problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
14.10. Prove that $C_{n+m}^{k}=C_{m}^{0} C_{n}^{k}+C_{m}^{1} C_{n}^{k-1}+\ldots+C_{m}^{k} C_{n}^{0}$ (Vandermonde). | 14.10. Compare the coefficients of $x^{k}$ on both sides of the equality $(x+1)^{m+n}=(x+1)^{m}(x+1)^{n}$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,518 |
14.11. Prove that $C_{2 n}^{n}=\left(C_{n}^{0}\right)^{2}+\left(C_{n}^{1}\right)^{2}+\ldots+\left(C_{n}^{n}\right)^{2}$. | 14.11. Apply the identity from problem 14.10 for $n=m=k$ and use the fact that $C_{n}^{i}=C_{n}^{n-i}$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,519 |
14.12. Prove that $C_{2 n}^{n+k}=\sum_{i} 2^{n-k-2 i} C_{n}^{k} C_{n-i}^{i+k}$. | 14.12. The coefficient of $x^{n-k}$ in the expansion of $(1+x)^{2 n}$ is $C_{2 n}^{n-k}=C_{2 n}^{n+k}$. On the other hand,
$$
(1+x)^{2 n}=\left(x^{2}+(1+2 x)\right)^{n}=\sum_{i} C_{n}^{i} x^{2 i}(1+2 x)^{n-i}=\sum_{i} C_{n}^{i} x^{2 i} \sum_{j} C_{n-i}^{j} 2^{j} x^{j}
$$
In the last expression, the coefficient of $x^... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,520 |
14.13. Prove that $C_{n-1}^{k-1} C_{n}^{k+1} C_{n+1}^{k}=C_{n-1}^{k} C_{n+1}^{k+1} C_{n}^{k-1}$. | 14.13. It is required to prove that
$$
\begin{aligned}
& \frac{(n-1)!}{(k-1)!(n-k)!} \cdot \frac{n!}{(k+1)!(n-k-1)!} \cdot \frac{(n+1)!}{k!(n-k+1)!}= \\
& =\frac{(n-1)!}{k!(n-k-1)!} \cdot \frac{(n+1)!}{(k+1)!(n-k)!} \cdot \frac{n!}{(k-1)!(n-k+1)!} .
\end{aligned}
$$
This equality is obvious. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,521 |
14.14. Prove that if $p+q=1$, then
$$
\sum_{0 \leqslant r \leqslant n / 2}(-1)^{r} C_{n-r}^{r} p^{r} q^{r}=\frac{p^{n+1}-q^{n+1}}{p-q}
$$ | 14.14. Let $S_{n}=\sum_{0 \leqslant r \leqslant n / 2}(-1)^{r} C_{n-r}^{r} p^{r} q^{r}$. In fact, the summation can be carried out over all $r$, since if $r>n / 2$, then $n-r<r$, and therefore $C_{n-r}^{r}=0$. Therefore,
$S_{n+1}-S_{n}=\sum(-1)^{r}\left(C_{n+1-r}^{r}-C_{n-r}^{r}\right) p^{r} q^{r}=\sum(-1)^{r} C_{n-r}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,522 |
14.15. Let $P(x)$ be a polynomial of degree $n$, and $P(x) = 2^x$ for $x = 1, 2, \ldots, n+1$. Compute $P(n+2)$. | 14.15. Answer: $P(n+2)=2^{n+2}-2$.
The expression $C_{m}^{x-1}=\frac{(x-1)(x-2) \ldots(x-m)}{m!}$ can be considered as a polynomial of degree $n$ in the variable $x$. Let
$$
f(x)=2\left(C_{x-1}^{0}+C_{x-1}^{1}+\ldots+C_{x-1}^{n}\right)
$$
It is clear that $C_{x-1}^{0}+C_{x-1}^{1}+\ldots+C_{x-1}^{n}=(1+1)^{x-1}=2^{x-... | 2^{n+2}-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,523 |
14.16. Prove that
$$
\sum_{m=1}^{n} m C_{n}^{m}=n 2^{n-1} \quad \text { and } \quad \sum_{m=1}^{n} m^{2} C_{n}^{m}=n(n+1) 2^{n-2}
$$ | 14.16. Differentiate the identity $(1+x)^{n}=\sum_{m=0}^{n} x^{m} C_{n}^{m}$ and multiply the resulting equation by $x$. We obtain $n x(1+x)^{n-1}=\sum_{m=1}^{n} m x^{m} C_{n}^{m}$; setting $x=1$ gives the first required identity. Repeating the differentiation with respect to $x$ and multiplying by $x$ again, we get
$... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,524 |
14.17. Prove that
$$
\sum_{m=1}^{n} \frac{1}{m+1} C_{n}^{m}=\frac{2^{n+1}-1}{n+1} \quad \text { and } \quad \sum_{m=1}^{n} \frac{(-1)^{m}}{m+1} C_{n}^{m}=\frac{1}{n+1}
$$ | 14.17. Let's write the identity $(1+t)^{n}=\sum_{m=0}^{n} t^{m} C_{n}^{m}$ and integrate it:
$$
\int_{0}^{x}(1+t)^{n} d t=\sum_{m=0}^{n} \int_{0}^{x} t^{m} C_{n}^{m} d t
$$
i.e.,
$$
\frac{(1+x)^{n+1}-1}{n+1}=\sum_{m=0}^{n} \frac{x^{m+1}}{m+1} C_{n}^{m}
$$
For $x=1$, we obtain the first required identity, and for $x... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,525 |
14.18. Prove that if $|x|<1$, then
$$
\left(\frac{1}{1-x}\right)^{n}=1+\sum_{k=1}^{\infty} C_{n+k-1}^{n-1} x^{k}
$$
See also problems $29.53,29.54$.
