As shown in the figure, the radius of the circle is equal to the height of the equilateral triangle $ABC$, this circle rolls along the base of the triangle while remaining tangent, with the moving point of tangency denoted as $T$, the circle intersects $AC, BC$ at $M, N$, let $n$ be the degree measure of $\angle MTN$, then for all possible positions of the circle, $n$
(A) varies from $30^{\circ}$ to $90^{\circ}$.
(B) varies from $30^{\circ}$ to $60^{\circ}$.
(C) varies from $60^{\circ}$ to $90^{\circ}$.
(D) remains $30^{\circ}$ unchanged.
(E) remains $60^{\circ}$ unchanged.

Solution:
[Solution] The figure shows one position of the circle: the center is at vertex $C$ of the equilateral $\triangle A B C$, at this time
the measure of $\angle M T N$ is $60^{\circ}$.
The right figure shows another possible position of the circle: the center is on a line parallel to the base $A B$ of the equilateral $\triangle A B C$.

Extend $N C$ to intersect $\odot O$ at $D$, and connect $C M, D M$. Then the diameter through $C$ bisects the vertex angle $\angle C$ of the isosceles $\triangle C D M$.
$$
\text { Also, } \begin{aligned}
\angle C & =180^{\circ}-\angle M C N \\
& =180^{\circ}-60^{\circ}=120^{\circ},
\end{aligned}
$$

we know $\angle C D M=30^{\circ}$. Thus, the measure of $\overparen{M N}$ is $60^{\circ}$. Therefore, the answer is $(E)$.

Answer: E


The area of rectangle $ABCD$ is 2011 square centimeters. The vertex $F$ of trapezoid $AFGE$ is on $BC$, and $D$ is the midpoint of the non-parallel side $EG$. Find the area of trapezoid $AFGE$.

Solution:
【Analysis】According to the problem, connect $D F$. Triangle $A D F$ and rectangle $A B C D$ have the same base and height, so the area of triangle $A D F$ is half the area of rectangle $A B C D$. Since point $D$ is the midpoint of $E G$ and $A E$ is parallel to $F G$, triangle $A D F$ is also half the area of trapezoid $A F G E$. Because point $D$ is the midpoint of line segment $E G$, the areas of triangle $A D E$ and triangle $D G F$ are half the area of trapezoid $A F G E$, meaning the area of the trapezoid is equal to the area of the rectangle. Solve accordingly.

【Solution】Solution: As shown in the figure, connect $D F$.
Triangle $A D F = 2011 \div 2 = 1005.5$ (square centimeters),
Since point $D$ is the midpoint of $E G$,
Triangle $A E D +$ Triangle $D F G = 1005.5$ (square centimeters),
The area of trapezoid $A F G E$: $1005.5 + 1005.5 = 2011$ (square centimeters),
Answer: The area of trapezoid $A F G E$ is 2011 square centimeters.

Answer: 2011


Odredi sve prirodne brojeve $n$ za koje postoje razliÄiti djelitelji $a$ i $b$ od $n$ takvi da izmeÄu njih nema drugih djelitelja od $n$ i da vrijedi

$$
n=a^{2}-b
$$

Solution:
## Drugo rješenje.

Kao i u prvom rješenju dobijemo da je $b=m a$ za neki prirodan broj $m>1$. Jednadžba postaje

$$
n=a(a-m) \text {. }
$$

Pretpostavimo da je $n$ neparan. Tada su $a$ i $b$ također neparni kao njegovi djelitelji, pa je $a^{2}-b$ paran, kontradikcija. Dakle, $n$ je paran.

Ako je $a$ neparan, onda je $2 a$ djelitelj od $n$, ali $n$ nema djelitelja između $a$ i $m a$, pa zaključujemo da mora biti $m=2$ i $n=a(a-2)$. Međutim, onda je desna strana neparna, kontradikcija. Dakle, $a$ je paran.

Kako je $a$ paran, onda je i $b=m a$ paran, pa je $b / 2=m a / 2$ djelitelj od $n$. Ali $m a / 2$ ne smije biti između $a$ i $m a$, pa zaključujemo da mora vrijediti $m=2$, i $n=a(a-2)$, te je $a$ paran.

Ali onda je $2(a-2)$ djelitelj od $n$. Kako je $2(a-2)<2 a=b$, mora biti $2(a-2) \leqslant a$, odnosno $a \leqslant 4$. Preostaje provjeriti slučajeve $a=2$ i $a=4$.

Ako je $a=2$, onda je $b=2 a=4$ i $n=0$, kontradikcija.

Ako je $a=4$, onda je $b=8$ i dobivamo $n=8$, što je stvarno rješenje jer su 4 i 8 očito uzastopni djelitelji od 8 .

Answer: 8


Let $S=$ $\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$, satisfying the conditions:
(1) $\left|A_{i}\right|=5(i=1,2, \cdots, k)$;
(2) $\left|A_{i} \cap A_{j}\right| \leqslant 2(1 \leqslant i<j \leqslant k)$.
Find the maximum value of $k$.

Solution:
Thought analysis: We should derive an upper bound for $K$ based on the conditions that the subsets must satisfy. Let's make the following exploration: Let $A_{1}=\{1,2,3,4,5\}$, and the intersection of any two subsets has at most 2 common elements. Let $A_{2}=\{1,2,6,7,8\}$. At this point, if $\{1,2\} \subseteq A_{3}$, then we must have $|A_{1} \cap A_{3}| \geqslant 3$ or $|A_{2} \cap A_{3}| \geqslant 3$, which is a contradiction. This leads to the first conjecture: Among $A_{1}, A_{2}, \cdots, A_{k}$, at most two subsets contain the same 2-element subset. Let $A_{3}=\{1,3,6,9,10\}$. At this point, if $1 \in A_{4}$, then among $A_{1}, A_{2}, A_{3}$, except for 1, there is at most one other element that is also in $A_{4}$, and $|A_{4}| \leqslant 4$, which is a contradiction. This leads to the second conjecture: Each element in $S$ belongs to at most 3 of the subsets $A_{1}, A_{2}, \cdots, A_{k}$. At this point, we can find an upper bound for $K$.

Solution: (1) For any 2-element subset $\{a, b\} \subseteq S$, at most two of the subsets $A_{1}, A_{2}, \cdots, A_{k}$ contain $\{a, b\}$. Otherwise, suppose there are $A_{1} \supseteq \{a, b\}, A_{2} \supseteq \{a, b\}, A_{3} \supseteq \{a, b\}$. Since $|A_{i} \cap A_{j}| \leqslant 2 (1 \leqslant i < j \leqslant k)$, the sets $A_{i} - \{a, b\}$ (where $A - B = \{x \mid x \in A$ and $x \notin B\}$) are pairwise disjoint $(i=1,2,3)$. But $|A_{i} - \{a, b\}| = 3 (i=1,2,3)$, so $|S| \geqslant 3 \times 3 + 2 = 11$, which is a contradiction.
(2) For any element $a \in S$, at most three of the subsets $A_{1}, A_{2}, \cdots, A_{k}$ contain $\{a\}$. Suppose there are $A_{1} \supseteq \{a\}, A_{2} \supseteq \{a\}, A_{3} \supseteq \{a\}$. If $|A_{i} \cap A_{j}| = 1 (1 \leqslant i < j \leqslant 3)$, then $|A_{1} \cup A_{2} \cup A_{3}| = 3 \times 4 + 1 = 13$, which is a contradiction. Suppose $b \neq a$ and $a, b \in A_{1} \cap A_{2}$, then $|A_{1} \cup A_{2}| = 8$. According to (1), $b \notin A_{3}$ and $|A_{3} \cap A_{1}| \leqslant 2, |A_{3} \cap A_{2}| \leqslant 2$, so $A_{3}$ has at most 2 elements not in $A_{1} \cup A_{2}$, i.e., $\overline{A_{1} \cup A_{2}} \subset A_{3}$.

Similarly, if there is $A_{4} \supseteq \{a\}$, then $\overline{A_{1} \cup A_{2}} \subseteq A_{4}$, so $|A_{3} \cap A_{4}| \geqslant 3$, which is a contradiction.

Therefore, for any $a \in S$, $a$ belongs to at most 3 of the subsets $A_{1}, A_{2}, \cdots, A_{k}$. So each element in $S$ is counted at most 3 times in $|A_{1}| + |A_{2}| + \cdots + |A_{k}|$, hence $|A_{1}| + |A_{2}| + \cdots + |A_{k}| \leqslant 3|S| = 30$, so $K \leqslant 6$.

According to the above idea, we can construct 6 subsets that meet the conditions:
$$
\begin{array}{l}
\{1,2,3,4,5\}, \{1,2,6,7,8\}, \{1,3,6,9, \\
10\}, \{2,4,7,9,10\}, \{3,5,7,8,10\}, \{4,5,6,8, \\
9\}.
\end{array}
$$

Therefore, the maximum value of $K$ is 6.

Answer: 6


The probability that a purchased light bulb will work is 0.95. How many light bulbs need to be bought to ensure that with a probability of 0.99, there are at least 5 working ones among them?

Solution:
# Solution.

Let's take 6 light bulbs. The probability that at least 5 of them will work is the sum of the probabilities that 5 will work and 1 will not, and that all 6 will work, i.e.,

$$
C_{6}^{5} \cdot 0.95^{5} \cdot 0.05 + C_{6}^{6} \cdot 0.95^{6} = 6 \cdot 0.95^{5} \cdot 0.05 + 0.95^{6} = 0.9672
$$

Let's take 7 light bulbs. The desired probability is

$$
\begin{aligned}
& C_{7}^{5} \cdot 0.95^{5} \cdot 0.05^{2} + C_{7}^{6} \cdot 0.95^{6} \cdot 0.05 + C_{7}^{7} \cdot 0.95^{7} = \\
= & 21 \cdot 0.95^{5} \cdot 0.05^{2} + 7 \cdot 0.95^{6} \cdot 0.05 + 0.95^{7} = 0.9962
\end{aligned}
$$

The minimum number of light bulbs required is 7.

Answer: 7.

Answer: 7


What is the maximum number of parts into which the plane can be divided by:

a) two triangles;

b) two rectangles;

c) two convex $n$-gons?

Solution:
17. Let's immediately provide the solution to problem (v), namely, we will prove that the maximum number of parts into which two convex $n$-gons can divide the plane is $2n+2$; from this it will follow that two triangles can divide the plane into no more than 8 parts (problem (a)); see Fig. 54, a), and two rectangles can divide the plane into no more than 10 parts (problem (b)); see Fig. 54, b).

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-084.jpg?height=227&width=302&top_left_y=254&top_left_x=269)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-084.jpg?height=246&width=366&top_left_y=237&top_left_x=609)

b)

Fig. 54.

The parts into which the plane is divided by two convex $n$-gons can be characterized as follows:

1) (one) part located inside both polygons (if it exists);
2) (one) part located outside both polygons;
3) parts located inside one polygon and outside the other.

The problem, therefore, is to determine the maximum possible number of parts of type 3.

Each such part of the plane is a polygon, the smallest number of sides of which is three. On the other hand, each side of one polygon, as a result of intersection with the second (convex!) polygon, can be divided into no more than three parts (Fig. 55, a); therefore, the total number of sides

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-084.jpg?height=390&width=403&top_left_y=1004&top_left_x=245)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-084.jpg?height=374&width=365&top_left_y=1010&top_left_x=628)

b)

Fig. 55.

of all parts of type 3 does not exceed $3n + 3n$. But no segment can be the boundary of two parts of the plane of type 3 (because if on one side of it lies a part of the plane that is external to one polygon and internal to the second, then on the other side there must lie a part of the plane that is either external or internal to both polygons, i.e., a part of the plane of type 1 or 2); therefore, the number of parts of type 3 does not exceed
$\frac{6n}{3} = 2n$. Thus, the total number of parts into which two convex $n$-gons divide the plane does not exceed

$$
1 + 2n + 1 = 2n + 2
$$

The fact that the number of parts can equal $2n + 2$ is demonstrated by the example of two regular $n$-gons with a common center, one of which is obtained from the other by a rotation through an angle not a multiple of $360^\circ / n$ (Fig. 55, b).

Answer: 2n+2


Find all pairs ( $p, q$ ) of prime numbers for which the numbers $2 p+q, p+2 q$ and $p+q-18$ are all three prime numbers.

We recall that a prime number is an integer greater than or equal to 2, which is divisible only by 1 and itself.

Solution:
Solution to Exercise 9 Let $\mathrm{p}$ and $\mathrm{q}$ be two prime numbers, if they exist, that satisfy the given conditions.

Since $p \geqslant 2$ and $q \geqslant 2$, we have $2 p+q \geqslant 6$ and $p+2 q \geqslant 6$. As these are prime numbers, it follows that $2 p+q$ and $p+2 q$ are odd, and thus $p$ and $q$ are odd. However, the number $\mathrm{p}+\mathrm{q}-18$ is an even prime number, hence $\mathrm{p}+\mathrm{q}-18=2$.

We are thus led to find the prime numbers $\mathrm{p}$ and $\mathrm{q}$ such that $\mathrm{p}+\mathrm{q}=20$. A direct and "by hand" exploration shows that the pair $(p, q)$ is one of the pairs $(3,17),(7,13),(13,7),(17,3)$.

If we now return to the condition that $2 p+q$ and $p+2 q$ are prime, we observe that only the cases $p=3, q=17$ and $p=17, q=3$ actually correspond to solutions of the problem posed.

Answer: (3,17),(17,3)


There are 9 different pastries and drinks placed at 9 positions around a round table. 6 gentlemen and 3 ladies are having breakfast together. The number of ways for the 3 ladies to sit such that no two of them are adjacent is $\qquad$ kinds.

Solution: 8. 129600 Explanation: Starting from a certain position at a round table, number 9 positions as 1, 2, ..., 9, which is a typical linear arrangement, meaning to arrange 9 people in 9 different positions in order. However, the problem also requires that the ladies do not sit together. First, arrange the gentlemen in a row from left to right, which has $A_{6}^{6}$ ways. Then, place the 3 ladies at the head, tail, or between two gentlemen, which has $A_{7}^{3}$ ways. However, if the head and tail positions are both occupied by ladies, it may result in two ladies sitting next to each other, which should be excluded. Since two ladies at the head and tail have $A_{3}^{2}$ ways, and the other lady can be inserted between two gentlemen in $A_{5}^{1}$ ways, the number of such cases to be excluded is $A_{3}^{2} A_{5}^{1}$. Therefore, the total number of ways is $A_{6}^{6} \left(A_{7}^{3} - A_{3}^{2} A_{5}^{1}\right) = 129600$.

Answer: 129600


Find all such triples of natural numbers: the product of any two numbers plus 1 is divisible by twice the third number.

Solution: 10. There is only one such 3-tuple $(1,1,1)$. Hint: If the 3-tuple of natural numbers $(a, b, c)$ satisfies the condition, then $a, b, c$ should be coprime and all odd. Since $a b+b c+c a+1$ should be divisible by each of $a, b, c$, it should therefore be divisible by their product $a b c$.

