If you had a 5-liter bowl and a 3-liter bowl, and an unlimited access to water, how would you measure exactly 4 liters?
Fill the 5-litre bowl and pour water to the 3-litre bowl, which you empty afterwards. From the 5-litre bowl pour the 2 remaining litres to the 3-litre bowl. Refill the 5-litre bowl and fill in the 3-litre bowl (with 1 litre), so there stay the 4 required litres in the 5-litre bowl.
Given three bowls: 8, 5 and 3 liters capacity, divide 8 liters in half (4 + 4 liters) with the minimum number of water transfers. Note that the 8-liter bowl is initially filled with 8 liters of water and the other two bowls are empty - that is all water you have.
pour 5 litres from the 8-litre to the 5-litre bowl,
pour 3 litres from the 5-litre to the 3-litre bowl,
pour these 3 litres back to the 8-litre bowl,
pour the remaining 2 litres from the 5-litre to the 3-litre bowl,
pour 5 litres from the 8-litre to the 5-litre bowl,
pour the missing 1 litre from the 5-litre to the 3-litre bowl (there should be 4 litres left in the 5-litre bowl),
pour the 3 litres back from the 3-litre to the 8-litre bowl (and that's it - in 8-litre bowl 4 litres).
Given three bowls: 7, 4 and 3 liters capacity. Only the 7-liter is full. Pouring the water the fewest number of times, make the quantities of 2, 2, and 3 liters.
Three numerals in each number stand for litres in each bowl:
700 - 340 - 313 - 610 - 601 - 241 - 223 (pouring 6 times)
How can you measure 6 liters of water using only 4 and 9-liter bowls?
First fill the 9-litre bowl. Then pour 4 litres to the 4-litre bowl (there are 5 litres in the 9-litre bowl afterwards) and pour out the water from the 4-litre bowl. And again pour 4 litres to the 4-litre bowl and empty it. Then pour the remaining litre to the 4-litre bowl but this time keep it there. Fill the 9-litre bowl to the top for the second time and pour water to fill the 4-litre bowl to the top. Thus the required 6 litres stay in the 9-litre bowl.
Measure exactly 2 liters of water if you have:
1. 4 and 5-liter glass
2. 4 and 3-liter glass
1st Fill the 5-litre bowl, pour water from it to fill the 4-litre bowl, which you empty afterwards. pour the remaining 1 litre to the 4-litre bowl. Refill the 5-litre bowl and pour water from it to fill the 4-litre bowl (where there is already 1 litre). Thus you are left with 2 litres in the 5-litre bowl.
2nd The same principle - this time from the other end. Fill the 3-litre bowl and pour all of the water to the 4-litre bowl. Refill the 3-litre bowl and fill the 4-litre bowl to the top. And there you have 2 litres in the 3-litre bowl.
Given three bowls: bowl A (8 liters capacity) filled with 5 liters of water; bowl B (5 liters capacity) filled with 3 liters of water; and bowl C (3 liters capacity) filled with 2 liters of water.
Can you measure exactly 1 liter, by transferring the water only 2 times?
1st Pour 1 litre from bowl A to bowl C. Thus 4 litres are left in the bowl A and bowl C is full (3 litres).
2nd Pour 2 litres from bowl C to bowl B. Doing that you have full bowl B (5 litres) and there is 1 litre left in bowl C.
You have 10 bags with 1000 coins each. In one of the bags, all coins are forgeries. A true coin weighs 1 gram; each counterfeit coin weighs 1.1 gram.
If you have an accurate scale, which you can use only once, how can you identify the bag with the forgeries? And what if you didn't know how many bags contained counterfeit coins?
If there is only 1 bag with forgeries, then take 1 coin from the first bag, 2 coins from the second bag ... ten coins from the tenth bag and weigh the picked coins. Find out how many grams does it weigh and compare it to the ideal state of having all original coins. The amount of grams (the difference) is the place of the bag with fake coins.
If there is more than 1 bag with forgeries, then there is lots of possible solution. I can offer you this one as an example: 1, 2, 4, 10, 20, 50, 100, 200, 500 and 1000.
A genuine gummy drop bear has a mass of 10 grams, while an imitation gummy drop bear has a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears and the others - imitation.
Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears?
