Solution

The ferry can be carried out as follows.
First one wife takes each of the other two wives in turn across the river. Then she returns and the husbands of the two wives already on the other side of the river cross. One of those husbands then crosses back over with his wife and then that man returns to the other side of the river with the other man on the same side of the bank. Finally, the wife who having already crossed was currently on the second side of the bank takes each of the other two wives in turn to cross back over the river.
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Source of the solution: the book “VO.Bugayenko., Lomonosov Tournaments., Competitions in Mathematics., MCNMO-CheRo, 1998”.

Answer

Yes it could be possible.

Hint

Solution

The first method. Let Gabby pour water from the full small jug into the large one, and then fill the small one and the she will pour water from that small jug into the large one until it is completely filled. Next, Gabby needs to empty the large jar and pour the remainder of what is in the small jar into it. If the small jar was 3 litres, then now in the large jar there will be 1 litre, otherwise there will be 3 litres. Now let Gabby again try to pour water from the full small jar into the big one. If it succeeds, then the small one was three-litre, if the water spills over the edge – four-litre.
The second method. If Gabby had a large jar holding 10 litres, then it would be enough to try to pour water into it from the small one three times. If the water spills over the edge, then the small one is 4 litres, if not, then it is only 3. With a five-litre jar, such a test is possible if Gabby empties the five-litre jar when it is filled.

Hint

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Solution

Nothing to translate

Hint

Solution

In paragraph $(b)$, the second player must make the amount of money on the table each time a multiple of four pounds. In other words, if the first player takes one pound, then the second takes three pounds, if the first takes two, then the second two, if the first takes three, then the second one. It is not difficult to verify that each time it possible to take the required amount from the table. Thus, before each turn of the first player on the table there is at least four pounds, and so the first player can not win with that turn.
Note that this strategy allows the second player to win if at the beginning of the game on the table there is any amount that is a multiple of four pounds.
In part a$)$ the first player should for his first move take two pounds, after which the amount will be equal to 12 pounds. Next, he must act as the second player in the solution to part b$)$.

Answer

a$)$ the first; b$)$ the second.

Hint

Solution

Let’s show how the first player should act in order to win regardless of the actions of the second player. His first move is to put the coin exactly in the middle of the table. After that, for any move of the second player, the first player must make the symmetric move to the second player relative to the center of the table. This move is possible, because after any move of the first player, the arrangement of coins on the table is symmetrical with respect to the center of the table. Hence, if there is a free space for the coin of the second player, then the place that is symmetric to it will also be free.

Answer

With the correct strategy, the first player wins.

Hint

Hint

If two residents of the island said the same thing, then they are either both liars or both knights.

Solution

It is clear that if two people made the same statement, then they are either both liars or both knights.
Since there is at least one liar and at least one knight on the island, either all the knights made the first statement, and all liars made the second, or vice versa. In the first case, there is both an even number of knights and liars, and in the second one, there is an odd number of both. So, the number of people on the island is even.

Answer

No there can not be.

Hint

If two residents of the island said the same thing, then they are either both liars or both knights.

Solution

For example, the fox can first take off 1 gram from the 5g and 11g pieces three times. One piece now has a mass of 2g and two pieces have masses of 8g. Now it is necessary to cut off and eat 1g off the 8g pieces six times.

Answer

Yes, the fox can.

Hint

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Hint

If every day no more than one car “rests”, then the total number of cars cannot exceed 7.

Solution

a) Five cars are not enough because the day when one of the cars “rests”, someone has no car to drive. Six cars are obviously enough.

b) If every day no more than one car “rests”, then the total number of cars cannot exceed the number of days in the week, that is, seven. But this, obviously, is not enough. So, on some day, two cars “rest”, and on this day, you need at least 8 more cars. Therefore, in total we have 10. We have proved that fewer than ten cars will not do.

Ten cars are enough: for example, every working day two cars “rest”.

Answer

a) 6; b) 10 cars.

Hint

If every day no more than one car “rests”, then the total number of cars cannot exceed 7.

Solution

Here is the desired algorithm:

The father goes with mother in 2 min,
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The dad goes back with a flashlight in 1 min,
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The child goes with the grandmother in 10 min,
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The mom goes back with the flashlight in 2 min,
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The dad and mom cross in 2 min.
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Only 17 minutes!
$\\$ $\\$ $\\$ $\\$
Ideology. We must get rid of the dogma that the flashlight should be back with the fastest moving person, that is, the dad. Then it is already easy to guess that you have to let grandmother and child go together.

