So that's it for weighty problems from Just Riddles and More! There are lots of variations of this type of problem - we hope you have enjoyed our selection of them! So, take your time and try to figure them out! There are 10 beads in each of the bags.
In four of the bags, the beads each weigh 10 grams. All the bags and beads look identical. The problem is that all the bags look identical and all the beads look identical. You can use a scale, but it has to be a single-tray scale, not a two-tray balance scale. Also, you may use the scale only once. How can you find out which bag has the lighter beads? How do you find out which bag contains low quality gold coins? You may use a scale only one time. Thanks to Jason Vuong for this variation!
The 9 marbles are all uniform in size, appearance and shape. You have access to a balance scale containing 2 trays - you may use the balance 2 times. You must determine which of the 9 marbles is the heavier one using the balance only 2 times. You have 9 gold coins. All 9 coins look exactly the same but one coin is a fake and is either lighter or heavier than the other 8 coins.
How do you find the fake gold coin? We think it is the most difficult! You have 12 identical-looking coins, one of which is counterfeit.
The counterfeit coin is either heavier or lighter than the rest. The only scale you have to use is a simple balance. Using the scale only three times Note: not loading, but using for balancingfind the counterfeit coin. Here are some really good, fun things to do for a little break. Take a few of the short tests.
See if you can follow directions. Can you solve rebus puzzles? Some of these are sure to bring a smile or two.If you become a registered user you can vote on this brain teaser, keep track of which ones you have seen, and even make your own.
Show all 12 comments. Puzzles Trivia Mentalrobics Games Community. Fun: 2. Logic Logic puzzles require you to think. You will have to be logical in your reasoning. There are 12 gold coins. One of the coins is fake. It is known that a fake coin weighs either slightly less or slightly more than a real coin.
Using a balance scale, how can you find the fake coin, and determine if it weighs less or more than a real coin, in only three weighings? Answer Divide the coins into three groups of 4 coins each.
First, weigh two of the groups. If they balance, then the fake coin is in the third group, so take three of the coins, and weigh them against three of the coins that we know for sure are real from the first 8 coins. If they balance, then the remaining coin is the fake one, but we still don't know if it is lighter or heavier, so weigh it against a real coin to determine that. But if it didn't balance, then we know that the fake coin is one of the three, and we also know if it is lighter or heavier, because we know which side had the real coins.
So weigh two of the coins against each other, if they balance then the third coin is fake, if they don't balance we know which coin is fake, because we already know if it is lighter or heavier. Now let's go back to the first weighing; if the two groups didn't balance, we know that the fake coin is in one of the groups, but we don't know which group.
Let's mark the group that was heavier as group A, and the group that was lighter group B. For the second weighing, weigh two coins of group A and one coin of group B, against the other two coins of group A and one coin of group B. If they balance, the fake coin must be one of the remaining two, and a weigh between them will tell us which one they are both from group B so we know that it's lighter.
If the second weigh didn't balance, look at the side that is heavier: Either the fake coin is one of the two group A coins, or it can the group B coin on the other side of the scale.Peg Solitaire is a puzzle game for one person, thought to be at least years old and of French origin. It is part of the Sequential Movement class of mechanical puzzles. It is related to the game Fox and Geese.
Beasley states that the origins of Solitaire are unknown, and that the earliest clear evidence of the game he found is this portrait of Anne Chabot de Rohan Princesse de Soubisewho livedengraved by Claude-Auguste Berey in - the French Wikipedia article on Berey says he lived circa George Bell has done a lot of analysis of Peg Solitaire in its various forms. George's website offers all kinds of information and is definitely worth a visit! Other statistics can be found at Jean-Charles Meyrignac's website euler.
The objective is to start with pegs in specific holes, and one or more vacant holes, then by legal jumping movements, remove pegs to achieve a goal configuration.
A game where one must start with a full board save one empty hole, and end with one peg, is called a " single vacancy to single survivor " problem. When the initial empty hole and the location of the final peg must be the same, it is called the " complement problem. Depending on the variant being played, sometimes diagonal jumps are permitted, or jumps in certain directions are prohibited.
In addition, the game can be played in reverse as noted by Leibniz - start with a single peg and then by "reverse jumps" re-populate the board, with the objective of filling a specific field of holes.
The shape of the board and the arrangement of holes varies as well. Solitaire is usually played on a hole cross-shaped board known as the "English" board though according to Beasley this board type probably appeared in Germany before appearing in England.
A standard notation labeling the holes on the English board assigned the letters abcdefg to the columns and the digits to the rows. The center hole is thus d4. The standard problem on the English board is to start with all holes filled save the central hole, and end with a lone peg there - this is called the Central Gameor the d4 complement problem.
Modern readers may be more familiar with the side-5 triangular variant offering diversion at tables in Cracker Barrel and other restaurants.
The portrait shows the Princess with the hole French board, but on this board it is impossible to start with the central hole vacant and end with the last peg there. Bell and Beasley explain, "The unsolvability of the problem 'start by vacating the central hole, play to leave a single man in this hole' on the hole board is a consequence of a property known as 'position class': the various positions possible on a Solitaire board can be divided into 16 different classes, and it is impossible to play from a position in one class to a position in another.
Example Solitaire Board Types from the Wikipedia article 1. Wiegleb, Germany45 holes 3. Asymmetrical20th century 4.
Find the Fake Coin
The English Board, 33 holes 5. Diamond, 41 holes 6.
Triangular side-515 holes Click here to open a window and run my Triangle Solitaire Solver. A puzzler improves at Solitaire by at first achieving the objective of leaving but a single peg in the correct hole rather than two or more pegs isolated across the board, or even one peg but in the wrong holeand then by reducing the number of moves required for the task to its minimum. The number of jumps will always be equal to or less than the number of initial pegs minus one, since every jump eliminates one peg - but the number of moves can be less if we count the same peg jumping more than one other peg in succession to be a single move.
