Solution

Add n to both parts. Let us prove that in the resulting equality both sides are equal to the number of pairs $(k, m)$ of natural numbers satisfying the inequality $k^m$ $≤$ n. Indeed, the integer part of the number a $≥$ 1 is equal to the number of natural numbers less than or equal to a. therefore$\\$

[$n^{1 / m}$] is the number of pairs $(k, m)$ for which $k^m$ $≤$ n,$\\$

[$log_{k}$n] is the number of pairs $(k, m)$ for which $k^m$ $≤$ n. $\\$

Points: 15.

Hint

Solution

For n = 1, the inequality becomes 0 = 0. For n$>$ 1, we prove that the sum of the fractional parts on each interval between two successive squares satisfies the inequality
$\\$ $\\$
From the inequality between the arithmetic mean and the mean square, we obtain for 0 ≤ a ≤ m.
$\\$ $\\$
Consequently,
$\\$ $\\$
Summing these inequalities for a = 0, 1, …, m-1 and the inequality $($obtained by dividing by 2 the two parts of $($2$)$ for a = m$)$, we arrive at inequality $($1$)$. Summing the inequality $($1$)$ over all m from 1 to n – 1, we obtain <
$\\$ $\\$
It remains to note that

Answer

See the solution above.

Hint

Solution

The sum of the ages of Alex, Beatrice, Victor, Gregory and Deborah should be equal to the sum of the ages of V, W, X, Y and Z. This means that Z is older than Deborah by $1+2+3+4=10$ years.

Answer

Z is older than Deborah by 10 years.

Hint

Solution

First, assume that $n$ and $m$ have different parities (one is odd and the other is even). For simplicity assume there is an odd number of columns and even number of rows. Since there are odd number of trees in each column and each row, if we sum all the trees column by column we will add together an odd number of odd numbers and we will get an odd number. However, if we added the trees in all the rows, we would get an even number, because we would add an even number of odd numbers together. Since these are just two ways of counting all the trees, we cannot get two different results, so it is impossible in that case. $\\$
However, if $n$ and $m$ are both even, it should be possible. Let us show how. $\\$
Let us remove the last column and the last row from the garden. In other words, we remove the last square of every row and the last square of every column. Now, place the trees randomly in the remaining squares, without thinking about parity or columns or rows. $\\$
Let us restore the last row and the last column now. Then, we can notice that for each row, we can find out if the tree should be planted in the last square of it or not basing in the number of trees already planted. If there is an odd number of trees in this particular row, we do not plant a tree in the last square, and if there is an even number of trees in that row, we do plant a tree in the last cell. We can repeat the same for the columns. $\\$
Now the question is – will the last, corner square agree. Is there a tree there or not? $\\$
We should follow the same strategy – if there is already an odd number of trees in the last row (without the corner), we do not plant the tree in the corner. If there is an even number of trees in the last row, we do plant the tree. However, the same should be true for the last column, but the last cell of the last column is also the corner – the same corner. $\\$
We can get two different answers and we should show that we won’t. To demonstrate that, we just need to show that the total number of trees planted in the last row without the corner has the same parity as the total number of trees planted in the last column without the corner. $\\$
A tree is planted somewhere in the last column if there is an even number of trees in its row. That means that the number of trees in the last column is equal to the number of rows in the smaller $(n-1) \times (m -1)$ garden (we created it when we removed the last row and the last column) with even number of trees. In the same way, the number of trees in the last row is equal to the number of columns in the smaller garden with even number of trees.

Let’s say there are $x$ columns with even number of trees and $y$ rows with even number of trees in the smaller garden. That means there are $(n-x-1)$ columns with odd number of trees and $(m-y-1)$ rows with an odd number of trees. Now, let’s count all the trees in the smaller garden in two ways, or rather, determine if there is an odd or even number of them. If we count by columns, we add $x$ even numbers and $(n-x-1)$ odd numbers. So there is a total odd number of trees only if $(n-x-1)$ is odd. Counting by rows, we add $y$ even numbers and $(m-y-1)$ odd numbers, so the total number of trees is od only if $(m-y-1)$ is odd. Therefore, the numbers $(m-y-1)$ and $(n-x-1)$ are either both even or both odd, and because the same is true about $m$ and $n$, we see that $y$ and $x$ have to be both even or both odd.
$\\$
Thus it is possible to plant trees like this. To find out the number of ways we can do it in, notice that the smaller garden can be planted in any way we like, and the rest just follows. For each square, we can choose whether we want to plant the tree there or not, so we have two possiblities. There are $(n-1) \times (m-1)$ squares, so there are $2^{(n-1) \times (m-1)}$ ways to do it.

Hint

Solution

Let us colour the chessboard normally and just look at the white diagonals. There are $15$ of them, $8$ in one direction and $7$ in the other direction. If there is an odd number of pieces on each diagonal, then if we add all the numbers of pieces from all the $15$ white diagonals, we would get an odd number, because $15$ is itself odd. However, that way we counted each of the fields of the chessboard exactly twice. Therefore we must have counted each of the pieces twice. Yet, we are getting an odd number. This is impossible and proves that such an arrangement is not possible.

Hint

Solution

Let us call these numbers $a_1, a_2, a_3, a_4, a_5, a_6$ and $a_7$. We know that each of the sums $a_1 + a_2 +a_3+a_4+a_5+a_6$, $a_1 + a_2 +a_3+a_4+a_5+a_7$, etc. up to $ a_2 +a_3+a_4+a_5+a_6+a_7$ is a multiple of $5$. There are $7$ such sums in total, one for each number “missing” from a sum. If we add all the seven of them together, we add each of the numbers $6$ times, because each of the numbers is only not present in one of the sums. Therefore the sum of the sums is $6 \times ( a_1 + a_2 +a_3+a_4+a_5+a_6+a_7)$. This is a sum of $7$ quantities, each of which was divisible by $5$, so itself it also must be divisible by $5$. But because $6$ and $5$ are respectively prime, it implies that the sum in the brackets, $a_1 + a_2 +a_3+a_4+a_5+a_6+a_7$, is divisible by $5$. However, if we just subtract any of the sums of $6$ numbers from this one, we will get a single number. Thus each of these numbers is a difference between two numbers divisible by $5$, so itself must be also divisible by $5$.

Hint

Solution

First, we can observe that there were no more that $3$ films picked. Otherwise we would have some $4$ judges voting each on a different film. So there were $3$ or fewer films picked by all the judges. Now suppose each movie was chosen by at most $3$ judges. That gives $3 \times 3 = 9$ judges in total at maximum. But we know there were $10$ of them, one more.

Hint

Solution

We can divide the cake into $16$ little $2 \times 2$ pieces. Each of these pieces should have $2$ roses on it. That means the total number of roses is $32$. However, if we can also cut the cake into $20$ $3 \times 1$ pieces and one $2 \times 2$ piece (see the picture). Each of the rectangular pieces has one rose on it, and the leftover square one has two. In total, there are $22$ roses, $10$ fewer than before. Thus, it is not possible.

Hint

Solution

All we need to do is notice that if we add together the numbers in the central column, central row and two diagonals, we will get a total of $4 \times 15 = 60$, because each column, row, and diagonal sums up to $15$. Adding those numbers, we see that we added each of the numbers only once, except the central number, which was included in every number triple, so we added it $4$ times. Therefore we added all the numbers in the square and the central number three additional times. $\\$
We can see that the sum of all the numbers in the magic square is $45$ (by adding three columns for example). Therefore the triple the number in the centre is $60-45=15$, so the number itself is indeed $5$.

Hint

Solution

a) Each child played $10$ rounds, but each round was played by two children. If we add the rounds together $10+10+10 = 30$ we will count each round twice, once from the perspective of one child, once from the perspective of the other. Thus the total number of rounds is a half of that, $15$. $\\$
b) If Lorenzo played more rounds with Isabella than with Francesca, he must’ve played more than a half of all his games with Isabella. That is, he played more than $5$ games with Isabella. That means Isabella played more than $5$ games with Lorenzo, which is also more than a half of all the games she played. However, that implies that Francesca played fewer than $5$ games with Lorenzo (since he played $10$ in total) and fewer than $5$ games with Isabella (for the same reason), which means she played fewer than $5+5=10$ games in total, which is not true, because we know she played exactly $10$. Therefore it is not possible that Lorenzo played more rounds with Isabella than with Francesca.

Hint

Solution

Counting the daggers row by row we must get a number that is larger than $6 \times 50 = 300$, because there are $6$ rows and there are more than $50$ daggers in each. However, counting the daggers by column gives that there must be fewer than $6 \times 50 = 300$ of them in total. Since it is impossible that there are simultaneously fewer than $300$ and more than $300$ daggers, the answer to this question is negative.

Hint

Solution

What we need to count here is the number of handshakes – clearly not all Venusians shook hands with all the Martians, so we can’t count the actual number of participants in two ways, but the number of handshakes is good enough. $\\$
Each fo the $24$ Martians made $6$ handshakes, so there were $24 \times 6 = 144$ handshakes. If there are $n$ Venusians, the total number of handshakes can also be calculated by multiplying $n$ by the number of handshakes each of the Venusians made, which is $8$. Therefore $8k = 144$, because the number of handshakes does not depend on the way we count it. After division we learn that $k = 18$, there were $18$ delegates from Venus at the meeting.

Hint

Solution

The bank’s balance is just a number – positive or negative. The sum of balances of two or more banks is the total balance of these banks. $\\$
There are $5$ triples of directors sitting next to each other. If we added all of them, we would get a positive number, because each of the triples has a positive balance. Now, how does that relate to the total balance of all the banks? To find it, we need to add all the balances of all banks. By adding the triples, however, we already added all the balances. Each of them was added three times to the sum, because each bank belonges to exactly three different triples. Therefore, the total balance of the banks is the sum of the balances of the triples divided by $3$. Which still must be a positive number. Thus the total balance of all the banks is positive.

Hint

Solution

There are $4$ roads in total, each of them goes through the capital. Besides the capital, each other fortress is on only one of the roads. One road has a total of $130$ cannons and the other three have $80$ cannons each. If we just sum it up: $130 + 80 + 80 + 80 = 370$ cannons, we would count the cannons in the capital $4$ times, while only counting the cannons in each of the other cities once. Thus the difference between our count and the actual total number of cannons is triple the number of cannons in the Grand Fortress, because we counted them three times too many (of the total four). The difference is $370-280 = 90$, so there are $90 \div 3 = 30$ cannons in the capital.

Hint

Solution

Let us count the total number of orcs in the formation. There are $4$ rows and the number of orcs in each of them is $7$. That means there are $4 \times 7 = 28$ orcs in total. Now suppose there is the same number of orcs in every column. Let us say this number is $k$. There are $10$ columns, so the total number of orcs is $k \times 10 = 10k$. We know that $10k = 28$, what gives $k = 2 \frac45$, which is not an integer. Thus it is impossible, because we cannot put $2 \frac45$ of an orc in each column.

Hint

Hint

Notice, nothing is said about whether musketeers are galloping in one or two directions.

Solution

The musketeers could go:

a) in different directions, towards each other;

b) in different directions, moving away from each other;

c) in one direction – Athos after Aramis;

d) in one direction – Aramis after Athos.

Accordingly, the problem leads to four different answers. For example, in case a) in an hour the distance will decrease by 9 leagues and leaving 11 leagues. The remaining answers can also be found.

Answer

11, 29, 21 or 19 leagues.

Hint

Notice, nothing is said about whether musketeers are galloping in one or two directions.

Hint

Try to calculate the sum of all of the numbers in the table.

Solution

If this were possible, the sum of all of the numbers in the table, counted “by rows,” would be positive, and “by columns” – negative, which is impossible.

Answer

No, it is not possible.

Hint

Try to calculate the sum of all of the numbers in the table.

Solution

Pick 5 students, one from each year taking part. Since the number of problems each created is different, these pupils between them created no fewer than $1 + 2 + 3 + 4 + 5=15$ problems. Hence the remaining 25 pupils between them created no more than $40 – 15 = 25$ problems. That is, each one must created exactly one problem. Therefore a total of 26 pupils created exactly one problem.

Such a scenario is indeed possible – for example where all the year 7’s created 1 problem, all the year 8’s – 2 problems and so on, and exactly 26 year 7’s took part, with one pupil from each of the other four years.

Answer

26 pupils.

Hint

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Solution

We find the sum of all the numbers of the table in two ways. Since the table contains M rows and the sum of the numbers of each row is 1, the whole sum is M. Similarly, considering the same sum over the columns, we get the result of K.

Answer

Source of the solution: V.O. Bugaenko “Lomonosov Tournament Competitions in Mathematics”. MCNMO-CheRo. 1998.

Hint

–

Solution

Solution 1

Amongst the given numbers there must be at least one positive number as if all 25 numbers were negative then their sum would also be negative which would contradict the conditions given. In a similar way we can prove that no fewer than 22 of the numbers in the group must be positive – but for the proof we need only one such number. Take this number and place it to one side. Consider the 24 remaining numbers. Split these into groups of 4 at random. All of these groups must be positive by definition. The sum of all 25 numbers is equal to the sum of all of the groups and the single known positive number. Therefore, since the sums of all 4 groups and the positive term is positive, the sum of all 25 numbers must also be positive.

Solution 2

If the sum of any 4 of the numbers is positive, then the sum of any 24 numbers must also be positive. There are only 25 combinations of 24 numbers. If we sum all 25 combinations, we get $24 \times$ the sum of all 25 numbers which is greater than 0 – each number is excluded from the group of 24 only once, so occurs in the sum 24 times. It follows therefore that the sum itself is also greater than 0.

Hint

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