Spike uses 51 gummy drop bears: from the 7 boxes he takes respectively 0, 1, 2, 4, 7, 13, and 24 bears.
The notion is that each box of imitation bears will subtract its number of bears from the total "ideal" weight of 510 grams (1 gram of missing weight per bear), so Spike weighs the bears, subtracts the result from 510 to obtain a number N, and finds the unique combination of 3 numbers from the above list (since there are 3 "imitation" boxes) that sum to N.
The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. To accomplish this, I put numbers into S one at a time in ascending order, starting with the obvious choice, 0. (Why is this obvious? If I'd started with k > 0, then I could have improved on the resulting solution by subtracting k from each number) Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. And most obviously all the new triples had to be distinct from any old triples; it was easy to find what the new triples were by adding the newest number to each old sum of pairs.
Now, in case you're curious, the possible weight deficits and their unique decompositions are:
3 = 0 + 1 + 2
5 = 0 + 1 + 4
6 = 0 + 2 + 4
7 = 1 + 2 + 4
8 = 0 + 1 + 7
9 = 0 + 2 + 7
10 = 1 + 2 + 7
11 = 0 + 4 + 7
12 = 1 + 4 + 7
13 = 2 + 4 + 7
14 = 0 + 1 + 13
15 = 0 + 2 + 13
16 = 1 + 2 + 13
17 = 0 + 4 + 13
18 = 1 + 4 + 13
19 = 2 + 4 + 13
20 = 0 + 7 + 13
21 = 1 + 7 + 13
22 = 2 + 7 + 13
24 = 4 + 7 + 13
25 = 0 + 1 + 24
26 = 0 + 2 + 24
27 = 1 + 2 + 24
28 = 0 + 4 + 24
29 = 1 + 4 + 24
30 = 2 + 4 + 24
31 = 0 + 7 + 24
32 = 1 + 7 + 24
33 = 2 + 7 + 24
35 = 4 + 7 + 24
37 = 0 + 13 + 24
38 = 1 + 13 + 24
39 = 2 + 13 + 24
41 = 4 + 13 + 24
44 = 7 + 13 + 24
Note that there had to be (7 choose 3) distinct values; they end up ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36, 40, 42, and 43.
This puzzle goes a step further from the previous one.
You have eight bags, each of them containing 48 coins. Five of those bags contain only true coins, the rest of them contain fake coins. A fake coin weighs 1 gram less than a real coin. You have an accurate scale, with the precision of up to 1 gram.
Weighing only once and using the minimum number of coins, how can you find the bags containing the fake coins?
Similar to the former brain teaser.
I take out 0 (no coin from the first bag), 1 (one coin from the second bag etc.), 2, 4, 7, 13, 24, 44 coins (from the last 8th bag). Each triple is unique enabling an easy way to identify the bags with fake coins (using only 95 coins).
One of twelve tennis balls is a bit lighter or heavier (you do not know which) than the others. How would you identify this odd ball if you could use an old two-pan balance scale only 3 times?
You can only balance one set of balls against another, so no reference weights and no weight measurements.
It is enough to use the pair of scales just 3 times. Let's mark the balls using numbers from 1 to 12 and these special symbols:
x? means I know nothing about ball number x;
x< means that this ball is maybe lighter then the others;
x> means that this ball is maybe heavier then the others;
x. means this ball is "normal".
At first, I lay on the left pan balls 1? 2? 3? 4? and on the right pan balls 5? 6? 7? 8?.
If there is equilibrium, then the wrong ball is among balls 9-12. I put 1. 2. 3. on the left and 9? 10? 11? on the right pan. If there is equilibrium, then the wrong ball is number 12 and comparing it with another ball I find out if it is heavier or lighter.
If the left pan is heavier, I know that 12 is normal and 9< 10< 11<. I weigh 9< and 10<.
If they are the same weight, then ball 11 is lighter then all other balls.
If they are not the same weight, then the lighter ball is the one up. If the right pan is heavier, then 9> 10> and 11> and the procedure is similar to the former text.
If the left pan is heavier, then 1> 2> 3> 4>, 5< 6< 7< 8< and 9. 10. 11. 12. Now I lay on the left pan 1> 2> 3> 5< and on the right pan 4> 9. 10. 11.
