Ten bags
There are 10 bags of coins. All the coins in one bag are fake. A genuine coin weighs 10 grams and a counterfeit coin weighs 9 grams. How can you identify a bag of counterfeit coins using just one weighing on a graduated scale?
Solution
First, you need to number all the bags from 1 to 10, then you need to take from each bag as many coins as its serial number (from 1 to 10). If all the coins were real, then the pile of coins would weigh 550 grams (1 + 2 + 3 ... + 10) * 10 = 550. If a bag of counterfeit coins has the number N (N = from 1 to 10), then the taken from the bags, the coins will weigh N grams less, therefore, the taken pile of coins will weigh N grams less. Those. By how many grams does a heap of weight differ from 550 grams, such a bag contains counterfeit coins.
Eight bags
You have 8 bags of coins, each containing 48 coins. Five bags contain real coins, and the rest contain counterfeit coins. Fake coins are 1 gram lighter than real ones. With one weighing on a precision scale, identify all bags of counterfeit coins using the minimum number of coins.
Solution
There is no need to get coins from the first bag (0), from the second bag you need to get one coin (1), from the third two (2), from the fourth - four (4), from the fifth - seven (7), from the sixth - thirteen (13), the seventh - twenty-four (24), the eighth - forty-four (44). Each three "piles" of coins, taken together, are unique in that they give a certain exact weight, allowing one to identify bags of counterfeit coins (95 coins in total). If all the coins in the proposed solution were real, then their total weight would be 95 cu. (0+1+2+4+7+13+24+44). Compare the scale reading with what it would ideally be if all the coins were real. The resulting difference (number of conventional units) will indicate the numbers of bags with counterfeit coins. For example, if the difference is 21, then the coins in the second, fifth and sixth bags are counterfeit, because It was from them that we took 21 coins (1+7+13).
Christmas balls
On New Year's tree There are three pairs of balls hanging: two white, two blue and two red. Externally the balls are identical. However, each pair has one light and one heavy ball. All light balls weigh the same, and so do all heavy balls. Using two cup scales, determine all the light and all the heavy balls.
Solution
Place one red and one white ball on the left scale and one blue and one white ball on the right scale. If equilibrium is achieved, then it is obvious that on each bowl there is one heavy and one light ball. Therefore, it is enough to compare two white balls to find out the answer to the question that interests us. However, if after the first weighing equilibrium is not achieved, then on the heavier side lies a heavy white ball. The next logical step would be to compare the weight of the red ball that has already been weighed and the one that has not yet been weighed. blue ball. After this, it will be clear to you which balls are light and which are heavy.
Nine bags
There are nine bags: eight with sand and one with gold. The bag of gold is a little heavier. You are given two weighings on the pan scales to find the bag of gold.
Solution
Divide the nine bags into three groups of three bags each. Weigh the two groups. This way you will find out which group contains the bag of gold. Now select 2 bags from the group that definitely contains a bag of gold and weigh them.
27 tennis balls
There are 27 tennis balls. 26 weigh the same, but the 27th is slightly heavier. What is the minimum number of weighings on a cup scale that guarantees that a heavy ball will be found?
Solution
It is enough to use the scales three times. Divide the 27 balls into 3 groups of 9 balls each. Compare two groups - the heavy ball will be in the group that outweighs. If the scales have reached balance, then the heavy ball is in the third group. Thus, we will define a group of 9 balls, one of which is the desired one. Divide this group into 3 subgroups, each with three balls. Similar to the first step, compare the weight of any two subgroups. Now compare two balls (two out of three, among which the one you are looking for must definitely be).
Shattered Weight
A merchant dropped a 40-pound weight and it broke into 4 unequal pieces. When these parts were weighed, it turned out that the weight of each of them (in pounds) was an integer. Moreover, these parts could be used to weigh any weight (representing a whole number) up to 40 pounds on a pan scale. How much did each part weigh?
Solution
The fragments weighed: 1 pound, 3 pounds, 9 pounds and 27 pounds, for a total of 40 pounds.
Nails in a bag
There are 24 kg of nails in a bag. How can you measure 9 kg of nails on a cup scale without weights?
Solution
One option: divide 24 kg into two equal parts of 12 kg, balancing them on the scales. Then divide 12 kg into two equal parts of 6 kg. After this, set aside one part and divide the other in the same way into parts of 3 kg. Finally, add these 3 kg to the six kilogram part. The result will be 9 kg of nails.
Each of the 10 bags contains 10 coins. Each coin weighs 10 g. But in one bag all the coins are counterfeit - not 10, but 11 g each. How can you determine in which bag (in the 1st, or in the 2nd, or in the 3rd) using just one weighing? m, etc.) are there counterfeit coins (all bags are numbered from 1 to 10)? The bags can be opened and any number of coins can be pulled out from each.
ANSWER
You need to take one coin out of the first bag, two from the second, three from the third, etc. (from the tenth bag - all ten coins). Next, all these coins should be weighed together once. If there were no counterfeit coins among them, i.e. all of them weighed 10 g, then their total weight would be 550 g. But since among the weighed coins there are counterfeit ones (11 g each), their total weight would be more than 550 g. Moreover, if it turns out to be 551 g, then the coins are counterfeit are in the first bag, because we took one coin from it, which gave an extra 1 g. If the total weight is 552 g, then the counterfeit coins are in the second bag, because we took two coins from it. If the total weight is 553 g, then the counterfeit coins are in the third bag, etc. Thus, with just one weighing it is possible to accurately determine which bag contains counterfeit coins.
Ten bags
There are 10 bags of coins. All the coins in one bag are fake. A genuine coin weighs 10 grams and a counterfeit coin weighs 9 grams. How can you identify a bag of counterfeit coins using just one weighing on a graduated scale?
First, you need to number all the bags from 1 to 10, then you need to take from each bag as many coins as its serial number (from 1 to 10). If all the coins were real, then the pile of coins would weigh 550 grams (1 + 2 + 3 ... + 10) * 10 = 550. If a bag of counterfeit coins has the number N (N = from 1 to 10), then the taken from the bags, the coins will weigh N grams less, therefore, the taken pile of coins will weigh N grams less. Those. By how many grams does a heap of weight differ from 550 grams, such a bag contains counterfeit coins.
Eight bags
You have 8 bags of coins, each containing 48 coins. Five bags contain real coins, and the rest contain counterfeit coins. Fake coins are 1 gram lighter than real ones. With one weighing on a precision scale, identify all bags of counterfeit coins using the minimum number of coins.
There is no need to get coins from the first bag (0), from the second bag you need to get one coin (1), from the third two (2), from the fourth - four (4), from the fifth - seven (7), from the sixth - thirteen (13), the seventh - twenty-four (24), the eighth - forty-four (44). Each three "piles" of coins, taken together, are unique in that they give a certain exact weight, allowing one to identify bags of counterfeit coins (95 coins in total). If all the coins in the proposed solution were real, then their total weight would be 95 cu. (0+1+2+4+7+13+24+44). Compare the scale reading with what it would ideally be if all the coins were real. The resulting difference (number of conventional units) will indicate the numbers of bags with counterfeit coins. For example, if the difference is 21, then the coins in the second, fifth and sixth bags are counterfeit, because It was from them that we took 21 coins (1+7+13).
Christmas balls
There are three pairs of balls hanging on the New Year tree: two white, two blue and two red. Externally the balls are identical. However, each pair has one light and one heavy ball. All light balls weigh the same, and so do all heavy balls. Using two cup scales, determine all the light and all the heavy balls.