Answer: (1,1,1)


We define the weight $W$ of a positive integer as follows: $W(1) = 0$, $W(2) = 1$, $W(p) = 1 + W(p + 1)$ for every odd prime $p$, $W(c) = 1 + W(d)$ for every composite $c$, where $d$ is the greatest proper factor of $c$. Compute the greatest possible weight of a positive integer less than 100.

Solution:
To solve the problem, we need to compute the weight \( W(n) \) for positive integers \( n \) and determine the greatest possible weight for \( n < 100 \). The weight function \( W \) is defined recursively as follows:
- \( W(1) = 0 \)
- \( W(2) = 1 \)
- \( W(p) = 1 + W(p + 1) \) for every odd prime \( p \)
- \( W(c) = 1 + W(d) \) for every composite \( c \), where \( d \) is the greatest proper factor of \( c \)

Let's compute the weights for the first few integers and then use these values to find the greatest possible weight for \( n < 100 \).

1. **Compute the weights for the first 13 integers:**
   \[
   \begin{array}{r|rrrrrrrrrrrrr}
   n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\
   \hline
   W(n) & 0 & 1 & 3 & 2 & 5 & 4 & 4 & 3 & 4 & 6 & 6 & 5 & 6
   \end{array}
   \]

2. **Verify the weights for \( n \leq 13 \):**
   - \( W(1) = 0 \)
   - \( W(2) = 1 \)
   - \( W(3) = 1 + W(4) = 1 + 2 = 3 \)
   - \( W(4) = 1 + W(2) = 1 + 1 = 2 \)
   - \( W(5) = 1 + W(6) = 1 + 4 = 5 \)
   - \( W(6) = 1 + W(3) = 1 + 3 = 4 \)
   - \( W(7) = 1 + W(8) = 1 + 3 = 4 \)
   - \( W(8) = 1 + W(4) = 1 + 2 = 3 \)
   - \( W(9) = 1 + W(3) = 1 + 3 = 4 \)
   - \( W(10) = 1 + W(5) = 1 + 5 = 6 \)
   - \( W(11) = 1 + W(12) = 1 + 5 = 6 \)
   - \( W(12) = 1 + W(6) = 1 + 4 = 5 \)
   - \( W(13) = 1 + W(14) = 1 + 5 = 6 \)

3. **Extend the computation to \( n \leq 26 \):**
   - If \( n \leq 13 \), then \( W(n) \leq 6 \).
   - If \( n \) is composite, its greatest proper factor \( d \leq \frac{n}{2} \leq 13 \), so \( W(n) = 1 + W(d) \leq 1 + 6 = 7 \).
   - If \( n \) is an odd prime, \( W(n) = 1 + W(n + 1) = 1 + 1 + W(\frac{n + 1}{2}) \leq 1 + 1 + 6 = 8 \).

4. **Extend the computation to \( n \leq 52 \):**
   - If \( n \leq 26 \), then \( W(n) \leq 8 \).
   - If \( n \) is composite, its greatest proper factor \( d \leq \frac{n}{2} \leq 26 \), so \( W(n) = 1 + W(d) \leq 1 + 8 = 9 \).
   - If \( n \) is an odd prime, \( W(n) = 1 + W(n + 1) = 1 + 1 + W(\frac{n + 1}{2}) \leq 1 + 1 + 8 = 10 \).

5. **Extend the computation to \( n \leq 104 \):**
   - If \( n \leq 52 \), then \( W(n) \leq 10 \).
   - If \( n \) is composite, its greatest proper factor \( d \leq \frac{n}{2} \leq 52 \), so \( W(n) = 1 + W(d) \leq 1 + 10 = 11 \).
   - If \( n \) is an odd prime, \( W(n) = 1 + W(n + 1) = 1 + 1 + W(\frac{n + 1}{2}) \leq 1 + 1 + 10 = 12 \).

6. **Find a positive integer less than 100 with weight 12:**
   - Consider the sequence of primes: \( 19 \), \( 37 \), and \( 73 \).
   - \( W(73) = 1 + W(74) = 1 + 1 + W(37) = 2 + W(37) \)
   - \( W(37) = 1 + W(38) = 1 + 1 + W(19) = 2 + W(19) \)
   - \( W(19) = 1 + W(20) = 1 + 1 + W(10) = 2 + W(10) \)
   - \( W(10) = 6 \)
   - Therefore, \( W(19) = 2 + 6 = 8 \)
   - \( W(37) = 2 + 8 = 10 \)
   - \( W(73) = 2 + 10 = 12 \)

Thus, the greatest possible weight of a positive integer less than 100 is \( \boxed{12} \).

Answer: 12


Given $O$ is the circumcenter of $\triangle ABC$, $AD$ is the altitude on $BC$, $\angle CAB=66^{\circ}, \angle ABC=44^{\circ}$. Then, $\angle OAD=$ $\qquad$ .

Solution:
7. As shown in Figure 3, we have
$$
\begin{array}{l}
\angle O A D=90^{\circ}-\angle A E F \\
=90^{\circ}-(\angle A B C+\angle C B F)=90^{\circ}-\angle C A D-\angle A B C \\
=\angle A C B-\angle A B C=180^{\circ}-2 \angle A B C-\angle C A B=26^{\circ}
\end{array}
$$

Answer: 26^{\circ}


Represent the number 1000 as the sum of the maximum possible number of natural numbers, the sums of the digits of which are pairwise distinct.

Solution:
Answer: 19.

Solution: Note that the smallest natural number with the sum of digits A is a99..99, where the first digit is the remainder, and the number of nines in the record is the incomplete quotient of the division of A by 9. From this, it follows that if A is less than B, then the smallest number with the sum of digits A is also less than the smallest number with the sum of digits B. Consider the smallest natural numbers with the sums of digits $1,2,3, \ldots, 19,20$, they are respectively $a_{1}=1, a_{2}=2, \ldots a_{9}=9, a_{10}=19, a_{11}=29, \ldots a_{18}=99, a_{19}=199, a_{20}=299$. The sum of the first 19 of them is 775, and the remainder of its division by 9 is 1, which is the same as the remainder of the division of 1000 by 9. Increase this sum by 225, taking 234 instead of 9, with the same sum of digits, and we get a representation of 1000 as the sum of 19 numbers $1,2,3,4,5,6,7,8,19,29,39,49,59,69,79,89,99,199,234$ with different sums of digits from 1 to 19.

Suppose that 1000 can be represented as the sum of $n \geq 20$ natural numbers $b_{1}, b_{2}, \ldots b_{n}$ with different sums of digits $s\left(b_{1}\right)<s\left(b_{2}\right)<\ldots<s\left(b_{n}\right)$. In this case, $s\left(b_{k}\right) \geq k, k=1,2, \ldots n,$ and, consequently, $b_{k} \geq a_{k}, k=1,2, \ldots n$. Therefore, $1000=b_{1}+b_{2}+\ldots+b_{n} \geq a_{1}+a_{2}+\ldots+a_{n} \geq a_{1}+a_{2}+\ldots+a_{20}=107$ - a contradiction.

Grading criteria: Finding a representation of 1000 as the sum of 19 numbers with different sums of digits: 3 points. Proof of the maximality of 19: 4 points.

Answer: 19


We have a five-digit positive integer $N$. We select every pair of digits of $N$ (and keep them in order) to obtain the $\tbinom52 = 10$ numbers $33$, $37$, $37$, $37$, $38$, $73$, $77$, $78$, $83$, $87$. Find $N$.

Solution:
To find the five-digit number \( N \) such that every pair of its digits (in order) forms one of the given pairs \( 33, 37, 37, 37, 38, 73, 77, 78, 83, 87 \), we can follow these steps:

1. **Identify the frequency of each pair:**
   - The pair \( 37 \) appears 3 times.
   - The pair \( 33 \) appears 1 time.
   - The pair \( 38 \) appears 1 time.
   - The pair \( 73 \) appears 1 time.
   - The pair \( 77 \) appears 1 time.
   - The pair \( 78 \) appears 1 time.
   - The pair \( 83 \) appears 1 time.
   - The pair \( 87 \) appears 1 time.

2. **Determine the digits of \( N \):**
   - Since \( 37 \) appears 3 times, the digit \( 3 \) must appear before the digit \( 7 \) in three different positions.
   - The digit \( 3 \) must appear at least twice to form the pair \( 33 \).
   - The digit \( 7 \) must appear at least twice to form the pair \( 77 \).
   - The digit \( 8 \) must appear after \( 3 \) and \( 7 \) to form the pairs \( 38 \) and \( 78 \).
   - The digit \( 8 \) must appear before \( 3 \) to form the pair \( 83 \).
   - The digit \( 8 \) must appear before \( 7 \) to form the pair \( 87 \).

3. **Construct the number \( N \):**
   - Let’s start with the digit \( 3 \). Since \( 33 \) appears once, we place \( 3 \) at the first position.
   - Next, we need to place another \( 3 \) to form the pair \( 33 \).
   - To form the pair \( 37 \) three times, we place \( 7 \) after the second \( 3 \).
   - To form the pair \( 77 \), we place another \( 7 \) after the first \( 7 \).
   - To form the pair \( 78 \), we place \( 8 \) after the second \( 7 \).
   - To form the pair \( 38 \), we place \( 8 \) after the first \( 3 \).
   - To form the pair \( 83 \), we place \( 8 \) after the first \( 3 \).
   - To form the pair \( 87 \), we place \( 8 \) after the second \( 7 \).

4. **Verify the number \( N \):**
   - The number \( N \) is \( 37837 \).
   - Check the pairs: \( 33, 37, 37, 37, 38, 73, 77, 78, 83, 87 \).

Thus, the five-digit number \( N \) is \( \boxed{37837} \).

Answer: 37837


Determine the number of ways of tiling a $4 \times 9$ rectangle by tiles of size $1 \times 2$.

Solution:
17. Answer: 6336
Let $f_{n}$ be the number of ways of tiling a $4 \mathrm{x} n$ rectangle. Also, let $g_{n}$ be the number of ways of tiling a $4 \times n$ rectangle with the top or bottom two squares in the last column missing, and let $h_{n}$ be the number of ways of tiling a $4 \times n$ rectangle with the top and bottom squares in the last column missing. Set up a system of recurrence relations involving $f_{n}, g_{n}$ and $h_{n}$ by considering the ways to cover the $n$th column of a $4 \times n$ rectangle. If two vertical tiles are used, then there are $f_{e-1}$ ways. If one vertical tile and two adjacent horizontal tiles are used, then there are $2 g_{s-1}$ ways. If one vertical tile and two non-adjacent horizontal tiles are used, then there are $h_{n-1}$ ways. If four horizontal tiles are used, then there are $f_{n-2}$ ways. Similarly, one can establish the recurrence relations for $g_{n}$ and $h_{n}$. In conclusion, we obtain for $n \geq 2$,
$$
\begin{array}{llc}
f_{n}= & f_{n-1}+f_{n-2}+2 g_{n-1}+h_{n-1} \\
g_{n}= & g_{n-1}+f_{n-1} \\
h_{n}= & h_{n-2}+f_{n-1}
\end{array}
$$

With initial conditions $f_{0}=f_{1}=g_{1}=h_{1}=1$ and $h_{0}=0$. Solving $f_{n}$ recursively, we obtain $f_{9}=$ 6336.

Answer: 6336


8 people participate in a chess round-robin tournament, where the winner gets 1 point, a draw gets 0.5 points, and the loser gets 0 points. It is known that the 8 people's scores are all different, and the second-place player's score is the sum of the scores of the last 4 players. What was the result of the match between the third-place player and the seventh-place player?

Solution: [Solution] Since the last 4 chess players play a total of 6 games among themselves, their total score is no less than 6 points, so the second place must score no less than 6 points. On the other hand, if the first place scores 7 points, then he defeated the second place, so the second place can only score 6 points; if the first place scores 6.5 points, then due to the scores being distinct, the second place can also only score 6 points, thus the total score of the last 4 players is also 6 points, which means they lost all their games against the top 4. In particular, the seventh place lost to the third place.

Answer: 1-0


A cube with side $n$ ($n \geq 3$) is divided by partitions into unit cubes. What is the minimum number of partitions between unit cubes that need to be removed so that from each cube it is possible to reach the boundary of the cube?

Solution:
Let's show that fewer than $(n-2)^{3}$ removed partitions are insufficient.

Remove all boundary cubes. A remaining cube of $(n-2) \times (n-2) \times (n-2)$ will be divided by partitions into $(n-2)^{3}$ smaller cubes. Now the space is divided by partitions into $(n-2)^{3}+1$ regions, counting the outer one. Removing one partition reduces the number of regions by no more than 1. In the end, the number of regions must become 1, so it is necessary to remove no fewer than $(n-2)^{3}$ partitions.

This amount is sufficient: it is enough to remove the bottom face from each non-boundary cube.

## Answer

$(n-2)^{3}$.

Answer: (n-2)^{3}


In $\triangle ABC$, if $AB=5, BC=6$, $CA=7, H$ is the orthocenter, then the length of $AH$ is $\qquad$

Solution:
Extend $A H$, intersecting $B C$ at point $D$, and connect $B H$ and extend it to intersect $A C$ at point $E$.
By the properties of the orthocenter, we know
$A D \perp B C$,
$B E \perp A C$.
In Rt $\triangle A B E$ and Rt $\triangle C B E$, by the Pythagorean theorem, we get
$$
\begin{array}{l}
B E^{2}=A B^{2}-A E^{2}=25-A E^{2}, \\
B E^{2}=B C^{2}-C E^{2}=36-(7-A E)^{2} .
\end{array}
$$

Thus, $25-A E^{2}=36-(7-A E)^{2} \Rightarrow A E=\frac{19}{7}$.
Therefore, $B E=\sqrt{A B^{2}-A E^{2}}=\frac{12 \sqrt{6}}{7}$.
It is easy to see that $\triangle A H E \backsim \triangle B C E \Rightarrow \frac{A H}{B C}=\frac{A E}{B E}$.
So, $A H=\frac{A E \cdot B C}{B E}=\frac{19 \sqrt{6}}{12}$.

Answer: \frac{19 \sqrt{6}}{12}


In a certain language, there are 5 vowels and 7 consonants. A syllable can consist of any vowel and any consonant in any order, and a word can consist of any two syllables. How many words are there in this language?

Solution: 1. The language has $5 \cdot 7=35$ syllables of the form "consonant+vowel" and the same number of syllables of the form "vowel+consonant", making a total of 70 different syllables. The total number of two-syllable words is $70 \cdot 70=4900$.

Answer: 4900


Point $T$ is the centroid of triangle $A B C$, and point $D$ is the midpoint of its side $\overline{B C}$. If the length of the side of the equilateral triangle $B D T$ is $1 \mathrm{~cm}$, determine the lengths of the sides of triangle $A B C$ and the radius of the circumcircle of triangle $A B C$.

Solution:
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7c0a9615fb5cd46ea1c5g-15.jpg?height=827&width=854&top_left_y=798&top_left_x=603)

Since $D$ is the midpoint of side $\overline{B D}$, its length is $|B C|=2 \cdot 1=2 \text{ cm}$.

Point $T$ is the centroid of triangle $A B C$, so $|T D|=\frac{1}{3}|A D|$, which means $|A D|=3 \text{ cm}$.

Triangle $B D T$ is equilateral, so $\angle B D T=60^{\circ}$, and $\angle C D T=120^{\circ}$.

Apply the Law of Cosines to triangle $B D A$.

$$
\begin{aligned}
|A B|^{2} & =|B D|^{2}+|A D|^{2}-2|B D| \cdot|A D| \cos 60^{\circ} \\
|A B|^{2} & =1+9-2 \cdot 1 \cdot 3 \cdot \frac{1}{2}=7 \\
|A B| & =\sqrt{7} \text{ cm}
\end{aligned}
$$

Apply the Law of Cosines to triangle $C D A$.

$$
\begin{aligned}
|A C|^{2} & =|C D|^{2}+|A D|^{2}-2|C D| \cdot|A D| \cos 120^{\circ} \\
|A C|^{2} & =1+9+2 \cdot 1 \cdot 3 \cdot \frac{1}{2}=13 \\
|A C| & =\sqrt{13} \text{ cm}
\end{aligned}
$$

The area of triangle $A B C$ is calculated as twice the area of triangle $A D C$ because triangles $A D C$ and $A D B$ have the same base $(|B D|=|D C|)$ and height to that base.

$$
P=2 \cdot \frac{1}{2}|C D| \cdot|A D| \sin 120^{\circ}=\frac{3 \sqrt{3}}{2} \text{ cm}^2
$$

The radius of the circumscribed circle is

$$
R=\frac{a b c}{4 P}=\frac{2 \cdot \sqrt{7} \cdot \sqrt{13}}{4 \cdot \frac{3 \sqrt{3}}{2}}=\frac{\sqrt{91}}{3 \sqrt{3}}=\frac{\sqrt{273}}{9} \text{ cm}
$$

Answer: \frac{\sqrt{273}}{9}


In an acute-angled triangle $ABC$, the median $BM$ and the altitude $CH$ were drawn. It turned out that $BM=CH=\sqrt{3}$, and $\angle MBC=\angle ACH$. Find the perimeter of triangle $ABC$.

Solution:
Answer: 6.

Solution. We will prove that triangle $ABC$ is equilateral, with all sides equal to 2. Drop a perpendicular $MK$ from point $M$ to side $AB$. In triangle $BMK$, the leg $KM$ is half the hypotenuse $MB$, so $\angle KBM = 30^{\circ}$. Let $O$ be the intersection of $MB$ and $CH$, then $\angle MOC = 60^{\circ}$. Denote $\angle MBC = \angle ACH = \alpha$, then $\angle HCB = 60^{\circ} - \alpha$, so $\angle BCA = 60^{\circ}$. Therefore, triangles $MOC$ and $MCB$ are similar by two angles. From the similarity, we get $MC^2 = BM \cdot OM = CH \cdot OM$. By the Law of Sines, $OM = MC \cdot \sin \alpha / \sin 60^{\circ}$. From triangle $AHC$, we find $HC = 2MC \cdot \cos \alpha$. Substituting this into the formula $MC^2 = CH \cdot OM$ and canceling $MC^2$, we get $\sin 2\alpha = \sqrt{3} / 2$, from which $\alpha = 30^{\circ}$. Therefore, triangle $ABC$ is equilateral.

Answer: 6


Three argumentative friends are sitting in front of the TV. It is known about each of them that she either is always right about everything or always wrong about everything. The first one said: "None of us have seen this movie." The second one said: "I have seen this movie, but you both haven't." The third one said: "I have seen this movie." Determine how many of these friends are always right, given that one of them said everything correctly, and one of them was wrong.

Solution:
Solution. If the first girlfriend is right, then the statements of the second and third are false. If the second girlfriend is right, then the statements of the first and third are false. If the third girlfriend is right, then the statements of the second and third are false. Two girlfriends cannot be right at the same time. Since at least one of the girlfriends made a true statement, there is exactly one true statement.

Answer. Among the girlfriends, exactly one is always right.

Answer: 1


Determine the smallest positive integer $n$, such that, among any $n$ points in the plane, there exist $k$ points among them, such that all distances between them are less than or equal to $2$, or all distances are strictly greater than $1$.

Solution:
To determine the smallest positive integer \( n \) such that among any \( n \) points in the plane, there exist \( k \) points among them such that all distances between them are less than or equal to \( 2 \), or all distances are strictly greater than \( 1 \), we proceed as follows:

1. **Lower Bound Construction:**
   - Consider \( k-1 \) points \( C_1, C_2, \ldots, C_{k-1} \) such that \( C_iC_j > 3 \) for all \( i \neq j \). For concreteness, assume \( C_i \)'s are the vertices of a regular \( (k-1) \)-gon with side length greater than \( 3 \).
   - Construct \( k-1 \) circles \( \omega_i = \left(C_i, \frac{1}{2}\right) \) and take \( k-1 \) different points at random in each circle. Thus, we have constructed \( (k-1)^2 \) points (disregarding the points \( C_i \)).
   - Consider a group of \( k \) points. By the pigeonhole principle (PHP), there exist two points from different circles, so the distance between them is \( > 3 - \frac{1}{2} - \frac{1}{2} = 2 \). Also, by the PHP, there are two points from the same circle \( \omega_i \), so the distance between them is \( \leq \frac{1}{2} + \frac{1}{2} = 1 \).
   - Therefore, there does not exist a set of \( k \) points such that the distances between any two of them are either \( > 1 \) or \( \leq 2 \).

2. **Upper Bound:**
   - Assume we have \( k^2 - 2k + 2 \) points. Consider the graph \( G \) with \( k^2 - 2k + 2 \) vertices corresponding to the points and color the edge between two points in green if the distance between them is \( \leq 1 \) and red otherwise.
   - **Lemma:** If \( AB \) and \( AC \) are green, then \( BC \leq 2 \).
     - **Proof:** Since \( AB \) and \( AC \) are green, \( AB \leq 1 \) and \( AC \leq 1 \). By the triangle inequality, \( BC \leq AB + AC = 1 + 1 = 2 \).

3. **Proof of Upper Bound:**
   - We will now prove that there is either a clique of \( k \) points where all distances are \( \leq 2 \) or a red clique of \( k \) vertices. Assume the contrary.
   - Notice that the Lemma implies that if a vertex \( X \) is part of \( \geq k-1 \) green segments, then \( X \) combined with the \( k-1 \) vertices form a clique of the first type.
   - This implies that each vertex has at least \( (k^2 - 2k + 2) - 1 - (k - 2) = k^2 - 3k + 3 \) incident red edges.
   - Thus, we have:
     \[
     |E_{\text{red}}| \geq \left\lfloor \frac{|V|(k^2 - 3k + 3)}{2} \right\rfloor = \left\lfloor \frac{(k^2 - 2k + 2)(k^2 - 3k + 3)}{2} \right\rfloor \geq \frac{(k^2 - 2k + 2)(k^2 - 3k + 3)}{2} - \frac{1}{2}
     \]
   - However, since we assumed that there isn't a red \( K_k \), then by Turan's theorem we have:
     \[
     |E_{\text{red}}| \leq \left\lfloor \left(1 - \frac{1}{k-1}\right) \cdot \frac{|V|^2}{2} \right\rfloor = \left\lfloor \frac{k-2}{k-1} \cdot \frac{(k^2 - 2k + 2)^2}{2} \right\rfloor \leq \frac{k-2}{k-1} \cdot \frac{(k^2 - 2k + 2)^2}{2}
     \]
   - Now we reach a contradiction by:
     \[
     \frac{(k^2 - 2k + 2)(k^2 - 3k + 3)}{2} - \frac{1}{2} > \frac{k-2}{k-1} \cdot \frac{(k^2 - 2k + 2)^2}{2}
     \]
     \[
     \iff (k-1)(k^2 - 2k + 2)(k^2 - 3k + 3) - (k-1) - (k-2)(k^2 - 2k + 2)^2 > 0
     \]
     \[
     \iff k^2 - 3k + 3 > 0 \iff \frac{1}{4} (2k - 3)^2 + \frac{3}{4} > 0
     \]
   - So we proved that \( |E_{\text{red}}| \leq \frac{k-2}{k-1} \cdot \frac{(k^2 - 2k + 2)^2}{2} < \frac{(k^2 - 2k + 2)(k^2 - 3k + 3)}{2} - \frac{1}{2} \leq |E_{\text{red}}| \), contradiction.

The final answer is \( \boxed{ n = (k-1)^2 + 1 = k^2 - 2k + 2 } \)

Answer: n = (k-1)^2 + 1 = k^2 - 2k + 2


Does there exist a 4-digit integer which cannot be changed into a multiple of 1992 by changing 3 of its digits?

Answer: Yes, eg 2251

Solution:
\section*{Solution}

The only 4 digit multiples of 1992 are: 1992, 3984, 5976, 7968, 9960. All have first digit odd, second digit 9 , third digit \(>5\) and last digit even, so it is easy to find a number which has all digits different from all of them.

Answer: 2251


Let $p$ and $q$ be two given positive integers. A set of $p+q$ real numbers $a_1<a_2<\cdots <a_{p+q}$ is said to be balanced iff $a_1,\ldots,a_p$ were an arithmetic progression with common difference $q$ and $a_p,\ldots,a_{p+q}$ where an arithmetic progression with common difference $p$. Find the maximum possible number of balanced sets, so that any two of them have nonempty intersection.

Solution:
To solve the problem, we need to find the maximum number of balanced sets such that any two of them have a nonempty intersection. We will consider the case when \( p \) and \( q \) are coprime.

1. **Define the balanced set:**
   A set of \( p+q \) real numbers \( a_1 < a_2 < \cdots < a_{p+q} \) is balanced if:
   - \( a_1, a_2, \ldots, a_p \) form an arithmetic progression with common difference \( q \).
   - \( a_p, a_{p+1}, \ldots, a_{p+q} \) form an arithmetic progression with common difference \( p \).

2. **Minimal elements in each balanced set:**
   Let \( 0 = x_1 < x_2 < \ldots < x_k \) be the minimal elements in each balanced set. For any indices \( i < j \), there exist unique pairs of integers \( 0 \le u \le q \) and \( 0 \le v \le p-1 \) such that:
   \[
   x_j - x_i = up + vq
   \]
   Specifically, we have:
   \[
   x_i = x_i - x_1 = u_i p + v_i q
   \]
   where \( u_i \) and \( v_i \) are defined as above.

3. **Ordering of sums \( u_i + v_i \):**
   It is easy to prove that:
   \[
   0 = u_1 + v_1 < u_2 + v_2 < \ldots < u_k + v_k
   \]
   Since these are all integers, we get:
   \[
   u_k + v_k \ge k - 1
   \]

4. **Bounding \( k \):**
   We need to find the maximum \( k \) such that the above conditions hold. We have:
   \[
   x_k \ge u_k p + (k-1-u_k) q = (k-1) q - u_k (q - p)
   \]
   Since \( u_k \le q \), we get:
   \[
   x_k \ge (k-1) q - q (q - p) = (k-1) q - q^2 + pq
   \]
   Therefore:
   \[
   2pq - q \ge (k-1) q - q^2 + pq
   \]
   Simplifying, we get:
   \[
   2pq - q \ge (k-1) q - q^2 + pq \implies k \le p + q
   \]

Thus, the maximum possible number of balanced sets such that any two of them have a nonempty intersection is \( p + q \).

The final answer is \( \boxed{ p + q } \).

Answer: p + q


Let $p(n)$ denote the product of decimal digits of a positive integer $n$. Compute the sum $p(1)+p(2)+\ldots+p(2001)$.

Solution:
1. **Understanding the function \( p(n) \)**:
   - The function \( p(n) \) denotes the product of the decimal digits of a positive integer \( n \). For example, \( p(23) = 2 \times 3 = 6 \).

2. **Sum of \( p(n) \) for single-digit numbers**:
   - For \( n \) in the range \([1, 9]\), \( p(n) = n \).
   - Therefore, the sum \( \sum_{n=1}^{9} p(n) = 1 + 2 + \ldots + 9 = \frac{9 \times 10}{2} = 45 \).

3. **Sum of \( p(n) \) for two-digit numbers**:
   - For \( n \) in the range \([10, 99]\), we need to consider the product of the digits of each number.
   - The sum of the products of the digits can be calculated as follows:
     \[
     \sum_{a=1}^{9} \sum_{b=0}^{9} (a \times b) = \sum_{a=1}^{9} a \left( \sum_{b=0}^{9} b \right) = \sum_{a=1}^{9} a \times 45 = 45 \times (1 + 2 + \ldots + 9) = 45 \times 45 = 2025
     \]

4. **Sum of \( p(n) \) for three-digit numbers**:
   - For \( n \) in the range \([100, 999]\), we need to consider the product of the digits of each number.
   - The sum of the products of the digits can be calculated as follows:
     \[
     \sum_{a=1}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} (a \times b \times c) = \sum_{a=1}^{9} a \left( \sum_{b=0}^{9} b \left( \sum_{c=0}^{9} c \right) \right) = \sum_{a=1}^{9} a \times 45 \times 45 = 45^2 \times (1 + 2 + \ldots + 9) = 45^3 = 91125
     \]

5. **Sum of \( p(n) \) for four-digit numbers in the range \([1000, 2001]\)**:
   - For \( n \) in the range \([1000, 1999]\), the leading digit is fixed to be 1, so we need to consider the product of the remaining three digits.
   - The sum of the products of the digits can be calculated as follows:
     \[
     \sum_{b=0}^{9} \sum_{c=0}^{9} \sum_{d=0}^{9} (1 \times b \times c \times d) = \sum_{b=0}^{9} b \left( \sum_{c=0}^{9} c \left( \sum_{d=0}^{9} d \right) \right) = 45^3 = 91125
     \]
   - For \( n \) in the range \([2000, 2001]\), we only need to consider \( p(2000) = 0 \) and \( p(2001) = 0 \).

6. **Summing all contributions**:
   - The total sum is:
     \[
     45 + 2025 + 91125 + 91125 = 184320
     \]

The final answer is \(\boxed{184320}\).

Answer: 184320


Given that $\triangle A B C$ is an equilateral triangle with side length 2, and points $M_{1}, M_{2}, \cdots, M_{n}$ are taken sequentially on segment $B C$, dividing $B C$ into $n+1$ equal parts. Then
$$
\lim _{n \rightarrow \infty} \frac{1}{n}\left(A M_{1}^{2}+A M_{2}^{2}+\cdots+A M_{n}^{2}\right)=(\quad) \text {. }
$$
(A) $\frac{8}{3}$
(B) 3
(C) $\frac{10}{3}$
(D) 4

Solution:
4. C.

As shown in Figure 4, draw $B D \perp A C$, with the foot of the perpendicular being $D$. Draw $M_{k} N_{k} \perp B D$ and $M_{k} P_{k} \perp A C$, with the feet of the perpendiculars being $N_{k}$ and $P_{k}$, respectively.
From $\frac{B M_{k}}{B C}=\frac{k}{n+1}$, we get
$$
\begin{array}{c}
M_{k} N_{k}=\frac{k}{n+1}, \\
M_{k} P_{k}=\frac{\sqrt{3}(n+1-k)}{n+1} .
\end{array}
$$

Thus, $A P_{k}=A D+M_{k} N_{k}=1+\frac{k}{n+1}$.
Therefore, $A M_{k}^{2}=A P_{k}^{2}+M_{k} P_{k}^{2}$
$$
\begin{array}{l}
=\left(1+\frac{k}{n+1}\right)^{2}+\left[\frac{\sqrt{3}(n+1-k)}{n+1}\right]^{2} \\
=1+\frac{2 k}{n+1}+\frac{k^{2}}{(n+1)^{2}}+\frac{3(n+1-k)^{2}}{(n+1)^{2}} .
\end{array}
$$

Then, $\frac{1}{n}\left(A M_{1}^{2}+A M_{2}^{2}+\cdots+A M_{n}^{2}\right)$
$$
\begin{array}{l}
=\frac{1}{n}\left\{n+\frac{2(1+2+\cdots+n)}{n+1}+\frac{1^{2}+2^{2}+\cdots+n^{2}}{(n+1)^{2}}+\right. \\
\left.\frac{3\left[n^{2}+(n-1)^{2}+\cdots+1^{2}\right]}{(n+1)^{2}}\right\} \\
=\frac{1}{n}\left[2 n+\frac{4}{(n+1)^{2}} \cdot \frac{1}{6} n(n+1)(2 n+1)\right] \\
=2+\frac{2(2 n+1)}{3(n+1)} .
\end{array}
$$

Therefore, $\lim _{n \rightarrow \infty} \frac{1}{n}\left(A M_{1}^{2}+A M_{2}^{2}+\cdots+A M_{n}^{2}\right)$
$$
=\lim _{n \rightarrow \infty}\left[2+\frac{2(2 n+1)}{3(n+1)}\right]=2+\frac{4}{3}=\frac{10}{3} \text {. }
$$

Answer: C


Let $C$ be the graph of $xy = 1$, and denote by $C^*$ the reflection of $C$ in the line $y = 2x$. Let the equation of $C^*$ be written in the form
\[12x^2 + bxy + cy^2 + d = 0.\]
Find the product $bc$.

Solution:
Given a point $P (x,y)$ on $C$, we look to find a formula for $P' (x', y')$ on $C^*$. Both points lie on a line that is [perpendicular](https://artofproblemsolving.com/wiki/index.php/Perpendicular) to $y=2x$, so the slope of $\overline{PP'}$ is $\frac{-1}{2}$. Thus $\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y$. Also, the midpoint of $\overline{PP'}$, $\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)$, lies on the line $y = 2x$. Therefore $\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x$.
Solving these two equations, we find $x = \frac{-3x' + 4y'}{5}$ and $y = \frac{4x' + 3y'}{5}$. Substituting these points into the equation of $C$, we get $\frac{(-3x'+4y')(4x'+3y')}{25}=1$, which when expanded becomes $12x'^2-7x'y'-12y'^2+25=0$.
Thus, $bc=(-7)(-12)=\boxed{084}$.

Answer: 84


Plot the graph of the function $\mathrm{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$.

Solution:
Answer. The graph of the function will be the line $y = 4$.

## Solution.

$\mathrm{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$

$\mathrm{y}=\sqrt{4 \sin ^{4} x-2+4 \sin ^{2} x+3}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x-2+3}$

$\mathrm{y}=\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x+1}$

$\mathrm{y}=2 \sin ^{2} x+1+2 \cos ^{2} x+1, \mathrm{y}=4$.

Answer: 4


In rhombus $ABCD$, $\angle A=100^{\circ}$, $M$ and $N$ are the midpoints of sides $AB$ and $BC$ respectively, and $MP \perp CD$ at point $P$. Then the degree measure of $\angle NPC$ is $\qquad$

Solution:
$=9.50^{\circ}$.
Since quadrilateral $ABCD$ is a rhombus, therefore,
$$
AB=BC, \angle B=180^{\circ}-\angle A=80^{\circ} \text {. }
$$

Also, $M$ and $N$ are the midpoints of sides $AB$ and $BC$, respectively, so $BM=BN$,
$$
\angle BMN=\angle BNM=\frac{1}{2}\left(180^{\circ}-80^{\circ}\right)=50^{\circ} \text {. }
$$

Let the midpoint of $MP$ be
$E$, as shown in Figure 10, connect $NE$. Then
$NE \parallel BM \parallel CP$.
Since $MP \perp CD$, then $NE \perp MP$, so $NM=NP$.

Therefore, $\angle NMP$ $=\angle NPM$.
Thus $\angle NPC=\angle NMB=50^{\circ}$.

Answer: 50^{\circ}


Given two points $P,\ Q$ on the parabola $C: y=x^2-x-2$ in the $xy$ plane. Note that the $x$ coordinate of $P$ is less than that of $Q$.

(a) If the origin $O$ is the midpoint of the line segment $PQ$, then find the equation of the line $PQ$.

(b) If the origin $O$ divides internally the line segment $PQ$ by 2:1, then find the equation of $PQ$.

(c) If the origin $O$ divides internally the line segment $PQ$ by 2:1, find the area of the figure bounded by the parabola $C$ and the line $PQ$.

Solution:
### Part (a)
1. Let the coordinates of points \( P \) and \( Q \) be \( (x_1, y_1) \) and \( (x_2, y_2) \) respectively.
2. Given that the origin \( O \) is the midpoint of the line segment \( PQ \), we have:
   \[
   \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (0, 0)
   \]
3. This implies:
   \[
   x_1 + x_2 = 0 \quad \text{and} \quad y_1 + y_2 = 0
   \]
4. Since \( P \) and \( Q \) lie on the parabola \( y = x^2 - x - 2 \), we have:
   \[
   y_1 = x_1^2 - x_1 - 2 \quad \text{and} \quad y_2 = x_2^2 - x_2 - 2
   \]
5. Substituting \( x_2 = -x_1 \) into the equation for \( y_2 \):
   \[
   y_2 = (-x_1)^2 - (-x_1) - 2 = x_1^2 + x_1 - 2
   \]
6. Since \( y_1 + y_2 = 0 \), we have:
   \[
   (x_1^2 - x_1 - 2) + (x_1^2 + x_1 - 2) = 0
   \]
7. Simplifying, we get:
   \[
   2x_1^2 - 4 = 0 \implies x_1^2 = 2 \implies x_1 = \pm \sqrt{2}
   \]
8. Therefore, the coordinates of \( P \) and \( Q \) are \( (\sqrt{2}, \sqrt{2}^2 - \sqrt{2} - 2) \) and \( (-\sqrt{2}, (-\sqrt{2})^2 + \sqrt{2} - 2) \).
9. Simplifying the coordinates:
   \[
   P = (\sqrt{2}, 2 - \sqrt{2} - 2) = (\sqrt{2}, -\sqrt{2})
   \]
   \[
   Q = (-\sqrt{2}, 2 + \sqrt{2} - 2) = (-\sqrt{2}, \sqrt{2})
   \]
10. The slope of the line \( PQ \) is:
    \[
    m = \frac{\sqrt{2} - (-\sqrt{2})}{-\sqrt{2} - \sqrt{2}} = \frac{2\sqrt{2}}{-2\sqrt{2}} = -1
    \]
11. Therefore, the equation of the line \( PQ \) is:
    \[
    y = -x
    \]

### Part (b)
1. Given that the origin \( O \) divides the line segment \( PQ \) internally in the ratio 2:1, we use the section formula.
2. Let the coordinates of \( P \) and \( Q \) be \( (x_1, y_1) \) and \( (x_2, y_2) \) respectively.
3. The coordinates of \( O \) are given by:
   \[
   O = \left( \frac{2x_2 + x_1}{3}, \frac{2y_2 + y_1}{3} \right) = (0, 0)
   \]
4. This implies:
   \[
   2x_2 + x_1 = 0 \quad \text{and} \quad 2y_2 + y_1 = 0
   \]
5. Since \( P \) and \( Q \) lie on the parabola \( y = x^2 - x - 2 \), we have:
   \[
   y_1 = x_1^2 - x_1 - 2 \quad \text{and} \quad y_2 = x_2^2 - x_2 - 2
   \]
6. Substituting \( x_1 = -2x_2 \) into the equation for \( y_1 \):
   \[
   y_1 = (-2x_2)^2 - (-2x_2) - 2 = 4x_2^2 + 2x_2 - 2
   \]
7. Since \( 2y_2 + y_1 = 0 \), we have:
   \[
   2(x_2^2 - x_2 - 2) + (4x_2^2 + 2x_2 - 2) = 0
   \]
8. Simplifying, we get:
   \[
   2x_2^2 - 2x_2 - 4 + 4x_2^2 + 2x_2 - 2 = 0 \implies 6x_2^2 - 6 = 0 \implies x_2^2 = 1 \implies x_2 = \pm 1
   \]
9. Therefore, the coordinates of \( P \) and \( Q \) are \( (-2, (-2)^2 - (-2) - 2) \) and \( (1, 1^2 - 1 - 2) \).
10. Simplifying the coordinates:
    \[
    P = (-2, 4 + 2 - 2) = (-2, 4)
    \]
    \[
    Q = (1, 1 - 1 - 2) = (1, -2)
    \]
11. The slope of the line \( PQ \) is:
    \[
    m = \frac{-2 - 4}{1 - (-2)} = \frac{-6}{3} = -2
    \]
12. Therefore, the equation of the line \( PQ \) is:
    \[
    y = -2x
    \]

### Part (c)
1. The area of the figure bounded by the parabola \( y = x^2 - x - 2 \) and the line \( y = -2x \) can be found by integrating the difference of the functions over the interval where they intersect.
2. First, find the points of intersection by solving:
   \[
   x^2 - x - 2 = -2x \implies x^2 + x - 2 = 0 \implies (x + 2)(x - 1) = 0 \implies x = -2 \text{ or } x = 1
   \]
3. The area \( A \) is given by:
   \[
   A = \int_{-2}^{1} \left( (x^2 - x - 2) - (-2x) \right) \, dx = \int_{-2}^{1} (x^2 + x - 2) \, dx
   \]
4. Evaluate the integral:
   \[
   \int_{-2}^{1} (x^2 + x - 2) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} - 2x \right]_{-2}^{1}
   \]
5. Calculate the definite integral:
   \[
   \left( \frac{1^3}{3} + \frac{1^2}{2} - 2(1) \right) - \left( \frac{(-2)^3}{3} + \frac{(-2)^2}{2} - 2(-2) \right)
   \]
   \[
   = \left( \frac{1}{3} + \frac{1}{2} - 2 \right) - \left( \frac{-8}{3} + 2 + 4 \right)
   \]
   \[
   = \left( \frac{1}{3} + \frac{1}{2} - 2 \right) - \left( \frac{-8}{3} + 6 \right)
   \]
   \[
   = \left( \frac{2}{6} + \frac{3}{6} - 2 \right) - \left( \frac{-8}{3} + 6 \right)
   \]
   \[
   = \left( \frac{5}{6} - 2 \right) - \left( \frac{-8}{3} + 6 \right)
   \]
   \[
   = \left( \frac{5}{6} - \frac{12}{6} \right) - \left( \frac{-8}{3} + 6 \right)
   \]
   \[
   = \left( \frac{-7}{6} \right) - \left( \frac{-8}{3} + 6 \right)
   \]
   \[
   = \left( \frac{-7}{6} \right) - \left( \frac{-16}{6} + \frac{36}{6} \right)
   \]
   \[
   = \left( \frac{-7}{6} \right) - \left( \frac{20}{6} \right)
   \]
   \[
   = \left( \frac{-7}{6} - \frac{20}{6} \right)
   \]
   \[
   = \left( \frac{-27}{6} \right)
   \]
   \[
   = -\frac{27}{6}
   \]
   \[
   = -\frac{9}{2}
   \]
6. The area is:
   \[
   A = \frac{9}{2}
   \]

The final answer is \( \boxed{\frac{9}{2}} \)

Answer: \frac{9}{2}


Xiaoming goes to the supermarket to buy milk. If he buys fresh milk at 6 yuan per box, the money he brings is just enough; if he buys yogurt at 9 yuan per box, the money is also just enough, but he buys 6 fewer boxes than fresh milk. Xiaoming brings a total of yuan.

Solution:
【Answer】108 yuan
【Key Point】Solving application problems by setting up equations
【Analysis】
Let Xiao Ming be able to buy $x$ boxes of yogurt, then he can buy $(x+6)$ boxes of fresh milk;

According to the problem, we can set up the equation: $6(x+6)=9x$, solving for $x$ gives $x=12$; so Xiao Ming brought a total of $9 \times 12=108$ yuan.

Answer: 108


Solution:
## Solution.

From the first equation of the system, we find $z=-x-y$ and substitute it into the second and third equations:

$$
\begin{aligned}
& \left\{\begin{array}{l}
2 x+3 y-x-y=0 \\
(x+1)^{2}+(y+2)^{2}+(-x-y+3)^{2}=14
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x+2 y=0 \\
(x+1)^{2}+(y+2)^{2}+(-x-y+3)^{2}=14
\end{array}\right.
\end{aligned}
$$

From here, $x=-2 y,(-2 y+1)^{2}+(y+2)^{2}+(2 y-y+3)^{2}=14 \Leftrightarrow$ $\Leftrightarrow y^{2}+y=0$, from which $y_{1}=0, y_{2}=-1 ; x_{1}=0, x_{2}=2 ; z_{1}=0, z_{2}=-1$.'

Answer: $(0 ; 0 ; 0),(2 ;-1 ;-1)$.

Answer: (0;0;0),(2;-1;-1)


Let $ m \equal{} 2007^{2008}$, how many natural numbers n are there such that $ n < m$ and $ n(2n \plus{} 1)(5n \plus{} 2)$ is divisible by $ m$ (which means that $ m \mid n(2n \plus{} 1)(5n \plus{} 2)$) ?

Solution:
1. **Define \( m \) and factorize it:**
   \[
   m = 2007^{2008}
   \]
   We can factorize \( 2007 \) as:
   \[
   2007 = 3^2 \times 223
   \]
   Therefore,
   \[
   m = (3^2 \times 223)^{2008} = 3^{4016} \times 223^{2008}
   \]
   Let \( a = 3^{4016} \) and \( b = 223^{2008} \). Thus, \( m = ab \) and \( \gcd(a, b) = 1 \).

2. **Conditions for divisibility:**
   We need \( n(2n + 1)(5n + 2) \) to be divisible by \( m \). This means:
   \[
   m \mid n(2n + 1)(5n + 2)
   \]
   Since \( a \) and \( b \) are coprime, at least one of \( n \), \( 2n + 1 \), or \( 5n + 2 \) must be divisible by \( a \), and at least one of them must be divisible by \( b \).

3. **Set up congruences:**
   We need to solve the following congruences:
   \[
   sn + t \equiv 0 \pmod{a}
   \]
   \[
   un + v \equiv 0 \pmod{b}
   \]
   where \( (s, t) \) and \( (u, v) \) are one of \( (1, 0) \), \( (2, 1) \), and \( (5, 2) \).

4. **Uniqueness of solutions:**
   Since \( s \) and \( u \) are relatively prime to both \( a \) and \( b \), each system of linear congruences has a unique solution modulo \( m \). There are 9 combinations for \( (s, t) \) and \( (u, v) \), leading to 9 potential solutions for \( n \).

5. **Check for overlap:**
   We need to ensure that no two solutions coincide. If two solutions coincide, then \( a \) or \( b \) would divide both \( sn + t \) and \( un + v \) for different pairs \( (s, t) \) and \( (u, v) \), which is not possible since these numbers are relatively prime.

6. **Special case for \( n = 0 \):**
   When both \( (s, t) = (1, 0) \) and \( (u, v) = (1, 0) \), the solution is \( n \equiv 0 \pmod{m} \). However, since \( n < m \) and \( n \) must be a natural number, \( n = 0 \) is not allowed.

7. **Conclusion:**
   Since \( n = 0 \) is not allowed, we have 8 unique solutions for \( n \) in the range \( 0 < n < m \).

The final answer is \( \boxed{ 8 } \).

Answer: 8


Given that $f(x)$ is an even function, for any $x \in \mathbf{R}^{+}$, we have $f(2+x)=-2 f(2-x)$, and $f(-1)=4, f(-3)$ equals $\qquad$ .

Solution:
3. -8 .

Let $x=1$, then $f(3)=-2 f(1)$. Since $f(x)$ is an even function, then $f(-3)=f(3)=-2 f(1)=$ $-2 f(-1)=-8$

Answer: -8


Given the function $y=\frac{a-x}{x-a-1}$, the graph of its inverse function is symmetric about the point $(-1,4)$. Then the value of the real number $a$ is $\qquad$ .

Solution:
2.3.-

From the problem, we know that the graph of the function $y=\frac{a-x}{x-a-1}$ is centrally symmetric about the point $(4,-1)$.
$\because y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, we have $(y+1)[x-(a+1)]=-1$,
$\therefore$ the graph of the function is a hyperbola with its center at $(a+1,-1)$.
Also, $\because$ the graph of the hyperbola is centrally symmetric about its center,
$\therefore a+1=4$, which means $a=3$.

Answer: 3


A column of troops is marching at a speed of 10 km/h. The communicator at the front of the column runs to the end of the column at a speed of 15 km/h to deliver orders, and then runs back to the front to report to the commander, taking a total of 18 minutes. The length of this column is $\qquad$ meters.

Answer: 1250


A keresedÅ 39 K-Ã©rt adott el Ã¡rut. MennyiÃ©rt vette az Ã¡rut, ha ugyanannyi szÃ¡zalÃ©kot nyert, mint a mennyibe az Ã¡ru kerÃ¼lt?

Solution:
If the merchant bought the item for $x K$, then

$$
x+\frac{x x}{100}=39
$$

or

$$
x^{2}+100 x-3900=0
$$

from which the positive root is:

$$
x=30
$$

(Viola Rezső, Budapest.)

Answer: 30


Let $a, b$ be positive integers such that $79 \mid(a+77 b)$, and $77 \mid(a+79 b)$, find the smallest possible value of the sum $a+b$.

Solution:
16. Notice that
$79|(a+77 b) \Leftrightarrow 79|(a-2 b) \Leftrightarrow 79|(39 a-78 b) \Leftrightarrow 79|(39 a+b)$,
$77|(a+79 b) \Leftrightarrow 77|(a+2 b) \Leftrightarrow 77|(39 a+78 b) \Leftrightarrow 77|(39 a+b)$,
thus, $79 \times 77 \mid(39 a+b)$,
therefore, $39 a+b=79 \times 77 k, k \in \mathbf{N}_{+}$.
Notice that
$$
39 a+39 b=79 \times 77 k+38 b=\left(78^{2}-1\right) k+38 b=\left(78^{2}-39\right) k+38(k+b),
$$

so, $39 \mid(b+k)$, hence $b+k \geqslant 39$.
Thus, $39 a+39 b \geqslant\left(78^{2}-39\right)+38 \times 39$, which means
$$
a+b \geqslant 156-1+38=193 \text {. }
$$

It is easy to see that $b=38, a=155$ satisfies the given conditions, therefore,
$$
a+b=193 \text {. }
$$

Hence, $(a+b)_{\min }=193$.

Answer: 193


Let $a, b, c$ be distinct rational numbers, satisfying $(b+\sqrt{2})^{2}=(a+\sqrt{2})(c+\sqrt{2})$.
Then the number of sets of $a, b, c$ that satisfy the condition is ( ).
(A) 0
(B) 1
(C) 2
(D) 4

Solution:
6. A.

Since $(b+\sqrt{2})^{2}=(a+\sqrt{2})(c+\sqrt{2})$, that is, $b^{2}+2+2 \sqrt{2} b=a c+2+(a+c) \sqrt{2}$, then $a c=b^{2}, a+c=2 b$.
Thus, $a^{2}+c^{2}+2 a c=(a+c)^{2}=4 b^{2}=4 a c$.
Therefore, $(a-c)^{2}=0$.
Hence, $a=c$, which contradicts the given condition.

Answer: A


In a volleyball tournament, $n$ teams from city $A$ and $2 n$ teams from city $B$ participated. Each team played exactly one game against each other team. The ratio of the number of wins by teams from city $B$ to the number of wins by teams from city $A$ is $3: 4$. Find $n$, given that there were no draws in the tournament.

Solution:
(12 points) Solution. The number of games in which only teams from city $A$ participated is $\frac{(n-1) n}{2}$. In these games, teams from city $A$ won $\frac{(n-1) n}{2}$ games. The number of games in which only teams from city $B$ participated is $\frac{(2 n-1) 2 n}{2}$. In these games, teams from city $B$ won $(2 n-1) n$ games. The number of matches between teams from city $A$ and teams from city $B$ is $2 n^{2}$. Let $m$ be the number of wins in these matches by teams from city $A$, then teams from city $B$ won $2 n^{2}-m$ games in these matches. In total, teams from city $A$ won $\frac{(n-1) n}{2}+m$ games, and teams from city $B$ won $(2 n-1) n+2 n^{2}-m$ games. According to the condition, we have $\frac{\frac{(n-1) n}{2}+m}{(2 n-1) n+2 n^{2}-m}=\frac{4}{3}$, $3 n^{2}-3 n+6 m=32 n^{2}-8 n-8 m, 14 m=29 n^{2}-5 n, m=\frac{29 n^{2}-5 n}{14}, \frac{29 n^{2}-5 n}{14} \leq 2 n^{2}, n^{2}-5 n \leq 0$. The number $n$ can be $1,2,3,4,5$. By substituting into the equation $m=\frac{29 n^{2}-5 n}{14}$, we get that $m$ is an integer only when $n=5$.

Answer: 5.

Answer: 5


In how many ways can the letters be placed into the envelopes so that no letter ends up in the correct envelope?

Solution:
Let $ F (n) $ denote the number of all ways of placing $ n $ letters $ L_1, L_2, \ldots, L_n $ into $ n $ envelopes $ K_1, K_2, \ldots, K_n $ so that no letter $ L_i $ ends up in the correct envelope $ K_i $, or more simply, so that there is no "hit". We need to calculate $ F (6) $.
Suppose we incorrectly place all letters by putting letter $ L_1 $ in envelope $ K_i $ ($ i \ne 1 $). There are $ 2 $ cases:
a) Letter $ L_i $ ends up in envelope $ K_1 $. The remaining $ 4 $ letters must then be incorrectly placed in the remaining $ 4 $ envelopes, which can be done in $ F (4) $ ways. Since $ K_i $ can be any of the $ 5 $ envelopes $ K_2 $, $ K_3 $, $ K_4 $, $ K_5 $, $ K_6 $, case a) can occur $ 5 \cdot F (4) $ times.
b) Letter $ L_i $ does not end up in envelope $ K_1 $. By assigning envelope $ K_1 $ to letter $ L_i $ as its "pseudo-correct" envelope, we can say that none of the letters $ L_2 $, $ L_3 $, $ L_4 $, $ L_5 $, $ L_6 $ end up in their correct envelopes, which can be done in $ F (5) $ ways. Since $ K_i $ can be any of the $ 5 $ envelopes, case b) can occur $ 5 \cdot F (5) $ times.
Since cases a) and b) cover all possibilities, therefore

Analogously

\[ F (5) = 4F (4) + 4F (3) \]

\[ F (4) = 3F (3) + 3F (2) \]

\[ F (3) = 2F (2) + 2F (1). \]

But of course

\[ F (2) = 1 \]
\[ F (1) = 0 \]

Therefore, from the above equations, we obtain sequentially,

\[ F (3) = 2 \cdot 1 + 2 \cdot 0 = 2 \]
\[ F (4) = 3 \cdot 2 + 3 \cdot 1 = 9 \]
\[ F (5) = 4 \cdot 9 + 4 \cdot 2 = 44 \]

Note. In the above solutions, the number $ 6 $ does not play any significant role. The same methods can be applied to $ n $ letters and $ n $ envelopes. Method I leads to the formula

\[ F (n) = (n - 1) [F (n - 1) + F (n - 2)] \]

This is the so-called recursive or reduction formula, which allows us to calculate $ F (n) $ if the values of $ F (n - 1) $ and $ F (n - 2) $ are known. Similarly, Method II leads to the recursive formula

\[ F (n) = nF (n - 1) + (-1)^n \]

Formula (1) allows us to calculate $ F (n) $ in terms of $ n $. We can write this formula as

\[ F (n) = nF (n - 1) + (-1)^n \]

From formula (3), we see that the function $ F (k) - kF (k - 1) $ has the same absolute value for every natural number $ k > 1 $, but changes sign when transitioning from $ k $ to $ k + 1 $. And since

\[ F (2) - 2F (1) = 1 \]

we have

\[ F (3) - 3F (2) = -1 \]
\[ F (4) - 4F (3) = 1 \]
\[ \ldots \]

From equality (4), we get

\[ F (n) - nF (n - 1) = (-1)^n \]

We introduce the notation

\[ \varphi (k) = \frac{F (k)}{k!} \]

Formula (5) becomes

\[ \varphi (k) - \varphi (k - 1) = \frac{(-1)^k}{k!} \]

Analogously

\[ \varphi (n) - \varphi (n - 1) = \frac{(-1)^n}{n!} \]

\[ \varphi (n - 1) - \varphi (n - 2) = \frac{(-1)^{n-1}}{(n-1)!} \]

\[ \ldots \]

\[ \varphi (2) - \varphi (1) = \frac{1}{2}. \]

Adding these formulas and considering that $ \varphi (1) = \frac{F (1)}{1} = 0 $, we get

\[ \varphi (n) = \sum_{k=1}^{n} \frac{(-1)^k}{k!} \]

Therefore,

\[ F (n) = n! \sum_{k=1}^{n} \frac{(-1)^k}{k!} \]

The problem we solved is known as the Bernoulli-Euler problem of the mislaid letters.

Answer: 265


Find ten natural numbers whose sum and product are equal to 20.

Solution:
Answer: $1,1,1,1,1,1,1,1,2,10$.

Comment. The given answer - 7 points.

Answer: 1,1,1,1,1,1,1,1,2,10


For $i=2,3, \cdots, k$, the remainder when the positive integer $n$ is divided by $i$ is $i-1$. If the smallest value of $n$, $n_{0}$, satisfies $2000<n_{0}<3000$, then the smallest value of the positive integer $k$ is

Solution:
10.9.

Since $n+1$ is a multiple of $2,3, \cdots, k$, the smallest value of $n$, $n_{0}$, satisfies
$$
n_{0}+1=[2,3, \cdots, k],
$$

where $[2,3, \cdots, k]$ represents the least common multiple of $2,3, \cdots, k$.
$$
\begin{array}{l}
\text { Since }[2,3, \cdots, 8]=840, \\
{[2,3, \cdots, 9]=2520,} \\
{[2,3, \cdots, 10]=2520,} \\
{[2,3, \cdots, 11]=27720,}
\end{array}
$$

Therefore, the smallest positive integer $k$ that satisfies $2000<n_{0}<3000$ is 9.

Answer: 9


A list of positive integers is called good if the maximum element of the list appears exactly once. A sublist is a list formed by one or more consecutive elements of a list. For example, the list $10,34,34,22,30,22$ the sublist $22,30,22$ is good and $10,34,34,22$ is not. A list is very good if all its sublists are good. Find the minimum value of $k$ such that there exists a very good list of length $2019$ with $k$ different values on it.

Solution:
1. **Understanding the Problem:**
   - We need to find the minimum value of \( k \) such that there exists a very good list of length 2019 with \( k \) different values.
   - A list is very good if all its sublists are good, meaning the maximum element in each sublist appears exactly once.

2. **Constructing a Very Good List:**
   - Consider the ruler sequence defined by \( \nu_2(n) + 1 \), where \( \nu_2(n) \) is the highest power of 2 dividing \( n \).
   - For example, \( \nu_2(1) = 0 \), \( \nu_2(2) = 1 \), \( \nu_2(3) = 0 \), \( \nu_2(4) = 2 \), etc.
   - The sequence \( \nu_2(1) + 1, \nu_2(2) + 1, \ldots, \nu_2(2019) + 1 \) will have values ranging from 1 to \( \nu_2(2019) + 1 \).

3. **Analyzing the Ruler Sequence:**
   - The maximum value of \( \nu_2(n) \) for \( n \leq 2019 \) is \( \nu_2(1024) = 10 \).
   - Therefore, the values in the sequence \( \nu_2(n) + 1 \) will range from 1 to 11.

4. **Verifying the Very Good Property:**
   - For any sublist of the ruler sequence, the maximum element appears exactly once because the sequence is constructed such that each value \( \nu_2(n) + 1 \) is unique for each \( n \).
   - This ensures that all sublists are good, making the entire list very good.

5. **Minimum Value of \( k \):**
   - Since the sequence \( \nu_2(n) + 1 \) uses values from 1 to 11, the minimum value of \( k \) is 11.

6. **Pigeonhole Argument for \( k \leq 10 \):**
   - If \( k \leq 10 \), then by the pigeonhole principle, some sublist would contain each value an even number of times, violating the condition for being good.
   - Therefore, \( k \leq 10 \) is not possible.

Conclusion:
\[
\boxed{11}
\]

Answer: 11


Graphically determine the intersection of the graphs of functions $f$ and $g$ with the rules $f(x)=3^{x}-1$ and $g(x)=x^{2}-2x$. Verify the solution by calculation.

Solution:
B2. We draw the asymptote $y=-1$, the intersection $z$ with the ordinate $(0,0)$, and calculate an additional point, for example, $(1,2)$, and then draw the graph of the exponential function. For the quadratic function, we calculate the roots $x_{1}=0$ and $x_{2}=2$ and the vertex $(1,1)$, and then draw its graph. We read off the intersection $(0,0)$. We verify the solution computationally: $f(0)=3^{0}-1=1-1=0$ and $g(0)=0^{2}-2 \cdot 0=0$.

Drawn asymptote $y=1$

1 point

Drawn intersection $z$ with the ordinate $(0,0)$ 1 point

Calculated and drawn roots of the quadratic function $x_{1}=0$ and $x_{2}=2$ $1+1$ points

Read off coordinates of the intersection $(0,0)$ 1 point

Computationally verified solutions

$1+1$ points

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-16.jpg?height=380&width=528&top_left_y=758&top_left_x=227)

Answer: (0,0)


Find all three-digit numbers that are 12 times the sum of their digits.

Solution:
According to the condition, the number is divisible by 3. Therefore, the sum of its digits is also divisible by 3, which means the number itself is divisible by 9.

In addition, it is divisible by 4. Therefore, we need to look for numbers that are divisible by 36. The sum of the digits of a three-digit number does not exceed 27, but the number 999 is not divisible by 12. Therefore, the sum of the digits of our number is no more than 18, and the number itself is no more than \(18 \cdot 12 = 216\). It remains to check the numbers 108, 144, 180, 216.

## Answer

108.

Answer: 108


In an isosceles triangle $ABC$, points $M$ and $N$ are located on the lateral sides $AB$ and $BC$ respectively.

Find the area of triangle $ABC$, given that $AM=5, AN=2\sqrt{37}, CM=11, CN=10$.

Solution:
Drop perpendiculars $M P$ and $N Q$ to the base $A C$; using the Pythagorean theorem, form an equation in terms of $x=A P$.

## Solution

Let $P$ and $Q$ be the projections of points $M$ and $N$ onto $A C$. Denote $A P=x$. Then, from the similarity of triangles $N Q C$ and $M P A$, we get that $Q C=2 x$.

$N C^{2}-C Q^{2}=A N^{2}-A Q^{2}$. Therefore, $A Q^{2}=148+4 x^{2}-100=48+4 x^{2}$.

Similarly, $C P^{2}=96+x^{2}$. Since $C P-A Q=(A C-x)-(A C-2 x)=x$, we have $\sqrt{96+x^{2}}-\sqrt{48+4 x^{2}}=x$. Solving this equation, we get $x=2$. Thus, $C P=10, A C=12, B C=15$. Using Heron's formula, $S_{A B C}=\sqrt{21 \cdot 6 \cdot 6 \cdot 9}=18 \sqrt{21}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3b3613da627e706c6878g-34.jpg?height=446&width=535&top_left_y=0&top_left_x=767)

## Answer

$18 \sqrt{21}$.

Answer: 18\sqrt{21}


Solution:
［Solution］Notice that $k!$ for $k \geqslant 5$ all have factors of 2 and 5, so their product is a multiple of 10.

Thus, the units digit of $S$ is the same as that of $1!+2!+3!+4!$, which is the units digit of $(1+2+6+24)$, which is 3.
Therefore, the answer is $(D)$.

Answer: D


Inside triangle \( A B C \) there are three circles \( k_{1}, k_{2}, k_{3} \) each of which is tangent to two sides of the triangle and to its incircle \( k \). The radii of \( k_{1}, k_{2}, k_{3} \) are 1, 4, and 9. Determine the radius of \( k \).

Solution:
18. Suppose that circles $k_{1}\left(O_{1}, r_{1}\right), k_{2}\left(O_{2}, r_{2}\right)$, and $k_{3}\left(O_{3}, r_{3}\right)$ touch the edges of the angles $\angle B A C, \angle A B C$, and $\angle A C B$, respectively. Denote also by $O$ and $r$ the center and radius of the incircle. Let $P$ be the point of tangency of the incircle with $A B$ and let $F$ be the foot of the perpendicular from $O_{1}$ to $O P$. From $\triangle O_{1} F O$ we obtain $\cot (\alpha / 2)=2 \sqrt{r r_{1}} /\left(r-r_{1}\right)$ and analogously $\cot (\beta / 2)=2 \sqrt{r r_{2}} /\left(r-r_{2}\right), \cot (\gamma / 2)=2 \sqrt{r r_{3}} /\left(r-r_{3}\right)$. We will now use a well-known trigonometric identity for the angles of a triangle:
$$ \cot \frac{\alpha}{2}+\cot \frac{\beta}{2}+\cot \frac{\gamma}{2}=\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2} \cdot \cot \frac{\gamma}{2} . $$
(This identity follows from $\tan (\gamma / 2)=\cot (\alpha / 2+\beta / 2)$ and the formula for the cotangent of a sum.) Plugging in the obtained cotangents, we get
$$ \begin{aligned} \frac{2 \sqrt{r r_{1}}}{r-r_{1}}+\frac{2 \sqrt{r r_{2}}}{r-r_{2}}+\frac{2 \sqrt{r r_{3}}}{r-r_{3}}= & \frac{2 \sqrt{r r_{1}}}{r-r_{1}} \cdot \frac{2 \sqrt{r r_{2}}}{r-r_{2}} \cdot \frac{2 \sqrt{r r_{3}}}{r-r_{3}} \Rightarrow \\ & \sqrt{r_{1}}\left(r-r_{2}\right)\left(r-r_{3}\right)+\sqrt{r_{2}}\left(r-r_{1}\right)\left(r-r_{3}\right) \\ & +\sqrt{r_{3}}\left(r-r_{1}\right)\left(r-r_{2}\right)=4 r \sqrt{r_{1} r_{2} r_{3}} . \end{aligned} $$
For $r_{1}=1, r_{2}=4$, and $r_{3}=9$ we get $(r-4)(r-9)+2(r-1)(r-9)+3(r-1)(r-4)=24 r \Rightarrow 6(r-1)(r-11)=0$. Clearly, $r=11$ is the only viable value for $r$.

Answer: 11


Calculate $\sqrt[3]{\frac{x}{2015+2016}}$, where $x$ is the harmonic mean of the numbers

$a=\frac{2016+2015}{2016^{2}+2016 \cdot 2015+2015^{2}}$ and $b=\frac{2016-2015}{2016^{2}-2016 \cdot 2015+2015^{2}}$.

The harmonic mean of two positive numbers $a$ and $b$ is the number $c$ such that $\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$.

Solution:
Solution:

Let's calculate

$\frac{1}{a}+\frac{1}{b}=\frac{2016^{2}+2016 \cdot 2015+2015^{2}}{2016+2015}+\frac{2016^{2}-2016 \cdot 2015+2015^{2}}{2016-2015}=$

$=\frac{(2016-2015)\left(2016^{2}+2016 \cdot 2015+2015^{2}\right)+(2016+2015)\left(2016^{2}-2016 \cdot 2015+2015^{2}\right)}{(2016+2015)(2016-2015)}$

$=\frac{2016^{3}-2015^{3}+2016^{3}+2015^{3}}{(2016+2015)(2016-2015)}=\frac{2 \cdot 2016^{3}}{4031}$, then

$\frac{1}{c}=\frac{1}{2} \cdot \frac{2 \cdot 2016^{3}}{4031}=\frac{2016^{3}}{4031}$, hence $x=\frac{4031}{2016^{3}}$

Finally, $\sqrt[3]{\frac{x}{2015+2016}}=\sqrt[3]{\frac{4031}{(2015+2016) \cdot 2016^{3}}}=\frac{1}{2016}$

Answer: $\frac{1}{2016}$.

Answer: \frac{1}{2016}


On a plane with a given rectangular Cartesian coordinate system, a square is drawn with vertices at points $(0 ; 0),(0 ; 59),(59 ; 59)$, and $(59 ; 0)$. Find the number of ways to choose two grid nodes inside this square (not including its boundary) such that at least one of these nodes lies on one of the lines $y=x$ or $y=59-x$, but neither of the chosen nodes lies on any line parallel to any of the coordinate axes.

Solution:
Answer: 370330

Solution. There are two possible cases.

1) Both selected nodes lie on the lines specified in the condition. Each of them contains 58 points inside the square, and there are no repetitions among them (the intersection point of the lines has non-integer coordinates). There are 116 ways to choose the first point, and 3 fewer ways to choose the second (all points are suitable except the first and the two points lying on the same horizontal or vertical line with the first). In this case, we counted each pair of points twice since we considered ordered pairs of points. Therefore, in this case, we get $\frac{116 \cdot 113}{2}=6554$ ways.
2) Exactly one of the selected nodes lies on the lines given in the condition. First, choose the node lying on one of the lines (116 ways). Now, let's count how many ways there are to choose the second node. In total, there are $58^{2}$ nodes marked in the square; from this, we need to exclude the nodes on the diagonals (116 nodes), as well as the nodes that lie on the same horizontal (56 nodes, considering the previously excluded diagonal nodes) or vertical (56 nodes) line with the chosen one. Therefore, the second node can be chosen in $58^{2}-116-112=$ 3136 ways, and the number of ways to choose a pair of nodes is $116 \cdot 3136=363776$.

Summarizing, we have $6554+363776=370330$ ways.

Answer: 370330


For what value of $a$ does the sum of the squares of the roots of the quadratic trinomial $x^{2}-(a-2) x-a-1$ take the smallest value?

Solution:
Solution. Based on Vieta's theorem, we have $x_{1}+x_{2}=$ $=a-2, x_{1} \cdot x_{2}=-a-1$. Then

$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(a-2)^{2}-2(-a-1)= \\
=a^{2}-2 a+6=(a-1)^{2}+5
\end{gathered}
$$

Since $(a-1)^{2} \geqslant 0$, $x_{1}^{2}+x_{2}^{2}$ takes its minimum value when $a=1$.

Remark 1. This problem can also be solved without using Vieta's theorem, by calculating the roots of the given quadratic trinomial using formulas. We have

158

$$
\begin{gathered}
x_{1}=\frac{a-2+\sqrt{ } a^{2}+8}{2}, x_{2}=\frac{a-2-\sqrt{ } a^{2}+8}{2}, \\
x_{1}^{2}+x_{2}^{2}=\left(\frac{a-2+\sqrt{ } a^{2}+8}{2}\right)^{2}+\left(\frac{a-2-\sqrt{ } a^{2}+8}{2}\right)^{2}= \\
=\frac{4 a^{2}-8 a+24}{4}=a^{2}-2 a+6=(a-1)^{2}+5 .
\end{gathered}
$$

Remark 2. The value of $a$ at which the quadratic trinomial $a^{2}-2 a+6$ obtained during the solution process takes its minimum value can also be found differently, by calculating the x-coordinate of the vertex of the corresponding parabola using the known formula. In our case, $a=\frac{2}{2}=1$.

Answer: 1


In the sequence $\left\{a_{n} \mid\right.$, $a_{7}=\frac{16}{3}$, when $n \geqslant 2$, $a_{n}=\frac{3 a_{n-1}-1}{7-a_{n-1}+4}$,
(1) Does there exist a natural number $m$, such that when $n>m$, $a_{n}<2$? If it exists, find the value of $m$; if not, explain the reason.
(2) When $n \geqslant 10$, prove that $\frac{a_{n-1}+a_{n}+1}{2}<a_{n}$.

Solution:
14. (1) From $a_{7}=\frac{16}{3}$ and the recursive formula, we can get $a_{8}-12, a_{4}=-8$.
$$
a_{n}-2=\frac{3 a_{n-1}+4}{7-a_{n-1}}-2=\frac{3\left(a_{n}-1-2\right)}{7-a_{n}-1} \text {, }
$$

Therefore, if $a_{n-1}2$.
In summary, there exists an $m$ that satisfies the condition, and $m=8$.
（2） $a_{10}=-\frac{4}{3}$, it is easy to prove that when $n \geqslant 10$, $a_{n} \geqslant-\frac{4}{3}$.
$$
\begin{array}{l}
a_{n-1}+a_{n+1}-2 a_{n} \\
=\left(\frac{7 a_{n}-4}{a_{n}+3}-a_{n}\right)+\left(\frac{3 a_{n}+4}{7-a_{n}}-a_{n}\right) \\
=\frac{-\left(a_{n}-2\right)^{2}}{a_{n}+3}+\frac{\left(a_{n}-2\right)^{2}}{7-a_{n}}=\frac{2\left(a_{n}-2\right)^{3}}{\left(7-a_{n}\right)\left(a_{n}+3\right)} .
\end{array}
$$

When $n \geqslant 10$, $a_{n}-20,7-a_{n}>0$, so $\frac{a_{n-1}+a_{n}+1}{2}<a_{n}$.

Answer: 8


Set $P=\left\{x \mid x=7^{3}+a \times 7^{2}+b \times 7+c, a, b, c$ are positive integers not exceeding 6 $\}$, if $x_{1}, x_{2}, \ldots, x_{n}$ are $n$ elements in set $P$ that form an arithmetic sequence, find the maximum value of $n$.

Solution:
Question 232, Solution: On the one hand, $7^{3}+7^{2}+7+1$, $7^{3}+7^{2}+7+2$, $7^{3}+7^{2}+7+3$, $7^{3}+7^{2}+7+4$, $7^{3}+7^{2}+7+5$, $7^{3}+7^{2}+7+6$, are six elements in set $P$ that form an arithmetic sequence, so $\mathrm{n}_{\max } \geq 6$.

On the other hand, all elements in set $P$ can be represented as four-digit numbers in base 7, with all digits being non-zero and the leading digit being 1. We can prove that any seven different elements in set $P$ do not form an arithmetic sequence.
Otherwise, if there exist seven numbers $a_{1}$, $a_{2}$, ..., $a_{7} \in P$ that form an arithmetic sequence with a common difference $d$. Note that $1 + 6d \leq a_{1} + 6d = a_{7} < 7^{4}$, so $d < 7^{4}$. Let $7^{\alpha} \| d$, then $0 \leq \alpha \leq 3$. Thus, the seven numbers $a_{1}$, $a_{1}+d$, $a_{1}+2d$, ..., $a_{1}+6d$ when represented in base 7, their $(\alpha+1)$-th digit from the right forms a complete residue system modulo 7, and thus there must be a number whose $(\alpha+1)$-th digit is 0, which means this number does not belong to $P$, leading to a contradiction.
In summary, the maximum value of $\mathrm{n}$ is 6.
Note: Similarly, we can prove the following proposition: Given a prime number $\mathrm{p}$, there exists a set $S$ such that any $\mathrm{p}$ different elements in set $S$ do not form an arithmetic sequence, and for any $a \in S$, there exist $a_{1}=a$, $a_{2}$, $a_{3}$, ..., $a_{p-1}$ as different elements in set $S$ that form an arithmetic sequence.

Proof: Define the set $S=\{x \mid x \in N, x$'s $p$-ary representation does not contain the digit $p-1\}$. We will prove that this set $S$ satisfies the given conditions.

First, for any $a \in S$, let the $p$-ary representation of $a$ be $a=\left(\overline{n_{k} n_{k-1} \ldots n_{0}}\right)_{p}$. Take $a_{i}=(i-1)p^{k+1}$, where $i=1$, $2$, ..., $p-1$, then $\left\{a_{i}\right\}_{1 \leq i \leq p-1}$ forms an arithmetic sequence, and in the $p$-ary representation of $a_{i}$, the leading digit is $i-1$, and the other digits are $n_{k}$, $n_{k-1}$, ..., $n_{0}$, none of which are $p-1$, so $a_{i} \in P$.
Next, we prove that set $S$ does not contain $p$ numbers that form an arithmetic sequence.
Otherwise, if there exist $p$ numbers $a_{1}$, $a_{2}$, ..., $a_{p} \in S$ that form an arithmetic sequence with a common difference $d$. Let $p^{\alpha} \| d$, $\alpha \in N$. Then the $p$ numbers $a_{1}$, $a_{1}+d$, $a_{1}+2d$, ..., $a_{1}+(p-1)d$ when represented in base $p$, their $(\alpha+1)$-th digit from the right forms a complete residue system modulo $p$, and thus there must be a number whose $(\alpha+1)$-th digit is $p-1$, which means this number does not belong to $S$, leading to a contradiction.
In summary, the set $S$ defined above satisfies the given conditions.

Answer: 6


Given a checkerboard $8 \times 8$, where the side length of each cell is one centimeter. A chess piece called Pythagoras, standing on cell $A$, attacks cell $B$ if the distance between the centers of cells $A$ and $B$ is five centimeters. What is the maximum number of non-attacking Pythagorases that can be placed on the board?

Answer: 32

Solution:
# Solution:

Notice that with a chessboard coloring, Pythagoras only attacks cells of the opposite color to the cell he is on.

Example: we color the board in a checkerboard pattern and place Pythagoruses on the cells of the color that is more prevalent.

Estimate: we mark the field as follows:

| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $A$ | $B$ | $C$ | $D$ | $E$ | $F$ | $G$ | $H$ |
| $I$ | $J$ | $K$ | $L$ | $M$ | $O$ | $P$ | $Q$ |
| $R$ | $S$ | $T$ | $U$ | $V$ | $W$ | $X$ | $Z$ |
| 5 | 6 | 7 | 8 | 1 | 2 | 3 | 4 |
| $E$ | $F$ | $G$ | $H$ | $A$ | $B$ | $C$ | $D$ |
| $M$ | $O$ | $P$ | $Q$ | $I$ | $J$ | $K$ | $L$ |
| $V$ | $W$ | $X$ | $Z$ | $R$ | $S$ | $T$ | $U$ |

On each pair of cells marked with the same symbol, there can be at most one Pythagoras.

## 8th Grade

#

Answer: 32


In quadrilateral $A B C D$, $\angle A=60^{\circ}, A B=1$, point $E$ is on side $A B$ such that $A E: E B=2: 1, P$ is a moving point on diagonal $A C$. Then the minimum value of $P E+P B$ is $\qquad$ .

Solution:
4. $\frac{\sqrt{7}}{3}$.

Since quadrilateral $A B C D$ is a rhombus, point $B$ and $D$ are symmetric with respect to diagonal $A C$. Therefore,
$$
P E+P B=P E+P D \geqslant D E \text {. }
$$

Equality holds if and only if $P$ is the intersection of $D E$ and $A C$.

Thus, the minimum value of $P E+P B$ is the length of segment $D E$.
As shown in Figure 6, draw $D F \perp A B$ at $F$.
From the given conditions, we easily know
$$
\begin{array}{c}
D F=\frac{\sqrt{3}}{2}, \\
A F=\frac{1}{2}, A E=\frac{2}{3}, \\
F E=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}, \\
D E=\sqrt{\left(\frac{1}{6}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\frac{\sqrt{7}}{3} .
\end{array}
$$

Therefore, the minimum value of $P E+P B$ is $\frac{\sqrt{7}}{3}$.

Answer: \frac{\sqrt{7}}{3}


Find the number of positive integers $n$ for which

(i) $n \leq 1991$;

(ii) 6 is a factor of $(n^2 + 3n +2)$.

Solution:
1. We start with the given conditions:
   - \( n \leq 1991 \)
   - \( 6 \) is a factor of \( n^2 + 3n + 2 \)

2. We need to find the number of positive integers \( n \) such that \( n^2 + 3n + 2 \) is divisible by \( 6 \).

3. First, we factorize the quadratic expression:
   \[
   n^2 + 3n + 2 = (n+1)(n+2)
   \]

4. For \( (n+1)(n+2) \) to be divisible by \( 6 \), it must be divisible by both \( 2 \) and \( 3 \).

5. Consider divisibility by \( 2 \):
   - Since \( n \) is an integer, either \( n \) is even or odd.
   - If \( n \) is even, \( n+1 \) is odd and \( n+2 \) is even.
   - If \( n \) is odd, \( n+1 \) is even and \( n+2 \) is odd.
   - Therefore, \( (n+1)(n+2) \) is always even, satisfying the divisibility by \( 2 \).

6. Consider divisibility by \( 3 \):
   - We need \( (n+1)(n+2) \equiv 0 \pmod{3} \).
   - This implies that at least one of \( n+1 \) or \( n+2 \) must be divisible by \( 3 \).
   - Therefore, \( n \) must not be congruent to \( 0 \pmod{3} \).

7. We now count the number of integers \( n \) from \( 1 \) to \( 1991 \) that are not multiples of \( 3 \):
   - The total number of integers from \( 1 \) to \( 1991 \) is \( 1991 \).
   - The number of multiples of \( 3 \) in this range is \( \left\lfloor \frac{1991}{3} \right\rfloor = 663 \).

8. Subtracting the number of multiples of \( 3 \) from the total number of integers gives:
   \[
   1991 - 663 = 1328
   \]

Conclusion:
The number of positive integers \( n \) for which \( n \leq 1991 \) and \( 6 \) is a factor of \( n^2 + 3n + 2 \) is \( \boxed{1328} \).

Answer: 1328


A positive integer $N$ is appended to the right of any positive integer (for example, appending 21 to the right of 35 yields 3521). If the new number is always divisible by $N$, then $N$ is called a âmagic numberâ. How many âmagic numbersâ are there among the positive integers less than 600?

Solution:
Number Theory, Divisors and Multiples.
(1) If $N$ is a one-digit number;

Then by the problem, we get $N \mid 10 m + N$, so $N \mid 10 m$;
Since $m$ is any positive integer; thus $N \mid 10$;
$10 = 2 \times 5$ has $(1+1) \times (1+1) = 4$ divisors: $1, 2, 5, 10$;
Therefore, $N = 1, 2, 5$.
(2) If $N$ is a two-digit number;

Then by the problem, we get $N \mid 100 m + N$, so $N \mid 100 m$;
Since $m$ is any positive integer; thus $N \mid 100$;
$100 = 2^2 \times 5^2$ has $(2+1) \times (2+1) = 9$ divisors: $1, 2, 4, 5, 10, 20, 25, 50, 100$;
Therefore, $N = 10, 20, 25, 50$.
(3) If $N$ is a three-digit number;

Then by the problem, we get $N \mid 1000 m + N$, so $N \mid 1000 m$;
Since $m$ is any positive integer; thus $N \mid 1000$;
$1000 = 2^3 \times 5^3$ has $(3+1) \times (3+1) = 16$ divisors:
$1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000$; Therefore, $N = 100, 125, 200, 250, 500$.
In summary, there are 12 "magic numbers" among the positive integers less than 600.

Answer: 12


Given $S$ as a binary string of $10^{4}$ bits containing only $0$ and $1$, a positive integer $k \leqslant 10^{4}$, a $k$-block of $S$ is a substring of $S$ consisting of $k$ consecutive bits. Two $k$-blocks $a_{1} a_{2} \cdots a_{k}=b_{1} b_{2} \cdots b_{k}$ are equal if and only if $a_{i}=b_{i}(i=1$, $2, \cdots, k)$. Consider all $10^{4}$-bit binary strings that contain at most seven different 3-blocks. Find the maximum number of different 10-blocks such a string can contain.

Solution:
12. The maximum value sought is 504, and the examples that achieve this value include all seven possible 3-blocks except 000.

Let $f(k)$ denote the maximum number of $k$-bit strings that can contain at most seven different 3-blocks.
Clearly, $f(1)=2, f(2)=4, f(3)=7$.
Next, we prove that
$$
f(k) \leqslant f(k-1)+f(k-2)+f(k-3),
$$

with equality holding when the seven possible 3-blocks do not include 000.
By recursion, we get $f(10)=504$.
First, construct a $10^4$-bit string that contains seven different 3-blocks and 504 10-blocks. Add a 1 after each of the 504 10-blocks, then combine these in some order to form a $504 \times 11$-bit string, and append $10^4 - 504 \times 11$ 1s to this string.

Since there are $2^3=8$ possible 3-blocks, assume that a certain 3-block: $abc$ does not appear in any $k$-bit string.

Note that there can be at most $f(k-1)$ strings that do not end in $c$; at most $f(k-2)$ strings that end in $xc$, where $x \neq b$; and at most $f(k-3)$ strings that end in $xbc$, where $x \neq a$.
Therefore, $f(k) \leqslant f(k-1)+f(k-2)+f(k-3)$.

Answer: 504


For a positive integer of three digits, we can multiply the three digits with each other. The result of this we call the digit product of that number. So, 123 has a digit product of $1 \times 2 \times 3=6$ and 524 has a digit product of $5 \times 2 \times 4=40$. A number cannot start with the digit 0. Determine the three-digit number that is exactly five times as large as its own digit product.

Solution:
B2. 175 We assume that the three-digit number consists of the digits $a, b$, and $c$, and is therefore equal to $100 a + 10 b + c$. The digit product of this number is $a \cdot b \cdot c$. Therefore, it must hold that
$100 a + 10 b + c = 5 \cdot a \cdot b \cdot c$. Since the right-hand side of this equation is a multiple of five, it must be that $c=0$ or $c=5$. If $c=0$, then $5 \cdot a \cdot b \cdot c=0$ and that cannot be, because we started with a number not equal to zero. Therefore, $c=5$. This means that $5 \cdot a \cdot b \cdot c$ is a multiple of 25 that ends in a 5, so it ends in 25 or 75, so $b=2$ or $b=7$. Moreover, it follows from the fact that $100 a + 10 b + c = 5 a b c$ ends in a 5 that $5 a b c$ is odd, so $a, b$, and $c$ are all odd, so $b=7$. Substituting the found values for $b$ and $c$ into the first equation, we get $100 a + 75 = 175 a$. The only solution to this equation is $a=1$. The only three-digit number that is exactly five times its digit product is therefore 175.

Answer: 175


Tangents are drawn from point $A$ to a circle with a radius of 10 cm, touching the circle at points $B$ and $C$ such that triangle $A B C$ is equilateral. Find its area.

Solution:
Solution: $\triangle A B C$ is equilateral, $A C, A B$ are tangents, $R=10, A B=$ ?

$\angle A C B=60^{\circ}, \angle A O C=90^{\circ}(A C \perp O C), \angle O C B=30^{\circ}$,

$B C=R \cos 30^{\circ} \cdot 2=R \sqrt{3}=10 \sqrt{3}, A B=B C=A C=10 \sqrt{3}$,

$S_{\triangle A B C}=\frac{A B^{2} \sqrt{3}}{4}=\frac{100 \cdot 3 \sqrt{3}}{4}=75 \sqrt{3}$.

Answer: $S_{\triangle A B C}=75 \sqrt{3}$.

Answer: 75\sqrt{3}


Given a package containing 200 red marbles, 300 blue marbles and 400 green marbles. At each occasion, you are allowed to withdraw at most one red marble, at most two blue marbles and a total of at most five marbles out of the package. Find the minimal number of withdrawals required to withdraw all the marbles from the package.

Solution:
28. Answer: 200 .

Since at most one red marble can be withdrawn each time, it requires at least 200 withdrawals.
On the other hand, 200 is possible. For example,
$$
150 \cdot(1,2,2)+20 \cdot(1,0,4)+10 \cdot(1,0,2)+20 \cdot(1,0,0)=(200,300,400) .
$$

Answer: 200


List all four-digit numbers that can be formed using the digits 0 and 9.

Solution: 6. $9000,9009,9090,9900,9099,9909,9990,9999$ (8 numbers).

Answer: 9000,9009,9090,9900,9099,9909,9990,9999


The real numbers $a$ and $b$, where $a>b$, are solutions to the equation $3^{2x}-10 \times 3^{x+1}+81=0$. What is the value of $20a^2+18b^2$?

Solution:
SOlUTION
198

In the equation $3^{2 x}-10 \times 3^{x+1}+81=0$, replace $3^{x}$ with $y$. The equation becomes $y^{2}-10 \times 3 \times y+81=0$. This factorises as $(y-3)(y-27)=0$ with solutions $y=3,27$. This means $3^{x}=3$ or $3^{x}=27$. The $x$-values are 1,3 respectively, so $a=3$ and $b=1$. The value of $20 a^{2}+18 b^{2}=20 \times 9+18 \times 1=198$.

Answer: 198


In Rt $\triangle A B C$, $\angle C=90^{\circ}, B C=5$, and the other two sides $A B$ and $A C$ are of integer lengths. Extend $A C$ to $D$, and make $\angle A B D=90^{\circ}$ to get $\triangle A B D$. Then the circumradius $R$ of $\triangle A B D$ is $\qquad$

Solution:
$2.7 \frac{1}{24}$.
By the Pythagorean theorem, we get $A B^{2}-A C^{2}=25$, which means $(A B+A C)(A B-A C)=25$.
Since the positive integer factors of 25 are $1,5,25$, we have
$$
\left\{\begin{array}{l}
A B+A C=25, \\
A B-A C=1 .
\end{array}\right.
$$

Solving, we get $A B=13, A C=12$.
By the projection theorem, we get
$$
A D=\frac{A B^{2}}{A C}=14 \frac{1}{12} \text {. }
$$

Therefore, $R=\frac{1}{2} A D=7 \frac{1}{24}$.

Answer: 7 \frac{1}{24}


Let $x$ be a natural number. Solve the equation

$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x} \cdots+\frac{1}{x}=3
$$

Solution:
Solution. Multiply both sides of the equation by $x$. We get

$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$

The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1$, and $a_{n}=1$. The progression has $(x-1)$ terms.

Since $S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n$, the equation can be rewritten as

$$
\frac{x-1+1}{2} \cdot(x-1)=3 x \quad \Rightarrow \quad x^{2}=7 x \quad \Rightarrow \quad x=7(x \neq 0)
$$

Answer. $x=7$

Answer: 7


The real solutions of the equation $x^{2}-6[x]+5=0$ are

Solution: 9. $1, \sqrt{7}, \sqrt{13}, \sqrt{19}, 5$ Let $[x]=n$, then $x^{2}-6 n+5=0, 6 n=x^{2}+5>0, n>0$. From $[x]=n \leqslant x$ we know, $n^{2}+5 \leqslant 6 n<(n+1)^{2}+5$, thus $n \in[1,5]$. Substituting $n=1,2,3,4,5$ respectively will do.

Answer: 1,\sqrt{7},\sqrt{13},\sqrt{19},5


The sum of the radii of all circles passing through point $A(1505,1008)$ and tangent to the lines $l_{1}: y=0$ and $l_{2}: y=\frac{4}{3} x$ is $\qquad$

Solution:
5. 2009 .

From $\tan 2 \theta=\frac{4}{3} \Rightarrow \tan \theta=\frac{1}{2}$, the center of the circle lies on the line $y=\frac{x}{2}$, and the center is $(2 r, r)$ (where $r$ is the radius of the circle).
$$
\begin{array}{l}
\text { Hence }(1505-2 r)^{2}+(1008-r)^{2}=r^{2} \\
\Rightarrow 4 r^{2}-8036 r+1505^{2}+1008^{2}=0 \\
\Rightarrow r_{1}+r_{2}=2009 .
\end{array}
$$

Answer: 2009


Solve the system of equations $\left\{\begin{array}{l}2 x+\sqrt{2 x+3 y}-3 y=5, \\ 4 x^{2}+2 x+3 y-9 y^{2}=32 .\end{array}\right.$

Solution:
Answer: $\left(\frac{17}{4} ; \frac{5}{2}\right)$.

Solution. Let $\sqrt{2 x+3 y}=u, 2 x-3 y=v$. Then the system becomes

$$
\left\{\begin{array} { l }
{ u + v = 5 , } \\
{ u ^ { 2 } v + u ^ { 2 } = 3 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=5-u \\
u^{2}(5-u)+u^{2}=32
\end{array}\right.\right.
$$

"Phystech-2016", mathematics, solution to ticket 3

From the second equation of the last system, it follows that $u^{3}-6 u^{2}+32=0$. By selecting the integer root $u=-2$ and factoring out the factor $(u+2)$ from the left side of the last equation, we get $(u+2)\left(u^{2}-8 u+16\right)=0$, from which $u=-2$ or $u=4$. The value $u=-2$ is not suitable. For $u=4$, we get $v=1$. Then

$$
\left\{\begin{array} { l }
{ 2 x + 3 y = 1 6 } \\
{ 2 x - 3 y = 1 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 4 x = 1 7 , } \\
{ 6 y = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{17}{4} \\
y=\frac{5}{2}
\end{array}\right.\right.\right.
$$

Answer: (\frac{17}{4};\frac{5}{2})


Let the edge length of a regular tetrahedron be $2 \sqrt{6}$, and let a sphere be constructed with the center $O$ of the tetrahedron as its center. The total length of the curves formed by the intersection of the sphere's surface with the four faces of the tetrahedron is $4 \pi$. Then the radius of the sphere $O$ is $\qquad$

Solution:
7. $\frac{\sqrt{5}}{2}$ or $\sqrt{5}$.

Let the radius of sphere $O$ be $R$.
If a face of the regular tetrahedron intersects the sphere as shown in Figure 3, then the circumference of the small circle is $\pi$, and the radius of the small circle is $\frac{1}{2}$.

Also, the distance from the center of the sphere to the face of the tetrahedron is 1, so
$$
\begin{aligned}
R & =\sqrt{1^{2}+\left(\frac{1}{2}\right)^{2}} \\
& =\frac{\sqrt{5}}{2} .
\end{aligned}
$$

If a face of the regular tetrahedron intersects the sphere as shown in Figure 4, let $D$ be the midpoint of $A C$.

According to the problem, the length of arc $\overparen{A B}$ is $\frac{\pi}{3}$.

Let the radius of the small circle $\odot O_{1}$ be $r$.
$$
\begin{array}{l}
\text { Then } \angle A O_{1} B=\frac{\pi}{3 r} . \\
\text { Also } \angle B O_{1} C=\frac{2 \pi}{3}, O_{1} D=\sqrt{2}, \\
\angle A O_{1} D=\frac{1}{2}\left(\angle B O_{1} C-\angle A O_{1} B\right) \\
=\frac{\pi}{3}-\frac{\pi}{6 r},
\end{array}
$$

Thus, $\cos \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)=\frac{\sqrt{2}}{r}$.
Let $f(r)=\cos \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)-\frac{\sqrt{2}}{r}$.
Then $f^{\prime}(r)=-\frac{\pi}{6 r^{2}} \sin \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)+\frac{\sqrt{2}}{r^{2}}>0$.
Therefore, the function $f(r)$ is monotonically increasing on the interval $(0,+\infty)$ and has at most one zero.
Since $f(2)=0$, the equation (1) has a unique solution 2.
Thus, $R=\sqrt{r^{2}+1}=\sqrt{5}$.
In summary, the radius of sphere $O$ is $\frac{\sqrt{5}}{2}$ or $\sqrt{5}$.

Answer: \frac{\sqrt{5}}{2}


Determine the coefficient of $x^{3}$ in the expanded form (after the powers have been carried out) of the expression

$$
(1+x)^{3}+(1+x)^{4}+\ldots+(1+x)^{1995}
$$

Solution:
Solution. According to the formula for the sum of a geometric progression, we get:

$$
(1+x)^{3}+(1+x)^{4}+\ldots+(1+x)^{1995}=\frac{(1+x)^{3}\left[(1+x)^{1993}-1\right]}{x}=\frac{(1+x)^{1996}-(1+x)^{3}}{x}
$$

Since the denominator contains $x$, it is sufficient to determine the coefficient of $x^{4}$ in the numerator. However, $x^{4}$ does not appear in $(1+x)^{3}$, so it is sufficient to determine the coefficient of $x^{4}$ in $(1+x)^{1996}$.

Thus, the desired coefficient is $\binom{1996}{4}$.

Answer: \binom{1996}{4}


A science and technology innovation competition sets first, second, and third prizes (all participants will receive an award), and the probabilities of winning the corresponding prizes form a geometric sequence with the first term $a$ and a common ratio of 2. The corresponding prizes form an arithmetic sequence with the first term of 700 yuan and a common difference of -140 yuan. The expected prize money for participating in this competition is $\qquad$ yuan.

Solution:
3. 500 .

Let the prize money obtained be $\xi$ yuan. Then $\xi=700,560,420$. From the problem, we know
$$
\begin{array}{l}
P(\xi=700)=a, P(\xi=560)=2 a, \\
P(\xi=420)=4 a .
\end{array}
$$

From $7 a=1$, we get $a=\frac{1}{7}$.
Therefore, $E \xi=700 \times \frac{1}{7}+560 \times \frac{2}{7}+420 \times \frac{4}{7}$ $=500$ (yuan).

Answer: 500


Suppose $a$, $b$, $c$ are real numbers such that $a+b+c = 0$ and $a^{2}+b^{2}+c^{2} = 1$.
Prove that $a^{2}b^{2}c^{2}\leq \frac{1}{54}$ and determine the cases of equality.

Solution:
1. Given the conditions:
   \[
   a + b + c = 0
   \]
   \[
   a^2 + b^2 + c^2 = 1
   \]
   We need to prove that \(a^2 b^2 c^2 \leq \frac{1}{54}\) and determine the cases of equality.

2. From the first condition, we have:
   \[
   c = -(a + b)
   \]
   Substituting \(c\) into the second condition:
   \[
   a^2 + b^2 + (-(a + b))^2 = 1
   \]
   Simplifying:
   \[
   a^2 + b^2 + (a + b)^2 = 1
   \]
   Expanding \((a + b)^2\):
   \[
   a^2 + b^2 + a^2 + 2ab + b^2 = 1
   \]
   \[
   2a^2 + 2b^2 + 2ab = 1
   \]
   Dividing by 2:
   \[
   a^2 + b^2 + ab = \frac{1}{2}
   \]

3. Let \(t = (a + b)^2\). Then:
   \[
   a^2 b^2 c^2 = (ab)^2 (a + b)^2 = (ab)^2 t
   \]
   From the equation \(a^2 + b^2 + ab = \frac{1}{2}\), we can express \(ab\) in terms of \(t\):
   \[
   a^2 + b^2 = t - 2ab
   \]
   Substituting into the equation \(a^2 + b^2 + ab = \frac{1}{2}\):
   \[
   t - 2ab + ab = \frac{1}{2}
   \]
   \[
   t - ab = \frac{1}{2}
   \]
   \[
   ab = t - \frac{1}{2}
   \]

4. Substituting \(ab\) back into \(a^2 b^2 c^2\):
   \[
   a^2 b^2 c^2 = (ab)^2 t = \left(t - \frac{1}{2}\right)^2 t
   \]
   Let \(f(t) = \left(t - \frac{1}{2}\right)^2 t\). We need to find the maximum value of \(f(t)\) for \(t\) in the interval \([0, \frac{2}{3}]\).

5. To find the critical points, we take the derivative of \(f(t)\):
   \[
   f(t) = t \left(t - \frac{1}{2}\right)^2
   \]
   \[
   f'(t) = \left(t - \frac{1}{2}\right)^2 + 2t \left(t - \frac{1}{2}\right)
   \]
   \[
   f'(t) = \left(t - \frac{1}{2}\right) \left(3t - \frac{1}{2}\right)
   \]
   Setting \(f'(t) = 0\):
   \[
   \left(t - \frac{1}{2}\right) \left(3t - \frac{1}{2}\right) = 0
   \]
   \[
   t = \frac{1}{2} \quad \text{or} \quad t = \frac{1}{6}
   \]

6. Evaluating \(f(t)\) at the critical points and endpoints:
   \[
   f\left(\frac{1}{6}\right) = \left(\frac{1}{6} - \frac{1}{2}\right)^2 \cdot \frac{1}{6} = \left(-\frac{1}{3}\right)^2 \cdot \frac{1}{6} = \frac{1}{9} \cdot \frac{1}{6} = \frac{1}{54}
   \]
   \[
   f\left(\frac{2}{3}\right) = \left(\frac{2}{3} - \frac{1}{2}\right)^2 \cdot \frac{2}{3} = \left(\frac{1}{6}\right)^2 \cdot \frac{2}{3} = \frac{1}{36} \cdot \frac{2}{3} = \frac{1}{54}
   \]

7. Therefore, the maximum value of \(f(t)\) is \(\frac{1}{54}\).

8. Equality holds if and only if \(t = \frac{1}{6}\) or \(t = \frac{2}{3}\). This corresponds to the cases:
   \[
   \{a, b, c\} = \left\{\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}\right\} \quad \text{or} \quad \{a, b, c\} = \left\{-\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right\}
   \]

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{54} } \)

Answer: \frac{1}{54}


Let $n$ be a positive integer with $d$ digits, all different from zero. For $k = 0,..., d - 1$, we define $n_k$ as the number obtained by moving the last $k$ digits of $n$ to the beginning. For example, if $n = 2184$ then $n_0 = 2184, n_1 = 4218, n_2 = 8421, n_3 = 1842$. For $m$ a positive integer, define $s_m(n)$ as the number of values $k$ such that $n_k$ is a multiple of $m.$ Finally, define $a_d$ as the number of integers $n$ with $d$ digits all nonzero, for which $s_2 (n) + s_3 (n) + s_5 (n) = 2d.$
Find  \[\lim_{d \to \infty} \frac{a_d}{5^d}.\]

Solution:
1. **Understanding the Problem:**
   - We are given a number \( n \) with \( d \) digits, all different from zero.
   - We define \( n_k \) as the number obtained by moving the last \( k \) digits of \( n \) to the beginning.
   - We need to find the number of values \( k \) such that \( n_k \) is a multiple of \( m \) for \( m = 2, 3, 5 \).
   - We define \( s_m(n) \) as the number of such \( k \) values.
   - We need to find \( a_d \), the number of \( d \)-digit numbers \( n \) for which \( s_2(n) + s_3(n) + s_5(n) = 2d \).
   - Finally, we need to find the limit \(\lim_{d \to \infty} \frac{a_d}{5^d}\).

2. **Analyzing \( n_k \):**
   - For \( n_k \) to be a multiple of 2, its last digit must be even.
   - For \( n_k \) to be a multiple of 3, the sum of its digits must be divisible by 3.
   - For \( n_k \) to be a multiple of 5, its last digit must be 0 or 5. However, since all digits are nonzero, the last digit must be 5.

3. **Conditions for \( s_2(n) + s_3(n) + s_5(n) = 2d \):**
   - Since \( s_5(n) \) counts the number of \( k \) values for which \( n_k \) is a multiple of 5, and the last digit of \( n_k \) must be 5, there must be exactly \( d \) such \( k \) values.
   - Similarly, for \( s_2(n) \) and \( s_3(n) \), we need to ensure that the conditions are met for \( d \) values of \( k \).

4. **Constructing \( n \):**
   - To satisfy \( s_5(n) = d \), one of the digits must be 5.
   - To satisfy \( s_2(n) = d \), half of the digits must be even (2, 4, 6, 8).
   - To satisfy \( s_3(n) = d \), the sum of the digits must be divisible by 3.

5. **Counting Valid \( n \):**
   - We need to count the number of \( d \)-digit numbers with the above properties.
   - The total number of \( d \)-digit numbers with nonzero digits is \( 9^d \).
   - The number of valid \( n \) is a fraction of \( 9^d \).

6. **Finding the Limit:**
   - We need to find the limit \(\lim_{d \to \infty} \frac{a_d}{5^d}\).
   - Since \( a_d \) is a fraction of \( 9^d \), we need to determine the fraction.

7. **Conclusion:**
   - Given the constraints and the properties of the digits, the fraction of valid \( n \) is \(\frac{1}{3}\).

The final answer is \(\boxed{\frac{1}{3}}\).

Answer: \frac{1}{3}


Find the positive integer $n\,$ for which
\[\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994\]
(For real $x\,$, $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$)

Solution:
Note that if $2^x \le a1994$. Thus, $k=8$.
So, $\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}$.
Alternatively, one could notice this is an [arithmetico-geometric series](https://artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series) and avoid a lot of computation.

Answer: 312


The G20 is an international economic cooperation forum, with 20 members from Asia, Europe, Africa, Oceania, and America. Among them, Asia has the most, Africa and Oceania have the same and the least, and the number of members from America, Europe, and Asia are consecutive natural numbers. Then the number of members from Asia in the G20 is . $\qquad$

Solution:
【Answer】Solution: According to the problem, the number of members in America is in the middle, and $20 \div 5=4$ (people), so America has 4 or 5 members,

When America has 4 members, Europe has 5 members, and Asia has 6 members, the total number of members in these three continents is $4+5+6=15$ (people), so the total number of members in Africa and Oceania is $20-15=5$ (people), and since the number of members in Africa and Oceania is the same, the sum of the members in these two continents must be an even number, so this situation does not meet the requirements and is discarded. When America has 5 members, Europe has 6 members, and Asia has 7 members, the total number of members in these three continents is $5+6+7=18$ (people),

So, the total number of members in Africa and Oceania is $20-18=2$ (people), and since the number of members in Africa and Oceania is the same,

The sum of the members in these two continents must be an even number, at this time, Africa and Oceania each have 1 member, so, Asia has 7 members, the answer is: 7.

Answer: 7