## 14.3. Binomial Theorem in Arithmetic | 14.18. We apply induction on $n$. For $n=1$, we obtain the usual formula for the sum of an infinite geometric series.
To move from $n$ to $n+1$, we need to prove the identity
$$
\left(1+\sum_{k=1}^{\infty} C_{n+k}^{n} x^{k}\right)(1-x)=1+\sum_{k=1}^{\infty} C_{n+k-1}^{n-1} x^{k}
$$
The coefficients of $x^{k}$ in the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,526 |
14.19. Prove that the numbers $(a+1)^{a^{n}}-1$ and $(a-1)^{a^{n}}+1$ are divisible by $a^{n+1}$ and not divisible by $a^{n+2}$. | 14.19. We will provide a proof only for the first number (the reasoning for the second number is analogous). For \( n=0 \), the statement is obvious. For \( n=1 \), we get \((1+a)^{a}-1 \equiv a \cdot a + C_{a}^{2} a^{2} \left(\bmod a^{3}\right)\). The number \( a^{2} + C_{a}^{2} a^{2} = a^{2} \left(\frac{a^{2} + a + 1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,527 |
14.20. Let $p_{1}=2, p_{2}=3, \ldots$ be the sequence of prime numbers. Prove that $p_{1} \ldots p_{k} \leqslant 4^{p_{k}}$.
## 14.4. Combinatorics in Arithmetic | 14.20. We will prove by induction on $n$ that if $p_{k} \leqslant n$, where $n \geqslant 2$, then $p_{1} \ldots p_{k} \leqslant 4^{n}$. For $n=2$, the required statement is obvious. Suppose it is proven for all numbers not exceeding $n-1$. To make the induction step, we will consider two cases separately.
Let $n=2 m$.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,528 |
14.21. How many four-digit numbers (from 0001 to 9999) exist where the sum of the first two digits is equal to the sum of the last two digits? | 14.21. Answer: 669. Let the sum of the first two digits be $n$, and the sum of the last two digits also be $n$. The value of $n$ ranges from 1 to 18. If the number of two-digit numbers whose digits sum to $n$ is $a_{n}$, then the desired number is $a_{1}^{2}+a_{2}^{2}+\ldots+a_{18}^{2}$. A two-digit number whose digits... | 669 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,529 |
14.22. How many pairs of integers $x, y$, lying between 1 and 1000, are there such that $x^{2}+y^{2}$ is divisible by $7 ?$ | 14.22. Answer: $142^{2}=20164$. The number $x^{2}+y^{2}$ is divisible by 7 if and only if both numbers $x$ and $y$ are divisible by 7. Indeed, the square of an integer when divided by 7 gives remainders of $0,1,2$, and 4. The number of integers between 1 and 1000 that are divisible by 7 is 142. Therefore, the desired n... | 20164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,530 |
14.23. How many natural numbers less than a thousand are there that are not divisible by 5 or 7? | 14.23. Answer: 686 numbers. First, we will cross out from the set of numbers $1,2, \ldots, 999$ the numbers that are multiples of 5; their quantity is $\left[\frac{999}{5}\right]=199$. Then, from the same set of numbers $1,2, \ldots, 999$, we will cross out the numbers that are multiples of 7; their quantity is $\left[... | 686 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,531 |
14.24. How many different integer solutions does the inequality $|x|+|y|<100$ have? | 14.24. Answer: 338350. The equations $|x|+|y|=0,|x|+|y|=1$, $|x|+|y|=2, \ldots,|x|+|y|=99$ have, respectively, $1,2^{2}, 3^{2}, \ldots$ $\ldots, 100^{2}$ integer solutions. Therefore, the desired number is $1^{2}+2^{2}+3^{2}+\ldots+100^{2}=\frac{100 \cdot 101 \cdot 201}{6}$ (see problem 9.12). | 338350 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 38,532 |
14.25. How many natural numbers $x$, less than 10000, exist for which $2^{x}-x^{2}$ is divisible by $7 ?$ | 14.25. Answer: 2857. The remainders of $2^{x}$ and $x^{2}$ when divided by 7 repeat with periods of 3 and 7, respectively, so the remainders of $2^{x}-x^{2}$ when divided by 7 repeat with a period of 21. Among the numbers $x$ from 1 to 21, the remainders of $2^{x}$ and $x^{2}$ when divided by 7 are equal for exactly 6 ... | 2857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,533 |
14.26. Prove that $\operatorname{GCD}\left(a_{1}, \ldots, a_{n}\right)=P_{\text {odd }} / P_{\text {even }}$, where $P_{\text {odd }}$ is the product of the LCM of all sets consisting of an odd number of different numbers $a_{1}, \ldots, a_{n}$, and $P_{\text {even }}$ is the product of the LCM of all sets consisting o... | 14.26. This problem is a generalization of problem 4.19, the solution of which shows that it is sufficient to consider the case when $a_{1}=p^{\alpha_{1}}, \ldots, a_{n}=p^{\alpha_{n}}$ and $\alpha_{1} \leqslant \alpha_{2} \leqslant \ldots \leqslant \alpha_{n}$.
