Solution

We denote the sum of the numbers by S, and the product by P. If S is equal to 3, 4 or 5, then P $<$ S and the sentence stated by Sage does not make sense. If S $≥$ 7, then among the variants of the sets which could make the sum S, there are the sets: $(1, 2, S-3)$ and $(2, 2, S-4)$. In both cases, P $>$ S, which contradicts the statement of Patricia. There remains the variant S = 6. In this case, P can be 4, 6 and 8. But Patricia said that her number is smaller than Sage’s. So Patricia named the numbers 1, 1 and 4 $($ after the words of Sage she understood that S ≠ 5, and rejected the option 1, 2, 2 $)$.

Answer

1, 1 and 4.

Hint

Solution

Let A be the first three-digit number, B the second. By the condition 1000A + B = 3AB, whence $(3A – 1)$B = 1000A. The numbers A and 3A – 1 are relatively prime, hence 1000 is divisible by 3A – 1. But A is a three-digit number, hence 3A – 1 is either a three-digit or four-digit number. There are 2 divisors of 1000 of this kind which are 500 and 1000. 3A – 1 can not equal 1000, since 1001 is not divisible by 3. So, 3A – 1 = 500, whence A = 167, B = 334.
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Compare with the tasks 107749 and 107760.

Answer

167 and 334.

Hint

Hint

The least intelligent girl should be pretty beautiful, and the least attractive should be quite smart.

Solution

Let assume that the ball was attended by three girls – Anna, Catherine and Natasha in order of increasing beauty. In this case, the order of increasing intelligence is: Catherine – Natasha – Anna $($ Anna being the most intelligent $)$. Then the young man dancing with Anna could have invited Catherine $($ who was more beautiful $)$ to dance the next dance with, then the one dancing with Natasha can next invite Anna to dance $($ who is smarter$)$, the one dancing with Catherine then asks Natasha $($ who is both more beautiful and more intelligent $)$.
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See also problem number 98238.

Answer

Yes, it could be possible.

Hint

The least intelligent girl should be pretty beautiful, and the least attractive should be quite smart.

Solution

Note first that you can not get an even number on the machine. Consequently, Peter will not be able to get the number 1982.

Lemma. After the first multiplication by 3, it is unprofitable to add 4 more than twice in a row. Proof of the lemma. By replacing the sequence of actions $ \times 3$, + 4, + 4, + 4 with +4,$ \times 3$, we will reduce costs. Now it is possible to restore the sequence of Peter’s actions starting from the last one: if the number is not divisible by 3, then it was obtained by adding a 4, and if it is divisible, then it was obtained either by a multiplication by 3 or only four additions were used. Thus, we obtain the sequence of actions optimal for Peter and it has the form: +4, + 4, + 4, + 4, + 4, × 3, + 4, + 4, × 3, + 4, × 3, + 4, + 4, × 3, + 4

Hint

Solution

We divide the weights into pairs in an arbitrary way. We put the first pair on different cups of the weighing scales arbitrarily. Then we will consistently place pairs of weights, putting each time the heavier weight on the lighter cup. Suppose that the weights a and a + α lie on the plates, where 0 $≤$ α $≤$ 20, and we put the weights b + β and b, where 0 $≤$ β $≤$ 20. Then the difference in the weights of the loads lying on the cups is | α – β | $≤$ 20.

Hint

Solution

In the first move, the rook stands on an arbitrary cell on the middle line running along the board. Then it moves diagonally each time away from the horse so that the horse is forced to move in the same direction.

Hint

Solution

a$)$ We take an arbitrary number ending in zeros, the square of which is a 12-digit number starting with 5, for example A = 750000 $(A^2 = 5625 \times 10^8)$. Since $(A – 1)^2$ = $A^2$ – 2A + 1, the preceding square is 562500000000 – $2 \times 750000$ + 1 = 562498500001. Therefore, there is no 12-digit square starting with the digits 562499.
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b$)$ Let $A^2$ be a 12-digit number starting with 1, that is, 1011 $≤$ $A^2$ $<$ $2 \times 10^11$. Neighbouring squares differ from $A^2$ by 2A – 1 and 2A + 1, and therefore the difference between adjacent squares is less than one million, so that in the series of consecutive squares between $10^11$ and $2 \times 10^11$ there will be numbers beginning with each set of 6 digits.
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c$)$ You can assign n + 1 digits to any n digit number so that the resulting number which has 2n + 1 digits is a complete square. Indeed, in the section from $10^{2n}$ to $10^{2n + 1}$ the difference between neighboring squares does not exceed
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Let us prove that k $(n)$ $≥$ n + 1. Consider the square preceded by $10^{2n}$:
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Thus, one can not assign n digits to a number so that a square is obtained. It is also clear that n – 2, n – 4, … numbers can not be assigned to it: the corresponding number is too close to $(10^{n-1})^2$, $(10^{n-2})^2$, … Therefore k $(n)$ ≠ n $($ and also n – 2, n – 4, …$)$.
$\\$ We also consider the square preceded by $(2 \times 10^{n-1})^2$:
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Thus, we can not assign to a number n – 1 $($ and also n – 3, n – 5, …$)$ digits to get a square, so that k $(n)$ ≠ n – 1 $(n – 3, n – 5, …)$.

Answer

a$)$ Not to any number; b$)$ to any number; c$)$ k $(n)$ = n + 1.

Hint

Solution

The thief is Brown. If we consider three possibilities – “Jones being the thief”, “Brown being the thief” and “Smith being the thief”, then we will see that only in the case where Brown is the thief are the conditions of the task fulfilled. If the thief is Smith, then both Brown and Jones told the truth. If the thief is Jones, then both Brown and Smith told one truth and one lie.

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

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

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.

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?

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

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!
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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

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

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

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

Nothing to translate

Hint

Solution

Evaluation. Suppose that among the rational numbers there is a 0 and it is written on the top row of the table. Suppose that x irrational and 50 – x rational numbers are written along the left side of the table. Then along the upper side there are written 50 – x irrational and x rational numbers. We note that the product of a nonzero rational number and an irrational number is irrational. Hence, in the table there are at least x $(x – 1)$ + $(50 – x)^2$ = $2x^2$ – 101x + $50^2$ irrational numbers. The parabola’s vertex y = $2x^2 – 101x + 50^2$ is at the point 25,25, therefore the minimum value of this function at an integer point is reached at x = 25 and is equal to $25 \times 24 + 25^2 = 1225$.
If 0 is replaced by a non-zero rational number, then the number of irrational numbers can only increase. Therefore in the table in any case there are no more than 2500 – 1225 = 1275 rational numbers.

Example, when the rational number is 1275. Along the left side, we put the numbers 1, 2, …, 24, 25, $\sqrt{2}, 2\sqrt{2}, …, 25 \sqrt{2}$, and along the upper side we place the numbers 0, 26, 27, …, 49, $26\sqrt{2}, 27\sqrt{2}, …, 50 \sqrt{2}$. Then only $25 \times 24$ + $25^2$ will be irrational = 1225 products of nonzero rational and irrational numbers.

Answer

1275 products.

Hint

Solution

Let f $(x)$ = 1/2 $ax^2$ + bx + c.These values have different signs, so one of the roots of the trinomial is located between $x_1$ and $x_2$.

Hint