If there is equilibrium, then the suspicious balls are 6< 7< and 8<. Identifying the wrong one is similar to the former case of 9< 10< 11< If the left pan is lighter, then the wrong ball can be 5< or 4>. I compare for instance 1. and 4>. If they weigh the same, then ball 5 is lighter the all the others. Otherwise ball 4 is heavier (is down).
If the left pan is heavier, then all balls are normal except for 1> 2> and 3>. Identifying the wrong ball among 3 balls was described earlier.
On a Christmas tree there were two blue, two red, and two white balls. All seemed same. However, in each color pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls.
Using a balance scale twice, identify the lighter balls.
Lay one red and one white ball on left pan and one blue and the other white ball on the right pan. If there is equilibrium, then it is clear that there is one heavier and one lighter ball on each side. That's why comparing white balls is enough to learn everything.
However, if at first weighing one side is heavier, then there must be a heavier white ball on that side. The next reasonable step is to compare the already weighed red ball and yet not weighed blue ball. After that, the character of each ball is clear, isn't it?
There are 9 similar balls. Eight of them weigh the same and the ninth is a bit heavier.
How would you identify the heavier ball if you could use a two-pan balance scale only twice?
Divide the 9 balls into 3 groups of 3. Weigh two groups. Thus you find out which group is the heavier ball in. Choose 2 balls from this group and compare their weights. And that's it.
Given 27 table tennis balls, one is heavier than the others.
What is the minimum number of weighings (using a two-pan balance scale) needed to guarantee identifying the heavy one? Of course, the other 26 balls weigh the same.
It is enough to use a pair of scales 3 times.
Divide the 27 balls to 3 groups, 9 balls in each. Compare 2 groups - the heavier one contains the ball. If there is equilibrium, then the ball is in the third group. Thus we know the 9 suspicious balls.
Divide the 9 balls to 3 groups of 3. Compare 2 groups, and as mentioned above, identify the group of 3 suspicious balls.
Compare 2 balls (of the 3 possibly heavier ones) and you know everything.
So we used a pair of scales 3 times to identify the heavier ball.
Suppose that the objects to be weighed may range from 1 to 121 pounds at 1-pound intervals: 1, 2, 3,..., 119, 120 and 121. After placing one such weight on either of two weighing pans of a pair of scales, one or more precalibrated weights are then placed in either or both pans until a balance is achieved, thus determining the weight of the object.
If the relative positions of the lever, fulcrum, and pans may not be changed, and if one may not add to the initial set of precalibrated weights, what is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects?
There are necessary at least 5 weights to bring into balance any of the 121 possible objects. And they weigh as follows: 1, 3, 9, 27, 81g.
Having 2 sandglasses: one 7-minute and the second one 4-minute, how can you correctly time 9 minutes?
Turn both sand-glasses. After 4 minutes turn upside down the 4-min sand-glass. When the 7-min sand-glass spills the last grain, turn the 7-min upside down. Then you have 1 minute in the 4-min sand-glass left and after spilling everything, in the 7-min sand-glass there will be 1 minute of sand down (already spilt). Turn the 7-min sand-glass upside down and let the 1 minute go back. And that's it. 4 + 3 + 1 + 1 = 9
A teacher of mathematics used an unconventional method to measure a 15-minute time limit for a test. He just used 7 and 11-minutehourglasses. During the whole time he turned sandglasses only 3 times (turning both hourglasses at once count as one flip).
Explain how the teacher measured 15 minutes.
When the test began, the teacher turned both 7min and 11min sand-glasses. After the 7min one spilt its last grain, he turned it upside down (the 11min one is still to spill sand for another 4 minutes). When the 11min sand-glass was spilt, he turned the 7min one upside down for the last time. And that's it.
Your job is to measure 45 minutes, if you have only two cords and matches to light the cords.
The two cords are twisted from various materials and so their different segments can burn at different rates.
Each cord burns from end to end in exactly one hour.
Start fire on both ends of one igniter cord and on one end of the second igniter cord. The very moment the first cord (where both ends burn) stops burning (that is after 30 minutes), start fire on the other end of the second cord (otherwise it would burn another 30 minutes). Thus the second igniter cord burns just 15 minutes from then. And that is all together 45 minutes.