The factor $p^{\alpha_{1}}$ appears only in $P_{\text {o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,534 |
14.27. Given 6 digits: $0,1,2,3,4,5$. Find the sum of all four-digit even numbers that can be written using these digits (the same digit can be repeated in a number). | 14.27. Answer: 1769580. We will separately calculate the sum of thousands, hundreds, tens, and units for the considered numbers. The first digit can be any of the five digits $1,2,3,4,5$. The number of all numbers with a fixed first digit is $6 \cdot 6 \cdot 3=108$, since the second and third places can be any of the s... | 1769580 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,535 |
14.28. In how many different ways can 1000000 be represented as a product of three natural numbers? Products that differ only in the order of the factors are considered the same.
## 14.5. Inequalities for binomial coefficients | 14.28. Answer: 139. Let the factors have the form $2^{a_{1}} 5^{b_{1}}, 2^{a_{2}} 5^{b_{2}}$, and $2^{a_{3}} 5^{b_{3}}$. Then $a_{1}+a_{2}+a_{3}=6$ and $b_{1}+b_{2}+b_{3}=6$. Here, the numbers $a_{i}$ and $b_{i}$ can be zero. If $a_{1}=k$, then for the decomposition $a_{2}+a_{3}=6-k$ we get $7-k$ options. Therefore, fo... | 139 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,536 |
14.29. Prove that $2^{n} \leqslant C_{2 n}^{n} \leqslant 2^{2 n}$.
## 14.6. Arithmetic of Binomial Coefficients | 14.29. On the one hand, $2^{2 n}=(1+1)^{2 n}=C_{2 n}^{0}+C_{2 n}^{1}+\ldots+C_{2 n}^{2 n} \geqslant C_{2 n}^{n}$. On the other hand,
$$
C_{2 n}^{n}=\frac{(2 n)!}{n!n!}=\frac{2 n}{n} \cdot \frac{2 n-1}{n-1} \cdot \ldots \cdot \frac{n+1}{1} \geqslant 2^{n}
$$
since $\frac{2 n-k}{n-k} \geqslant 2$ for $k=0,1, \ldots, n-... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,537 |
14.30. Let $p$ be a prime number and $1 \leqslant k \leqslant p-1$. Prove that $C_{p}^{k}$ is divisible by $p$. | 14.30. By definition $C_{p}^{k}=\frac{p(p-1) \ldots(p-k+1)}{1 \cdot 2 \cdot \ldots \cdot k}$. The numerator is divisible by $p$, while the denominator is not divisible by $p$. Therefore, the integer that results from dividing the numerator by the denominator is divisible by $p$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,538 |
14.31. Let $p$ be a prime number. Prove that for all natural $m \leqslant p-1$ the number $(-1)^{m} C_{p-1}^{m}-1$ is divisible by $p$. | 14.31. The identity $C_{p}^{k}=C_{p-1}^{k-1}+C_{p-1}^{k}$ shows that
$$
\sum_{k=0}^{m}(-1)^{k} C_{p}^{k}=(-1)^{m} C_{p-1}^{m}
$$
i.e., $\sum_{k=1}^{m}(-1)^{k} C_{p}^{k}=(-1)^{m} C_{p-1}^{m}-1$. If $1 \leqslant m \leqslant p-1$, then according to problem 14.30, each term $(-1)^{k} C_{p}^{k}$ is divisible by $p$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,539 |
14.32. Prove that if $p$ is a prime number, then $C_{n p}^{n} \equiv$ $\equiv n\left(\bmod p^{2}\right)$. | 14.32. The identity $(1+x)^{n p}=\left((1+x)^{n}\right)^{p}$ shows that
$$
C_{n p}^{n}=\sum_{k_{1}+\ldots+k_{n}=p} C_{p}^{k_{1}} \ldots C_{p}^{k_{n}}
$$
According to problem 14.30, the number $C_{p}^{k}$ is divisible by $p$ if $0<k<p$. Therefore, in the given sum, all terms are divisible by $p^{2}$, except for those ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,540 |
14.33. Let $p$ be an odd prime number, $n=p^{\alpha-2}$, where $\alpha \geqslant 2$. Prove that if $m \geqslant 2$, then $C_{n}^{m}$ is divisible by $p^{\alpha-m}$. | 14.33. For $m=2$, it is necessary to prove that $C_{n}^{m}=\frac{n(n-1)}{2}$ is divisible by $n$; this is obvious.