Place one red and one white ball on the left scale and one blue and one white ball on the right scale. If equilibrium is achieved, then it is obvious that on each bowl there is one heavy and one light ball. Therefore, it is enough to compare two white balls to find out the answer to the question that interests us. However, if after the first weighing equilibrium is not achieved, then on the heavier side lies a heavy white ball. The next logical step is to compare the weight of the red ball that has already been weighed and the blue ball that has not yet been weighed. After this, it will be clear to you which balls are light and which are heavy.
Nine bags
There are nine bags: eight with sand and one with gold. The bag of gold is a little heavier. You are given two weighings on the pan scales to find the bag of gold.
Divide the nine bags into three groups of three bags each. Weigh the two groups. This way you will find out which group contains the bag of gold. Now select 2 bags from the group that definitely contains a bag of gold and weigh them.
27 tennis balls
There are 27 tennis balls. 26 weigh the same, but the 27th is slightly heavier. What is the minimum number of weighings on a cup scale that guarantees that a heavy ball will be found?
It is enough to use the scales three times. Divide the 27 balls into 3 groups of 9 balls each. Compare two groups - the heavy ball will be in the group that outweighs. If the scales have reached balance, then the heavy ball is in the third group. Thus, we will define a group of 9 balls, one of which is the desired one. Divide this group into 3 subgroups, each with three balls. Similar to the first step, compare the weight of any two subgroups. Now compare two balls (two out of three, among which the one you are looking for must definitely be).
Shattered Weight
A merchant dropped a 40-pound weight and it broke into 4 unequal pieces. When these parts were weighed, it turned out that the weight of each of them (in pounds) was an integer. Moreover, these parts could be used to weigh any weight (representing a whole number) up to 40 pounds on a pan scale. How much did each part weigh?
The fragments weighed: 1 pound, 3 pounds, 9 pounds and 27 pounds, for a total of 40 pounds.
Nails in a bag
There are 24 kg of nails in a bag. How can you measure 9 kg of nails on a cup scale without weights?
One option: divide 24 kg into two equal parts of 12 kg, balancing them on the scales. Then divide 12 kg into two equal parts of 6 kg. After this, set aside one part and divide the other in the same way into parts of 3 kg. Finally, add these 3 kg to the six kilogram part. The result will be 9 kg of nails.
Ten hats
There are ten numbered hats on the table. Each hat contains ten gold coins. One of the hats contains counterfeit coins. A real coin weighs 10 grams, and a fake one only 9. Scales with a scale in grams are provided to help. How to determine which hat contains counterfeit coins using the scale for only one weighing? The scales can weigh no more than 750 grams.
We take 1 coin from the first hat, 2 from the second, 3 from the third, etc. We weigh all this and subtract the result from the ideal weight (in our case, 55 × 10 = 550 grams). The resulting number will match the number of the hat with counterfeit coins.
81 coins
There are 81 coins of the same denomination. One of them is counterfeit and it is lighter than a real coin. How can you find this coin using four weighings on a pan scale?
It is necessary each time to divide the entire volume of coins into 3 equal piles and weigh 2 of them. If the heaps are equal in weight, then the desired coin is in the third heap, but if one of the two heaps is lighter, then the counterfeit coin is in it. Next, the found heap must again be divided into 3 parts and any 2 weighed. In the first weighing, heaps of 27 coins are measured, in the second weighing, heaps of 9 coins are measured, in the third weighing, heaps of 3 coins are measured, and in the fourth weighing, one is placed on the scales. coin.
Puzzle scales
In the two pictures, the scales are in equilibrium. How many pears do you think should be used to balance the six oranges on the third scale?
The first scale shows that 2 apples + 1 orange weigh the same as one pear. The second scale shows that 2 apples + 2 oranges = 6 apples, i.e. 2 oranges equal 4 apples or 1 orange = 2 apples. Based on the data of the first and second scales, we find that 1 pear is equal to 4 apples or 2 oranges. Hence, 6 oranges will be balanced by 3 pears.
In the two pictures, the scales are in equilibrium. How many pears do you think should be used to balance out two apples and one orange?
According to the data of the second scale, it is clear that an apple is equal to a pear and an orange. If we substitute these data on the first scales, we find that two oranges are equal to one orange and two pears, therefore, one orange is equal to two pears. Substituting two pears instead of an orange on the second scale, we find that an apple is equal to three pears. Hence, to balance the third scale, 8 pears are needed.
In the two pictures, the scales are in equilibrium. How many pears do you think should be used to balance two apples and two oranges?
It is necessary to increase the fruit on the first scale three times, you get 12 pears + 3 apples = 15 oranges. On the second scale we know the weight of 3 apples = 3 oranges and 6 pears, let’s transfer them instead of 3 apples to the first scale. We get: 18 pears = 12 oranges or 3 pears = 2 oranges. Next, multiply scales B by 2. We get: 6 apples = 6 oranges + 12 pears. If we replace 6 oranges with the equivalent in pears, we get: 6 apples = 21 pears or 2 apples = 7 pears. Thus, 2 apples + 2 oranges = 7 pears + 3 pears = 10 pears.
How many oranges are needed to balance the scales in the last picture? Items can only be delivered to the right side of the scale.
To balance the scales you will need 5 oranges.
Sugar in bags
There are two bags, one empty and the other containing 9 kg of sugar. How to distribute sugar into bags in the proportion of 2 kg in one bag and 7 kg in the other in 3 weighings on a cup scale using 50g and 200g weights?
1. It is necessary to weigh the sugar into bags into 2 equal parts of 4.5 kg each. 2. Divide the sugar in one bag again into halves, 2.25 kg each, and scatter them into bags (one bag will contain 2.25 kg, and the other will contain 6.75 kg). 3. Using two weights totaling 250g, separate 250g of sugar from the 2.25kg bag and transfer it to another bag. As a result, one bag will contain 7 kg, the other 2 kg of sugar.
4 coins
There are 4 coins, one of which is counterfeit and it differs from the genuine ones in weight either more or less. How to identify a counterfeit coin after 2 weighings on a cup scale?
Let's put coins 1 and 2 on the scales: 1) if they are not balanced, then remove the second one and put the third one in its place. If the scales are in balance, then coin 2 is counterfeit. If the scales do not balance, then coin 1 is counterfeit. 2) the scales are balanced, then we remove coin 2 and put coin 3 in its place. If the scales are balanced, then the counterfeit coin is 4. If the scales are not balanced, then the counterfeit coin is 3.
Two weights
There are standard scales with cups and two weights: 10 and 2 kg. How can you use them to weigh 3 kg of plums?
Initially, weigh 2 kg of plums. Then we divide them equally among the scales so that the scales are balanced. 1 kg of plums received. Name 1 kg and a 2 kg weight, you can measure any desired amount, including 3 kg.
68 coins
There are 68 coins, all of them different in weight. How to find the lightest and heaviest in 100 weighings?
We weigh all the coins in pairs, putting the light ones in one pile, the heavy ones in another, for a total of 34 weighings. In the first pile, we weigh in turn all the coins with the lightest one at the moment, i.e. if a lighter one is found, then the next coins are weighed with it, and so on 33 times. With the right pile - the same thing, but we only identify the heaviest coin, also 33 weighings. Total - exactly 100 weighings.
Damaged scales
Among the 100 identical-looking coins, there are several counterfeit ones. All counterfeit coins weigh the same, all real ones do the same, and a counterfeit coin is lighter than a real one. There are also scales (with two bowls without a pointer), each bowl only holds one coin. At the same time, the scales are slightly damaged: if the coins are of different weights, the heavier coin outweighs, but if they are the same, any cup can outweigh. How can you find at least one counterfeit coin using these scales?
Divide the coins into 33 piles of 3 coins + 1 coin. We weigh each trio among themselves, we get 3 inequalities, as a result of which we see, either each coin will weigh less than the other two once, or will weigh less than the other two twice. 1>2 (the following options are possible: n=n, f=f, 2-fake) 1<3 (н=н, ф=ф, 1- фальшивка)
2>3 (n=n, f=f, 3-fake) this is possible if all three coins have the same weight among themselves, that is, we put any one of them aside 1<2(н=н,ф=ф,1-ф)
1<3(н=н,ф=ф,1-ф)
2>3(n=n,f=f,3-f) 1 is more likely to be fake, so we put it aside. And we do this with each of the 33 piles, as a result we will set aside 11 +1 coins that did not end up in any of the piles. We again divide these 12 coins into 4 piles of 3 coins each, do the same manipulations, as a result we get 4 coins, divide them into 1 pile + 1, the coin from the pile that turns out to be lighter is set aside again and compared with a single coin. The one that is lighter will be false.
80 coins
There are 80 coins, one of which is counterfeit, and it is lighter than the others. In what minimum number of weighings on a scale without weights can you find a counterfeit coin?
A counterfeit coin can be identified in 4 weighings. The algorithm is as follows. First weighing: put 27 coins on the bowls. In the case of equilibrium, the false one is among the remaining 26. If one bowl is lighter, then there are 27 false ones among those lying on it. Second weighing: we put 9 coins from the number of “suspects” on both bowls and reason in a similar way. In the third weighing we will put 3 coins on the bowls, and in the fourth - one coin each. As you can see, here the division is not in half, but into three, if possible, equal parts.
Sage
When the ruler of the country decided to reward an intelligent man for a good deed, he wanted to take as much gold as an elephant weighs. But how do you weigh an elephant? There were no such scales in those days. What could you come up with in such a situation?
The sage did this: he placed the elephant in the boat, then marked the water level on the side. When the elephant was taken out of the boat, all that remained was to place the gold there.
Five items
Five objects of different weights must be arranged in descending order of their weight. You can only use the simplest scales without weights, which only allow you to determine which of the two objects being compared in weight is heavier. How should one proceed to solve the problem in an optimal way, that is, so that the number of weighings is minimal? How many weighings will have to be done?
The first weighing is to compare any two of the five given items. Let A be the lighter object and B the heavier object. Then we write the result of the first weighing in the form A Then compare the other two objects and denote the lighter one as D and the heavier one as E: D Let's denote the fifth item C. The third weighing is to compare objects B and E. Both possibilities arising here lead to similar reasoning, so we will limit ourselves to considering case B By the fourth weighing we compare the fifth object C with object B. It is necessary to distinguish two cases: a) B b) C In the first case (B A Let us compare (this will require a fifth weighing) objects C and E. Here it is also necessary to distinguish between two possible cases: E If A In case A In the second case (C A Let's compare items A and C (fifth weighing). In both possible cases (A Since we have exhausted all possible cases, the proof ends here.
Two scales
There are 9 identical coins, one of which is counterfeit and for this reason is lighter than the others. We have two scales without weights, which allow us to compare the weight of any group of coins. However, some of the available scales are crude; they cannot distinguish a counterfeit coin from a real one. Their accuracy does not allow them to detect differences in weight. But other scales are accurate. But which scales are rough and which are accurate is unknown. In this situation, how can you identify a counterfeit coin using three weighings?
Let's put four coins for each cup on scales No. 1. If one group of coins outweighs, then the rest is clear - these scales are accurate, and we know 4 coins, among which one is counterfeit. Let the scales be in balance. Let us denote by A the ninth coin and add to it coins B and C - one from each four. We put the remaining two triplets of coins on scales No. 2. The worst option is equilibrium again. Then on scales No. 2 we compare coins B and C. In the case of equilibrium, coin A will be counterfeit.
2000 balls
There are 6 weights weighing 1, 2, 3, 4, 5 and 6 g. They are marked accordingly. However, there is reason to believe that one mistake was made when marking the weights. How can you determine whether the markings on the weights are correct using two weighings on a cup balance, where you can compare the weights of any group of weights?
We place weights marked 1, 2 and 3 g on one pan of the scale, and 6 g on the other. Equilibrium means that an error in marking is possible only within groups 1-2-3 and 4-5. During the second weighing, we place weights of 3 and 5 g on one bowl, and 6 and 1 g on the other. If the first bowl is overweight, then there is no error in the markings.
8 coins
There are 8 seemingly identical coins. One of them is fake and is known to be lighter than the real one. How can you find a counterfeit coin with just two weighings on a pan scale?
We divide the coins into three piles of 3, 3 and 2 coins. We weigh the piles containing three coins. If the weight is the same, then we weigh 2 coins from the third pile among themselves and identify the counterfeit (lighter) one. If one group of three coins is lighter than the other, then there is a counterfeit coin there. We leave the lighter group of three coins and put two coins on the scales and proceed according to the previous algorithm: if the weight is the same, then the third one is counterfeit, and if not, then the one that is lighter.
Saladin's Puzzle
This story happened a long time ago, back during the Crusades. One of the knights was captured by the Muslims and appeared before their leader, Sultan Saladin, who announced that he would free the prisoner and his horse if he received a ransom of 100 thousand gold coins. “Oh, great Saladin,” the knight, who did not have a penny to his name, then addressed the Sultan, “you are depriving the last hope. In my homeland, a wise and resourceful captive is given a chance to be released. If he solves a given puzzle, he is released on all four sides, if not, the ransom amount doubles!”
“So be it,” replied Saladin, who himself loved puzzles. “Listen. They will give you twelve gold coins and simple scales with two cups, but without weights. One of the coins is counterfeit, but it is not known whether it is lighter or heavier than the real ones. You must find it in just three weighings. If you don’t complete the task before morning, you’ll have to blame yourself!” Could you get out?
It is necessary to divide 12 coins into 4 piles of 3 coins each. Let's put 2 heaps on the scales (one at a time in different bowls). Then two cases are possible: 1) If the scales are not in balance, then the counterfeit coin is in one of these heaps. We remove the lighter pile and put a third one in its place. If the scales are in balance, then the counterfeit coin is in the pile removed from the scales. If the scales are not in balance, then the counterfeit coin is in the heavier pile. (2 weighings have been done so far). 2) If the scales are in balance after the first weighing, then remove any pile and put a third in its place. If the scales are in balance, then the counterfeit coin is in the fourth pile. If the scales are not in balance, then the counterfeit coin is in the third pile. (2 weighings have been done so far). After finding a pile of 3 coins, then we next determine which of the 3 coins is counterfeit: you need to put 2 coins in the third weighing and if they are in balance, then the third coin is counterfeit. If they do not balance, then instead of a lighter coin you need to put a third one. If the scales balance, then the counterfeit coin is removed. If they do not balance, then the heavier coin is counterfeit.
20 pounds of tea
How to weigh 20 pounds of tea into 10 boxes of 2 pounds each in nine weights, having only 5 and 9 pound weights, using a regular cup scale?
1) Place a 5-pound weight on one pan of the scale and a 9-pound weight on the other. Then balance the scale by pouring 4 pounds of tea into a bowl with a 5-pound weight. 2) Remove the weights from the scales, leave 4 pounds in one pan, and balance the scale by adding another 4 pounds to the second. 3) Weigh 4 pounds again. 4) And again 4 pounds. Thus, after four weighings, the remainder will also be 4 pounds. 5-9) Divide 4 pounds in half, balancing the scales.
101 coins
Among 101 identical coins, one is counterfeit and differs in weight. How can you use a cup scale without weights to determine in two weighings whether a counterfeit coin is lighter or heavier? There is no need to find a counterfeit coin.
We weigh 50 and 50 coins: 1) Equality: We take the remaining coin and put it in the left pile instead of one of the ones there 1.1 The left pile is heavier => the counterfeit coin is heavier 1.2 The left pile is lighter => the counterfeit coin is lighter 2) Inequality: We take the heavier pile and split it into two piles of 25 coins. 2.1 The weight of the piles is the same => the counterfeit coin is lighter 2.2 The weight of the piles is not the same => the counterfeit coin is heavier
Baron Munchausen's problem
Baron Munchausen has eight outwardly identical weights weighing 1 g, 2 g, 3 g, ..., 8 g. He remembers which weight weighs how much, but Count Sclerosis does not believe him. Will the baron be able to carry out one weighing on a cup scale, as a result of which the weight of at least one of the weights will be unambiguously established?
7+8=1+2+3+4+5, 6 remains.
2N coins
There are 2N numbered coins, and: all real coins weigh the same, all counterfeit coins also weigh the same, the counterfeit coin is lighter than the real one. coins with numbers from 1 to N are real, and coins with numbers from N+1 to 2N are fake. Of these two statements, the judge knows only the first, and the expert knows both. How can an expert convince a judge of the truth of the second statement after three weighings on cup scales without weights?
a: N=7
b: N=9
Problem "a" was proposed at one of the All-Union Mathematical Olympiads in the 1970s. Since then, N=7 (and in general, N=2^K-1 for K weightings) has been considered unimprovable. And yet, this is not the case. Improvement (problem "b") was invented by S. Tokarev in 1997.
a) 1) The expert weighs coins 1 and 8. (1 > 8) The judge is convinced that the 8 is false. 2) The expert weighs 1+8 and 9+10. (1+8 > 9+10) The judge is convinced that 9+10 is easier than one fake and one real. Therefore, he concludes that both 9 and 10 are false. 3) The expert weighs 1+8+9+10 and 11+12+13+14. Likewise, the judge can make a judgment about all coins 11-14. Note that exactly one real coin is needed. b) Preliminary action: the expert groups the coins into the following three piles: A (1, 2; 10, 11); B (3, 4, 5; 12, 13, 14); B (6, 7, 8, 9; 15, 16, 17, 18); Each pile contains an equal number of real and counterfeit coins, this is known to the expert, but this will be proven to the judge as a result of weighing. 1) Real coins from pile A and fake coins from pile B are placed on the left pan of the scale, and fake coins from pile A and real coins from pile B are placed on the right pan. The right pan is heavier than the left one. 2) Real coins from pile B and fake coins from pile C are placed on the left pan of the scale, and fake coins from pile B and real coins from pile C are placed on the right pan. The right pan is heavier than the left one. 3) Real coins from pile B and counterfeit coins from piles A and B are placed on the left pan of the scale, and fake coins from pile B and real coins from piles A and B are placed on the right pan. The right pan is heavier than the left one. Let x denote the difference in weights of real and counterfeit coins in pile A, i.e. (1+2) -(10+11), y - the same for pile B, that is (3+4+5)-(12+13+14), z - (6+7+8+9)- (15+16+17+18). Our weighings proved the following three inequalities to the judge: y > x; z > y; x+y > z. Since x,y,z are integers, strict inequalities can be replaced with non-strict ones: y >= x+1 z >= y+1 x+y >= z+1. Hence: x+y >= y+2 => x >= 2; x+y >= x+3 => y >= 3; 2z >= x+y+3 >= z+4 => z >= 4. On the other hand, it is obvious that the difference between K real coins and K unknown coins cannot be greater than K, and equality occurs only when all unknown coins are counterfeit. This proves everything the judge needs... Note that in this case, 9 real coins are not needed! How many of them are actually needed? Think... An even more interesting problem is for four weighings. The algorithm from problem a) allows an expert to prove that 15 coins are fake. A generalization of Tokarev's algorithm allows us to improve this estimate to 27.
