Riddle about counterfeit coins. Puzzles for weighing 10 bags of gold coins

Ten bags

There are 10 bags of coins. In one bag, all the coins are counterfeit. The genuine coin weighs 10 grams and the counterfeit coin weighs 9 grams. How to determine a bag of counterfeit coins using one weighing on a scale with divisions?

Solution

First you need to number all the bags from 1 to 10, then you need to take as many coins from each bag 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 the bag of fake coins has the number N (N = 1 to 10), then the taken from the bags, the coins will weigh N grams less, therefore, the pile of coins taken will weigh N grams less. Those. by how many grams the pile differs in weight from 550 grams, such a bag contains counterfeit coins.

eight bags

You have 8 bags of coins, 48 ​​coins each. Five of the bags contain real coins, and the rest are counterfeit. Fake coins are 1 gram lighter than real ones. With a single weighing on an accurate balance, identify all bags of counterfeit coins using the minimum number of coins.

Solution

It is not necessary to get coins from the first bag (0), from the second bag it is necessary to get one coin (1), from the third two (2), the fourth - four (4), the fifth - seven (7), the sixth - thirteen (13), on the seventh, twenty-four (24), on the eighth, forty-four (44). Every three "piles" of coins taken together are unique in that they give a certain exact weight that allows you to identify bags of counterfeit coins (95 coins are used in total). If all the coins in the proposed solution were real, then their total weight would be 95 c.u. (0+1+2+4+7+13+24+44). Compare the reading on the scale with what it would ideally be if all the coins were real. The resulting difference (the number of conventional units) will indicate the numbers of bags with counterfeit coins. For example, if the difference is 21, then the counterfeit coins are in the second, fifth, and sixth bags, because it is from them that we took 21 coins (1+7+13).

Christmas balls

Three pairs of balls hang on the Christmas tree: two white, two blue and two red. Externally, the balls are the same. However, in each pair there is one light and one heavy ball. All light balls weigh the same among themselves, and so do all heavy balls. Using two weighings on a balance pan, determine all light and all heavy balls.

Solution

Place one red and one white ball on the left pan of the scale, and one blue and one white ball on the right pan. If equilibrium is reached, 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 our question. However, if balance is not achieved after the first weighing, then on the heavier side lies a heavy white ball. The next logical step is to compare the weight of the already weighed red ball and the unweighed blue ball. After that, 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 slightly heavier. You are given two weighings on a pan balance to find a bag of gold.

Solution

Divide the nine bags into three groups of three bags each. Weigh two groups. Thus, you will find out in which of the groups the bag of gold is. Now select 2 bags from the group where exactly there is a bag of gold and weigh them.

27 tennis balls

There are 27 tennis balls. 26 weigh the same and 27 is slightly heavier. What is the minimum number of weighings on a pan balance that guarantees finding a heavy ball?

Solution

It is enough to use the scales three times. Divide 27 balls into 3 groups of 9 balls each. Compare the 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, three balls each. Similar to the first step, compare the weights of any two subgroups. Now compare two balls (two out of three, among which there must be exactly the one you are looking for).

Cracked weight

The merchant dropped a 40-pound weight, and it split into 4 unequal parts. When these parts were weighed, it turned out that the weight of each of them (in pounds) is an integer. What's more, any weight (which is an integer) up to 40 pounds could be weighed on a pan balance using these parts. How much did each piece weigh?

Solution

The fragments weighed: 1 lb, 3 lb, 9 lb and 27 lb, for a total of 40 lb.

Nails in a bag

There are 24 kg of nails in a bag. How can you measure 9 kg of nails on a pan balance without weights?

Solution

One option: divide 24 kg into two equal parts of 12 kg, balancing them on the scales. Then also divide 12 kg into two equal parts of 6 kg. After that, set aside one part, and divide the other in the same way into parts of 3 kg. Finally, to the six-kilogram part, add these 3 kg. The result is 9 kg of nails.

psychological exercises for training

Riddle about counterfeit coins

In front of you are 10 open bags of coins in sufficient quantity (say, each bag contains 100 coins). In one bag are counterfeit coins that weigh 2 grams each. In the remaining nine bags, the coins are real, 1 gram each. Coins do not differ from each other in anything other than weight. It is impossible to determine the weight by hand. In front of you are electronic scales. How to determine in one (!!!) weighing which bag contains counterfeit coins? No tricks are accepted: coins cannot be dipped into the water, thrown from the ninth floor, poured one at a time at an equal pace and counted as one weighing, and so on. Just one weigh-in. It is necessary to determine the fake bag using only electronic scales.

Answer to the riddle:

We have 10 bags and they are open. First, we number the bags of coins. Next, we put a different number of coins from each bag on the scales. From the first 1 coin, from 2 - two coins, from 3 - three coins, from 4 - four coins, from 5 - five coins, from 6 - six coins, from 7 - seven coins, from 8 - eight coins, from 9 - nine coins, out of 10 - ten coins. We calculate the total amount if all the coins were normal (not fake): 1+2+3+4+5+6+7+8+9+10=55. And then we look at the scoreboard of electronic scales - we draw a conclusion from how much the amount will differ from the ideal one. For example, if the scales show the amount of 58 grams, then these extra 3 grams came to us from 3 bags, then there are fake coins in it.

19.09.2012

Alexei

in my opinion, you can do this. number the bags and put one coin in one line from each in the order of numbering the bags. then take the coins in turn and see the difference in weight))) the coin will immediately be visible to the cat 200 grams. one weighing - after all, we put the coins only once on the scales - and then we just took off one coin at a time)))
17.11.2013

Elena

stylish challenge!
26.02.2014

Gennady

Aleksey, each coin withdrawal is a measurement, but it is necessary for one weighing!
13.06.2014

Maksim

Gennady is right, Alexey's method does not fit the condition of the problem))
07.09.2014

just put the bags in turn, a bag in which 10 coins of 1 gram will weigh 10 grams, and when we put a bag of fake coins, it will weigh 2
.
01.07.2015

Anna

and why exactly from the third bag, maybe the 5th or another
20.09.2015

cap

each withdrawal of a coin is a measurement, but it is necessary for one weighing! so every time you put a coin on the scales - this is also a measurement ..
29.10.2015

Sergey

I struggled with this riddle a couple of years ago for 3 days, until at 3 o'clock in the morning I came up with a solution)))
29.11.2015

Vladimir

everything is correct. it's just that the scales turn on only when all the coins are already on them
06.12.2015

Elena

I've known this puzzle since childhood... it's simple and complex at the same time.
08.12.2015

Kanamat

From the first one from strict two and so on from 10-10 by how much what amount of weight more in the bag are fake
25.07.2017

Alexander

Such a riddle was in the film about Colombo. He figured it out, of course.

Each of the 10 bags contains 10 coins. Each coin weighs 10 g. But in one bag all the coins are fake - not 10, but 11 g. m, etc.) are counterfeit coins (all bags numbered from 1 to 10)? The bags can be opened and any number of coins can be pulled out of each.

ANSWER

One coin must be pulled out from the first bag, two from the second, three from the third, and so on. (from the tenth bag - all ten coins). Then all these coins should be weighed together once. If there were no counterfeit coins among them, i.e. all of them would weigh 10 g each, then their total weight would be 550 g. But since there are fake coins (11 g each) among the weighed coins, their total weight will be more than 550 g. Moreover, if it turns out to be 551 g, then fake coins 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 fake 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, and so on. Thus, with only one weighing, it is possible to determine exactly which bag contains counterfeit coins.