Let $m \geqslant 3$. In the expression $C_{n}^{m}=\frac{n(n-1) \ldots(n-m+1)}{1 \cdot 2 \cdot \ldots \cdot m}$, the numerator is divisible by $n=p^{\alpha-2}$, and the highest power of $p$ that divides th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,541 |
14.34. Let $p$ be a prime number. Prove that if $p^{\alpha}$ divides $C_{n}^{k}$, then $p^{\alpha} \leqslant n$.
See also problems $21.31,21.32$.
## 14.7. The Inclusion-Exclusion Principle | 14.34. It is clear that $C_{n}^{k}=\frac{n(n-1) \ldots(n-k+1)}{k!}$. The prime number $p$ divides the numerator of this fraction. Let $p^{r}$ be the highest power of $p$ that divides at least one of the numbers from $n-k+1$ to $n$ inclusive; let this number be $m$ (if there are several such numbers, we choose any one o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,542 |
14.35. Given $N$ objects and a set of properties $P_{1}, \ldots, P_{n}$. Let $N_{i}$ be the number of objects having property $P_{i}$, $N_{i j}$ be the number of objects having properties $P_{i}$ and $P_{j}$, and so on. Prove that the number of objects not having any of the given properties is
$$
N-\sum N_{i}+\sum_{i_... | 14.35. First solution. An element that does not possess any of the given properties contributes only to the first term $N$. An element that possesses exactly one property contributes to the first two terms; its total contribution is $1-1=0$. An element that possesses exactly $k \geqslant 2$ properties contributes to th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,543 |
14.36. Let $n=p_{1}^{a_{1}} \ldots p_{k}^{a_{k}}$ be the prime factorization of the number, $\varphi(n)$ the number of integers from 1 to $n$,
coprime with $n$. Prove that
$$
\varphi(n)=n\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right) \ldots\left(1-\frac{1}{p_{k}}\right)
$$ | 14.36. Let the objects be the numbers from 1 to $n$, and the property $P_{i}$ be that the number is not divisible by $p_{i}$. The number of numbers from 1 to $n$ that are divisible by $p_{i}$ is $n / p_{i}$, those divisible by both $p_{i}$ and $p_{j}$ is $\frac{n}{p_{i} p_{j}}$, and so on. Therefore, by the principle o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,544 |
14.37. a) Prove that the number of permutations $a_{1}, \ldots, a_{n}$ of the numbers $1, \ldots, n$, for which $a_{i} \neq i$ for all $i$, is
$$
n!\left(1-1+\frac{1}{2!}-\frac{1}{3!}+\ldots+\frac{(-1)^{k}}{k!}+\ldots+\frac{(-1)^{n}}{n!}\right)
$$
b) Prove that the proportion of such permutations among all permutatio... | 14.37. a) Let's use the principle of inclusion and exclusion. Specifically, let the objects be the permutations of the numbers $1, \ldots, n$; their number is $n!$. The property $P_{i}$ is that $a_{i}=i$. Then $N_{i_{1} \ldots i_{k}}=(n-k)!$ is the number of permutations that leave $k$ numbers $i_{1}, \ldots, i_{k}$ in... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,545 |
14.38. Let $m$ and $n$ be non-negative integers. Prove that the number $\frac{m!(2 n+2 m)!}{(2 m)!n!(n+m)!}$ is an integer.
## 14.9. Catalan Numbers
Let $c_{0}=1, c_{1}=c_{0} c_{0}=1, c_{2}=c_{0} c_{1}+c_{1} c_{0}=2, c_{3}=c_{0} c_{2}+c_{1} c_{1}+c_{2} c_{0}=5$ and generally $c_{k}=c_{0} c_{k-1}+c_{1} c_{k-2}+\ldots+... | 14.38. Let $A(m, n)=\frac{m!(2 n+2 m)!}{(2 m)!n!(n+m)!}$. Then $A(m, 0)=1$ and $A(0, n)=C_{2 n}^{n}$. Moreover,
$4 A(m, n-1)+A(m-1, n)=\frac{(m-1)!(2 n+2 m-2)!}{(2 m-2)!(n-1)!(n+m-1)!}\left(\frac{4 m}{2 m(2 m-1)}+\frac{1}{n}\right)$.
Noting that $\frac{2 n+2 m-1}{n(2 m-1)}=\frac{(2 n+2 m-1)(2 n+2 m) m}{n(2 m-1) 2 m(n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,546 |
14.39. Prove that the number of different ways to cut a convex $n$-gon into triangles with non-intersecting diagonals is $c_{n-2}$ (Euler). | 14.39. Let the desired number be $d_{n}$. It is clear that $d_{3}=1$ and $d_{4}=2$. For convenience, let $d_{2}=1$. We will check that $d_{5}=d_{2} d_{4}+d_{3} d_{3}+d_{4} d_{2}$. Fix one side of the pentagon. Suppose this pentagon is cut by non-intersecting diagonals into triangles. Exactly one of these triangles is a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,547 |
14.40. A string contains $n+1$ letters. It is required to place $n$ pairs of round brackets in such a way that inside each pair of brackets there are either two adjacent letters, or a letter and an adjacent bracketed expression, or two adjacent bracketed expressions*. Prove that the number of different ways to place th... | 14.40. There is a one-to-one correspondence between the ways of placing parentheses for \( n+1 \) letters and the ways of cutting a \( (n+2) \)-gon into triangles with non-intersecting diagonals. Figure 14.2 shows how this is done in a specific case.