Dungeon Escape
The king, his son the prince and his daughter the princess were in the dungeon of a high tower. They weighed 195, 105 and 90 pounds respectively. The food was lifted to them in two baskets attached to the ends of a long rope. The rope was thrown over a beam driven under the roof itself. It turned out that when one basket was on the ground, the second was at the level of the window in the prisoners' cell. These baskets remained the only hope for salvation. Naturally, as soon as one basket became heavier than the other, it sank. However, if the difference in weight exceeded 15 pounds, the basket would plummet downward. The only thing that would help the prisoners escape from captivity was the 75-pound cannonball in the cell - they could try to use it as a counterweight. How did the prisoners manage to escape?
1. The princess descends, using the cannonball as a counterweight. 2. The princess, having reached the ground, does not get out of the basket. The prince takes the place of the core and descends, using the princess as a counterweight. 3. The princess rises up and, together with the king, puts the cannonball into the basket. 4. The prince sits in the lowered basket with the cannonball, which allows the king to be lowered. 5. When the king is on the ground, the prince with the cannonball is on top. The prince gets out of the basket and the basket with the cannonball goes down. 6. The princess sits in an empty basket near the dungeon and descends to the ground. 7. The prince pulls the cannonball out of the raised basket and descends himself, using the princess as a counterweight. 8. The princess lowers the cannonball in an empty basket, and she sits down in the raised one and descends, using the cannonball as a counterweight.
1999 coins
There is a set of 1999 coins. It is known that 1410 of them are fake. A counterfeit coin differs in weight by 1 g from a genuine one, and some counterfeit coins may be lighter and others heavier than genuine ones. We have cup scales that can show the difference in weight. How to determine the authenticity of any coin from a set in one weighing?
We weigh all the coins except this one and look at the difference in weight. Let us denote the weight of a normal coin as N, then all coins will weigh either 1998*N+2x (where 0=<705) - в данном случае наша монета настоящая, либо 1998*N+(2x-1) (где 0=<705) - в этом случае наша монета фальшивая.
Polypropylene bag 10 kg with handle
We offer premium quality 10 kg polypropylene bags for small wholesale products at competitive prices. This is a modern environmentally friendly packaging used in the food industry, wholesale and retail trade, and agriculture. The container is made from primary polypropylene, a synthetic material with high consumer properties.
White PP bag 10 kg with a handle is designed for packaging products with a bulk structure: sugar, salt, flour, starch, cereals, pasta, legumes, seeds, tea, coffee. The cut-out handle makes it easy to transport and carry by hand. The PromTrust catalog presents polypropylene food bags of high-quality weaving, manufactured in accordance with GOST.
The products are safe in contact with food, do not emit hazardous substances, and do not absorb odors. The packaging meets sanitary and hygienic requirements, which is confirmed by certificates of the State Sanitary and Epidemiological Supervision.
Scope of application of polypropylene bags 10 kg
A 10 kg polypropylene bag is suitable for packaging, storing and transporting dry bulk cargo. Can be used for food and non-food products. The container protects the contents from moisture, dust, pollution, solar radiation, temperature changes, and insect damage. The product is poured through the bottom or top (depending on the model) and the packaging is sewn up. Bag stitching threads LSh-210, machine GK-9 and other models are suitable for stitching.
Advantages of polypropylene bags 10 kg
The material is characterized by impact resistance, withstands repeated bending and friction. The packaging is suitable for long-term storage of products in warehouse conditions. Product advantages:
- chemical inertness;
- dense structure;
- a light weight;
- ease of use;
- resistance to low, high temperatures, UV radiation;
- breathability;
- resistance to decay, bacteria, alkalis, organic solvents;
- dielectric properties;
- does not collapse in boiling water;
- reusable and recyclable;
- economical price
Thanks to the rough texture, the packaging does not slip. The container is not damaged during transportation, preventing production losses.
Buy polypropylene bags 10 kg in Moscow with delivery
At the PromTrust company you can buy a 10 kg polypropylene bag wholesale, large wholesale and retail. We deliver orders in Moscow, the Moscow region and send them to the regions, pickup is possible. The bags are compressed and supplied in packs of 500 pieces.