Hint

Solution

Errors in the calculation of red balls could only have been done by one of them, and the other two correctly counted the number of red balls. Therefore, there were 2 red balls. In the calculation of the red balls, C was mistaken, so he confused the red with the orange, and the yellow and green ones were correct. We get that there were then 8 yellow balls, and 9 green ones. All the remaining balls were orange. The total number of balls being considered was correct – 23. So, there were 4 orange balls.

Answer

Reds – 2, orange – 4, yellow – 8, green – 9.

Hint

Hint

What are the second and fourth digits of the quotient? What are the first and last digits of the quotient equal to?

Solution

From the arrangement of the asterisks in the rebus one can at once conclude that the second and fourth digits of the quotient are zeros. The divisor multiplied by 8 gives a two-digit number, and when multiplied by the first and fifth digits of the quotient, it is a three-digit number, then the first and last digits of the quotient are 9, and in all the quotient is equal to 90809. We find the divisor. This is a two-digit number that, when multiplied by 8, gives a two-digit number, and when multiplied by 9, it gives a three-digit number. Hence it is easy to conclude that the divisor is 12. And the whole example has the form – 1089708/ 12 = 90809.

Hint

What are the second and fourth digits of the quotient? What are the first and last digits of the quotient equal to?

Hint

Try in five attempts to determine to which of the six suitcases the first key fits.

Solution

The standard wrong solution: “Each of the six suitcases we try to open with each of the six keys, all attempts $6 \times 6$ = 36”. You can find a match between the keys and the suitcases in fewer attempts.
Take the first key and in turn try to open the suitcases. If one of the suitcases opens then its fine, we set aside this suitcase with this key. If, however, none of the first 5 suitcases open, then this key necessarily corresponds to the sixth suitcase. What happened? We used no more than five attempts; we had 5 keys and 5 suitcases.
Again, take one key and open all the bags in a row. To determine which suitcase corresponds to this key, you need four attempts. And so on. In total, 5 + 4 + 3 + 2 + 1 = 15 attempts are needed. And if there were 10 suitcases, the number of attempts needed would be 9 + 8 + … + 2 + 1 = 45.

Hint

Try in five attempts to determine to which of the six suitcases the first key fits.

Solution

Let EIGHT = x, then $EIGHT^2$ – EIGHT = x $(x – 1)$. But the difference $EIGHT^2$ – EIGHT ends in five zeros. Since x and x – 1 are consecutive numbers, one of them is divisible by 32, and the other is odd and is divisible by 3125 $(100000 = 32 \times 3125)$. Note that the second digit of the number $(I)$ is zero $($ see the multiplication scheme $)$. Five-digit numbers, which are multiples of 3125 and satisfy the condition I = 0, are 40625 and 90625. The consecutive numbers are 40624 and 40626, 90624 and 90626. Only 90624 is divisible by 32. So the desired number is 90624 or 90625. The first one does not fit, since 4⊃ does not end in 4.

Answer

EIGHT = 90625.

Hint

Solution

Pay attention to the last phrase. We rewrite the example in the following form
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$\\$
Obviously, $x_4$ = $x_8$ = $x_{15}$, $x_{11}$ = 2. In addition, it is clear that $x_1$ = 1 or $x_1$ = 2, otherwise $2x_1$ is greater than 5, which contradicts the condition.
1$)$ Let $x_1$ = 1, then $x_9$ = 2, $x_5$ = 3, then the doubled multiplier is in the range of 3402 to 3492, or the multiplier is between 1701 and 1746. Let’s check three possible options, given that the sum of the digits for both factors is the same. 1701 × 27, 1711 × 28, 1721 × 29 shows us that no option is suitable.
2$)$ Let $x_1$ = 2. Then $x_9$ = 4, $x_5$ = 1, $x_2$ = 2, with $x_6$ $≤$ 5, otherwise $x_6$ + 4 $>$ 9, which contradicts the hypothesis, but then 5 $≤$ $x_4$ $≤$ 7. Checking the three possible variants 2221 × 25, 2231 × 26, 2241 × 27, we find that only the second case is suitable.

Hint