. What is obtained will be called a correct bracket structure. Prove that the number of correct bracket structures with $n$ pairs of parentheses is $c_{n}$.
A Dyck path of $2 n$ ste... | 14.41. A correct bracket structure of $n$ brackets is characterized by having $n$ left brackets, $n$ right brackets, and in any initial segment, the number of right brackets does not exceed the number of left brackets. Clearly, for $n=1$ and 2, the number of correct bracket structures is 1 and 2, respectively. The requ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,549 |
14.42. a) Prove that the number of different Dyck paths of $2 n$ steps is $c_{n}$.
b) Prove that the number of different Dyck paths of $2 n$ steps is $\frac{1}{n+1} C_{2 n}^{n}$. | 14.42. a) A one-to-one correspondence between Dyck paths and correct bracket structures from problem 14.41 can be established by associating the segment $(1,1)$ with a left parenthesis and the segment $(1,-1)$ with a right parenthesis.
b) Extend the Dyck path (at the end) by one segment $(1,-1)$. Fix one of the $n+1$ ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,550 |
14.43. a) Prove that the number of different sequences $a_{1}, a_{2}, \ldots, a_{2 n}$, for which $a_{i}= \pm 1, a_{1} \geqslant 0$, $a_{1}+a_{2} \geqslant 0, \ldots, a_{1}+a_{2}+\ldots+a_{2 n-1} \geqslant 0$ and $a_{1}+a_{2}+\ldots+a_{2 n}=0$, is $c_{n}$.
b) A cashier, who has no money at the initial moment, sells ti... | 14.43. a) A one-to-one correspondence can be established between such sequences and Dyck paths by associating the number $a_{i}$ with the segment $\left(1, a_{i}\right)$.
b) Let $a_{i}=1$ if the $i$-th person in the queue has 50 rubles, and $a_{i}=-1$ if 100. The cashier will be able to give change to everyone if and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,551 |
14.44. A chess rook moves from the bottom-left corner of an $n \times n$ board to the top-right corner. In doing so, it only makes moves to the right and upwards. Prove that the number of different paths on which the rook never lands on the main diagonal, except for the starting and ending positions, is $c_{n-2}$. | 14.44. The first and last moves of the rook are uniquely determined. Therefore, we can consider the paths of the rook without the first and last moves. A one-to-one correspondence between such paths and Dyck paths of $2(n-2)$ steps is established in an obvious way. | c_{n-2} | Combinatorics | proof | Yes | Yes | olympiads | false | 38,552 |
14.46. The vertices of a regular $(2 n+1)$-gon are marked with zeros and ones. There are $n+1$ zeros and $n$ ones in total. We will consider two such sets of markings to be the same if one can be obtained from the other by rotating the polygon.
a) Prove that the number of different sets of markings is $\frac{1}{2 n+1}... | 14.46. a) The number of ways to mark $n$ vertices of a $(2n+1)$-gon with ones (and put zeros in the remaining places) is $C_{2n+1}^{n}$. Since the number $2n+1$ is not divisible by $n$, no set of markings can transform into itself under a rotation of the $(2n+1)$-gon. The number of rotations of the $(2n+1)$-gon that ma... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,554 |
14.47. Let $\frac{1}{1-x-y+2xy}=\sum_{p, q=0}^{\infty} a_{p, q} x^{p} y^{q}$. Prove that $(-1)^{n} a_{2 n, 2 n+2}=\frac{1}{n+1} C_{2 n}^{n}$ - the Catalan number $c_{n}$.
### 14.10. Elements of Probability Theory | 14.47. Using the formula from problem 14.18, we get
$$
\begin{aligned}
& \frac{1}{1-x-y+2xy} = \frac{1}{(1-x)(1-y)} \cdot \frac{1}{1+\frac{xy}{(1-x)(1-y)}} = \\
& = \sum_{k=0}^{\infty} \frac{(-1)^{k} x^{k} y^{k}}{(1-x)^{k+1}(1-y)^{k+1}} = \sum_{i, j, k=0}^{\infty} (-1)^{k} \binom{k+i}{k} \binom{k+j}{k} x^{k+i} y^{k+j}... | (-1)^{n}\frac{1}{n+1}\binom{2n}{n} | Combinatorics | proof | Yes | Yes | olympiads | false | 38,555 |
14.48. Which sum of the numbers rolled is more likely when throwing two dice: 9 or $10?$ | 14.48. A n s w e r: 9. The numbers 9 and 10 can be obtained in two different ways: $9=3+6=4+5$ and $10=4+6=5+5$. However, we need to consider the order in which the numbers fall on the dice. Therefore, the number 9 can be obtained in four different ways, while the number 10 can only be obtained in three: $9=3+6=6+3=4+5... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,556 |
14.50. The two players are of equal strength, i.e., they have equal chances of winning each game. They agreed that the prize would go to the one who first wins 6 games. They had to stop the game after the first player won 5 games, and the second player won 3. In what ratio should the prize be fairly divided? | 14.50. Answer: $7: 1$. Let the players play three more games, i.e., the game is fictitiously continued even after the first player wins a prize. The second player will receive a prize if and only if he wins all three games. Since all $2^{3}=8$ outcomes of these three games are equally likely, the second player will rec... | 7:1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,558 |
14.51. In a drawer, there are red and black socks. If two socks are randomly pulled out of the drawer, the probability that both are red is $1 / 2$.