For a given configuration, an interesting mathematical and computational challenge is to determine this minimum, and the correct sequence of moves - indeed, to determine if a solution exists at all for a given starting configuration and objective.
The shortest solution to the central game on the English board requires 18 moves, shown by Ernest Bergholt in and proved in Asymmetrical20th century. Triangular side-515 holes. Click here to open a window and run my Triangle Solitaire Solver. Cafe Puzzle - Gordon Bros. This is a vintage plastic peg solitaire set. In the 's, Setko aka the Set Screw and Mfg.Step 1: we name all the bag of gold coins as 1, 2, Find out the total weight.
Step 4: Subtract the total of step 2 from total of step 3.
Conclusion: If step 4 results 1 gram, then bag 1 is the low quality coins, if step 4 results 2 grams, then bag 2 is the one, if step 4 results 3 grams, then bag 3 is the one Thanks to Jason Vuong for this variation!
Perhaps an easier way to look at it?! Read on and see what you think: 1. I would put one coin on the scale from one of the bags. If the change in weight on the scale is 10 grams, I 'd add a coin from another bag. I'll keep on adding a coin from each bag until the change in the total weight after adding a coin is by 9 grams and the bag that coin comes from is the low quality one.
I just made it look big and in 3 different steps to make it sound easier but in practicality it's really simple and I believe even a 3 year old kid can do it. Thanks Jay!! The first coin you place on the scale will give a reading, this will count for one use of it, so you would not be able to place another coin on it because this will give you a second reading, resulting in a second use of the scale.
Some of our visitors came up with some nice variations for the solution and we have posted them. Here are some really good, fun things to do for a little break. Take a few of the short tests. See if you can follow directions. Can you solve rebus puzzles? Some of these are sure to bring a smile or two. Can you read this?Form a triangle with the six coins. Now you have 3 rows with 3 coins in each row! Yes it is possible! OFind the 10 Rows - trust me they are there!
The dots are only there as place markers. Draw a regular five-pointed star. Place a coin at each of the points and each of the intersections. How many times does Justin Bieber say 'baby' in his song 'Baby'? Is best defined as the total weight of persons gear equipment stores fuel and motor assembly found on a vessel?
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The 9 Coins Problem
Wiki User Bringing in the third-dimension was cute, but lazy. Here is a 2-D solution: Related Questions. How do you arrange 96 coins in 12 rows? Carefully arrange 12 rows with 8 coins in each row. Arrange 9 coins in 3 rows with 4 coins in each row? How could you arrange 6 coins so that you have 3 coins with 3 coins in each row?
Previous Question: A certain street has buildings. A sign-maker is contracted to number the houses from 1 to How many zeroes will he need? It requires a maximum of 4 weighings. Divide the 9 balls into sets of 3 each. Let them be set1, set2 and set3. Weigh set1 and set2 against each other. If they weigh equal, then set3 contains odd coin. Otherwise, weigh set1 against set3. If they weigh equal, set2 contains odd coin. If in both cases, they do not weigh equal, then set1 contains odd coin.
Now, we have identified set with defective coin in maximum 2 weighings. Repeat this process on individual coins of the found set. This requires a max of 2 more weighings. Hence, the odd coin can be found in a maximum of 4 weighings at all times.
Read Solution 8 :. Actually, it is not given whether the ball is overweight or underweight. It can be either. So, in first iteration, we cannot find identify the group containing the odd ball. We need two iterations for that. So, if there needs only 'n' iterations to find which ball is overweight or underweight, and is giventhen it requires 2n iterations for finding which ball is odd.
One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh L2 against L3. If both are equal, L4 is the odd coin and is lighter.
If L2 is light, L2 is the odd coin and is lighter. If L3 is light, L3 is the odd coin and is lighter. Weight H1 against H2? If both are equal, there is some error. If H1 is heavy, H1 is the odd coin and is heavier. If H2 is heavy, H2 is the odd coin and is heavier. Weight H3 against H4? If both are equal, L1 is the odd coin and is lighter. If H3 is heavy, H3 is the odd coin and is heavier.Difficulty Popularity Two friendsTorres and Lampard, meet after a long time.
Torres: Hey, how are you man? Lampard: Not bad, got married and I have three kids now. Torres: That's awesome. How old are they? Lampard: The product of their ages is 72 and the sum of their ages is the same as your birth date.
Torres: Cool. But I still don't know. Lampard: My eldest kid just started taking piano lessons. Torres: Oh now I get it. How old are Lampard's kids? The product of their ages is So what are the possible choices? That could be anything from 1 to 31 but the fact that Torres was unable to find out the ages, it means there are two or more combinations with the same sum.
From the choices above, only two of them are possible now. The answer is 3, 3 and 8. Difficulty Popularity Outside a room there are three light switches. One of switch is connected to a light bulb inside the room.
You are allowed to set each switch the way you want it and then enter the room note: you can enter the room only once Your task is to then determine which switch controls the bulb?? Set the first switches on for abt 10min, and then switch on the second switch and then enter the room. Three cases are possible 1. Difficulty Popularity A worker is to perform work for you for seven straight days. Difficulty Popularity Tanu lives on the 13th floor takes the elevator down to the ground floor every morning and goes to her office.
In the evening, when she comes back on a rainy dayor if there are other people around in the elevator, she goes to her 13th floor directly. Otherwise, she goes to the 1oth floor and walks up three flights of stairs to his apartment. Whats the reason for her unusual behavior? The women is a of short stature. She can't reach the upper elevator buttons, but can push is with his umbrella. Difficulty Popularity Four friends need to cross a dangerous bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one.
The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins.