that pile into 3 piles of 3 coins, then compare two of those. Now you'll know which of the 3-coin piles contains the lighter coin. Now just weigh two of the remaining coins and you'll find the right coin.

[BAndrew]

Make 3 piles of 9 coins each. Compare two of them and you'll know which of the piles contains the lighter coin. Divide that pile into 3 piles of 3 coins, then compare two of those. Now you'll know which of the 3-coin piles contains the lighter coin. Now just weigh two of the remaining coins and you'll find the right coin.

Υou are very close! Specify what happens if the scale balances in each case and you are good to go.

[Oscar House]

Make 3 piles of 9 coins each. Compare two of them and you'll know which of the piles contains the lighter coin. Divide that pile into 3 piles of 3 coins, then compare two of those. Now you'll know which of the 3-coin piles contains the lighter coin. Now just weigh two of the remaining coins and you'll find the right coin.

Υou are very close! Specify what happens if the scale balances in each case and you are good to go.

Right. If there's three piles to compare and two of the piles are the same weight (i.e. the scale balances), that means the lighter coin is in the third, unweighed coin pile and you'll just continue dividing that pile into three smaller piles until you find it.