a) What is the smallest number of socks that can be in the drawer?
b) What is the smallest number of socks that can be in the drawer if it is known that the number of bl... | 14.51. a) Answer: 4. Let there be $m$ red socks and $n$ black socks in the box. The probability that the first selected sock is red is $\frac{m}{n+m}$. Given that the first selected sock is red, the probability that the second selected sock is also red is $\frac{m-1}{n+m-1}$. Therefore, the probability that both socks ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,559 |
14.52. On each face of a die, one of the numbers from 1 to 6 is written, but some numbers may be written multiple times. We will say that die $X$ wins over $Y$ if, when these dice are rolled simultaneously, the number on $X$ is greater than the number on $Y$ with a probability greater than $1 / 2$. Can three dice $A, B... | 14.52. Answer: yes, they can. For example, let the following numbers be written on the faces of the dice:
$$
\begin{array}{ll}
\text { on die } A & 1,4,4,4,4,4 \text {; } \\
\text { on die } B & 2,2,2,5,5,5 \text {; } \\
\text { on die } C & 3,3,3,3,3,6 .
\end{array}
$$
$B$ wins against $A$ if $B$ rolls a 5 or $B$ ro... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,560 |
15.1. Let $a_{1}, a_{2}, \ldots, a_{k}$ be fixed numbers and $u_{n+k}=$ $=a_{1} u_{n+k-1}+\ldots+a_{k} u_{n}$. Prove that
$$
u_{1}+u_{2} x+u_{3} x^{2}+u_{4} x^{3}+\ldots=\frac{P_{k-1}(x)}{1-a_{1} x-a_{2} x^{2}-\ldots-a_{k} x^{k}}
$$
where $P_{k-1}(x)$ is a polynomial of degree not higher than $k-1$. | 15.1. In the product
$$
\left(u_{1}+u_{2} x+u_{3} x^{2}+u_{4} x^{3}+\ldots\right)\left(1-a_{1} x-a_{2} x^{2}-\ldots-a_{k} x^{k}\right)
$$
the coefficient of $x^{n+k-1}, n \geqslant 1$, is $u_{n+k}-a_{1} u_{n+k-1}-\ldots-a_{k} u_{n}=0$. Therefore, this product represents a polynomial $P_{k-1}(x)$ of degree not higher ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,561 |
15.2. Let $a_{1}, a_{2}, \ldots, a_{k}$ be fixed numbers and $u_{n+k}=$ $=a_{1} u_{n+k-1}+\ldots+a_{k} u_{n}$. Suppose that $x_{1}, \ldots, x_{k}$ are distinct roots of the equation $x^{k}=a_{1} x^{k-1}+a_{2} x^{k-2}+\ldots+a_{k}$. Prove that then $u_{n}=c_{1} x_{1}^{n-1}+\ldots+c_{k} x_{k}^{n-1}$ for some fixed number... | 15.2. According to problem 10.36, one can choose numbers $c_{1}, \ldots, c_{k}$ such that
$$
\begin{aligned}
u_{1} & =c_{1}+\ldots+c_{k} \\
u_{2} & =c_{1} x_{1}+\ldots+c_{k} x_{k} \\
\ldots & \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
u_{k} & =c_{1} x_{1}^{k-1}+\ldots+c_{k} x_{k}^{k-1}
\end{aligned}
$$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,562 |
15.3. Prove that $F_{n+k}=F_{n-1} F_{k}+F_{n} F_{k+1}$. | 15.3. We apply induction on $k$. For $k=1$ we get $F_{n+1}=$ $=F_{n-1}+F_{n}$, and for $k=2$ we get $F_{n+2}=F_{n-1}+2 F_{n}=F_{n-1}+F_{n}+F_{n}=$ $=F_{n+1}+F_{n}$. The base of induction is proved. Now suppose that $F_{n+k-2}=F_{n-1} F_{k-2}+F_{n} F_{k-1}$ and $F_{n+k-1}=F_{n-1} F_{k-1}+F_{n} F_{k}$. Then $F_{n+k}=$ $=... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,563 |
15.4. a) Prove that the numbers $F_{n}$ and $F_{n+1}$ are coprime.
b) Prove that the GCD $\left(F_{m}, F_{n}\right)=F_{d}$, where $d=$ GCD $(m, n)$. | 15.4. a) Suppose that the numbers $F_{n}$ and $F_{n+1}$ are divisible by an integer $d>1$. Then $F_{n-1}=F_{n+1}-F_{n}$ is also divisible by $d$, and so on. In the end, we get that $F_{2}=1$ is divisible by $d$.
b) Let's use the formula $F_{n+k}=F_{n-1} F_{k}+F_{n} F_{k+1}$ (problem 15.3). Set $m=n+k$. For $m>k$, we ge... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,564 |
15.5. Prove that if $m>2$, then $F_{n}$ is divisible by $F_{m}$ if and only if $n$ is divisible by $m$. | 15.5. The number $F_{n}$ is divisible by $F_{m}$ if and only if $\text{GCD}\left(F_{n}, F_{m}\right)=F_{m}$. On the other hand, according to problem 15.4, GCD $\left(F_{n}, F_{m}\right)=F_{\text{GCD}(n, m)}$. Thus, we obtain the following condition: GCD $(n, m)=m$ (here we use the fact that $m>2$). The obtained equalit... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,565 |
15.6. Prove that
$$
F_{n}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right)
$$
(Binet). | 15.6. According to problem $15.2 F_{n}=c_{1} x_{1}^{n}+c_{2} x_{2}^{n}$, where $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}=x+1$, and $c_{1}$ and $c_{2}$ are some constants. Solving the quadratic equation, we get $x_{1,2}=\frac{1 \pm \sqrt{5}}{2}$. The constants $c_{1}$ and $c_{2}$ we find by using the fact... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,566 |
15.7. Prove that for any natural $n$ the number
$$
C_{n}^{1}+5 C_{n}^{3}+25 C_{n}^{5}+125 C_{n}^{7}+\ldots
$$
is divisible by $2^{n-1}$. | 15.7. Binet's formula (problem 15.6) shows that
$$
2^{n-1} F_{n}=C_{n}^{1}+5 C_{n}^{3}+25 C_{n}^{5}+125 C_{n}^{7}+\ldots
$$ | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,567 |
15.8. a) Let \( f_{1}(x)=1, f_{2}(x)=x \) and \( f_{n}(x)=x f_{n-1}(x)+f_{n-2}(x) \) for \( n \geqslant 3 \). Prove that
\[
f_{n+1}(x)=x^{n}+C_{n-1}^{1} x^{n-2}+C_{n-2}^{2} x^{n-4}+C_{n-3}^{3} x^{n-6}+\ldots
\]
b) Prove that
\[
F_{n+1}=1+C_{n-1}^{1}+C_{n-2}^{2}+C_{n-3}^{3}+\ldots
\] | 15.8. a) The polynomials $f_{1}(x)$ and $f_{2}(x)$ have the given form. Therefore, it is sufficient to check that polynomials of the given form satisfy the specified recurrence relation. If we look at the coefficients of the powers of $x$ separately, this recurrence relation reduces to the basic identity for binomial c... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,568 |
15.9. Prove that
$$
\frac{1}{1-x-x^{2}}=F_{1}+F_{2} x+F_{3} x^{2}+F_{4} x^{3}+\ldots
$$ | 15.9. It is clear that $\left(1-x-x^{2}\right)\left(F_{1}+F_{2} x+F_{3} x^{2}+F_{4} x^{3}+\ldots\right)=F_{1}+$ $+\left(F_{2}-F_{1}\right) x+\left(F_{3}-F_{2}-F_{1}\right) x^{2}+\left(F_{4}-F_{3}-F_{2}\right) x^{3}+\ldots$ At the same time, $F_{1}=1$, $F_{2}-F_{1}=0, F_{3}-F_{2}-F_{1}=0, F_{4}-F_{3}-F_{2}=0, \ldots$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,569 |
15.10. Prove that for any natural number $n$ there exists a Fibonacci number that is divisible by $n$.
untranslated portion:
Докажите, что для любого натурального числа $n$ найдётся число Фибоначчи, делящееся на $n$.
For the translated portion:
Prove that for any natural number $n$ there exists a Fibonacci number th... | 15.10. Consider pairs of remainders of division by \( n \) of consecutive Fibonacci numbers \( F_{k} \) and \( F_{k+1} \) for \( k=1,2, \ldots, n^{2}+1 \). The number of different pairs of remainders is \( n^{2} \), so among the considered pairs of remainders, there will be two identical pairs, i.e., the numbers \( F_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,570 |
15.12. Prove that if $a^{2}-a b-b^{2}= \pm 1$, where $a$ and $b$ are natural numbers, then $a=F_{n+1}$ and $b=F_{n}$ for some $n$. | 15.12. The equality $a^{2}=b^{2}+a b \pm 1$ shows that $a \geqslant b$, and $a=b$ only in the case when both numbers are equal to 1. Therefore, we will assume that $a>b$. For the pair $b, a-b$, the required equality also holds, since $b^{2}-(a-b) b-(a-b)^{2}=-\left(a^{2}-a b-b^{2}\right)$. Therefore, after several such... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,572 |
15.13. Find all solutions in natural numbers of the equation $C_{n}^{m-1}=C_{n-1}^{m}$, where $C_{n}^{m}=\frac{n!}{m!(n-m)!}$ is the binomial coefficient. | 15.13. The considered equation is equivalent to the equation $m n=(n-m)(n-m+1)$. Let $d=$ GCD $(m, n), n=d a$ and $m=d b$. After reducing by $d$, we get the equation $a b d=(a-b)((a-b) d+1)$. The number $d$ is coprime with $(a-b) d+1$, and the number $a b$ is coprime with $a-b$, so $a b=(a-b) d+1$ and $d=a-b$. Substitu... | n=F_{2k}F_{2k+1},=F_{2k}F_{2k-1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,573 |
15.14. a) Prove that $F_{2 n+1} F_{2 n-1}=F_{2 n}^{2}+1$.
b) Prove that $F_{2 n+1}^{2}+F_{2 n-1}^{2}+1=3 F_{2 n+1} F_{2 n-1}$. | 15.14. a) We apply induction on $n$. For $n=1$, the equality is obvious. Suppose that the equality $F_{2 n+1} F_{2 n-1}=F_{2 n}^{2}+1$ has been proven for some $n$. Then
\[
\begin{aligned}
& F_{2 n+2}^{2}+1=\left(F_{2 n+1}+F_{2 n}\right)^{2}+1=F_{2 n+1}^{2}+2 F_{2 n+1} F_{2 n}+F_{2 n}^{2}+1= \\
& \quad=F_{2 n+1}^{2}+2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,574 |
15.15. For which natural numbers $a$ and $b$ is the number $a^{2}+b^{2}+1$ divisible by $a b?$ | 15.15. We can assume that $a \geqslant b$. According to problem 15.14 b), the pair of numbers $(a, b)=\left(F_{2 n+1}, F_{2 n-1}\right), n \geqslant 1$, has the required property; the pair of numbers $(a, b)=(1,1)$ also does. We will show that no other pairs of natural numbers $(a, b)$, where $a \geqslant b$, have this... | F_{2n+3},F_{2n+1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,575 |
15.16. Find all pairs of natural numbers $a$ and $b$, for which $a^{2}+1$ is divisible by $b$, and $b^{2}+1$ is divisible by $a$. | 15.16. The formula $F_{2 n+1}^{2}+F_{2 n-1}^{2}+1=3 F_{2 n+1} F_{2 n-1}$ from problem 15.14 b) shows that the pair ( $F_{2 n-1}, F_{2 n+1}$ ) has the required property. We will show that no other pair of natural numbers $(a, b)$, where $a \leqslant b$, has this property. We will use induction on $b$. First, note that i... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,576 |
15.17. Prove that any natural number $n$ can be uniquely represented in the form $n=F_{k_{1}}+F_{k_{2}}+\ldots$ $\ldots+F_{k_{m}}$, where $k_{1}>k_{2}+1, k_{2}>k_{3}+1, \ldots, k_{m-1}>k_{m}+1, k_{m}>1$. | 15.17. Let $F_{k_{1}}$ be the largest Fibonacci number not exceeding $n$, i.e., $F_{k_{1}} \leqslant n < F_{k_{1}+1}$. Then $0 \leqslant n - F_{k_{1}} < F_{k_{1}+1} - F_{k_{1}} = F_{k_{1}-1}$. Therefore, if we choose $F_{k_{2}}$ as the largest Fibonacci number not exceeding $n - F_{k_{1}}$, then $F_{k_{2}} < F_{k_{1}-1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,577 |
15.18. Prove that a Fibonacci number with an odd index cannot be divisible by a prime number of the form $4k+3$.
## 15.3. Fibonacci Numbers and the Euclidean Algorithm | 15.18. Let in problem $15.3 k=n-1$. As a result, we get $F_{2 n-1}=F_{n-1}^{2}+F_{n}^{2}$. Therefore, if $F_{2 n-1}$ is divisible by a prime number $p$ of the form $4 k+3$, then both numbers $F_{n-1}$ and $F_{n}$ are divisible by $p$ (problem 31.2). In this case, the number $F_{n-2}=F_{n}-F_{n-1}$ is also divisible by ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,578 |
15.19. Let $a$ and $b$ be natural numbers, with $a > b$ and $\operatorname{GCD}(a, b) = d$. Prove that if the Euclidean algorithm applied to $a$ and $b$ stops after $n$ steps, then $a \geqslant d F_{n+2}$ and $b \geqslant d F_{n+1}$. | 15.19. We apply induction on $n$. If $n=1$, then $b=d=d F_{2}$ and $a \geqslant 2 d=d F_{3}$. Suppose the required statement is proved for some $n \geqslant 1$. Consider numbers $a>b$, for which the Euclidean algorithm stops after $n+1$ steps. On the first step, the pair $(a, b)$ is replaced by the pair $(b, c)$, where... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,579 |
15.20. a) Prove that $F_{n+5}>10 F_{n}$ for $n \geqslant 2$.
b) Prove that if $F_{n+1}<10^{k}$, then $n \leqslant 5 k$. | 15.20. a) For $n=2$ and 3, the required inequality is easily verified, so we will assume that $n>3$. We will use the identity from problem 15.3: $F_{n+5}=F_{(n-2)+7}=F_{n-3} F_{7}+F_{n-2} F_{8}=13 F_{n-3}+21 F_{n-2}>$ $>10 F_{n-3}+20 F_{n-2}=10 F_{n}$.
b) We will use induction on $k$. For $k=1$, it is sufficient to no... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,580 |
15.22. What is the number of subsets of the set $\{1,2,3, \ldots, n\}$ that do not contain two consecutive numbers? | 15.22. Answer: $F_{n+2}$. Let the desired number be $x_{n}$. For $n=1$ there are subsets $\varnothing$ and $\{1\}$, so $x_{1}=2$. For $n=2$ there are subsets $\varnothing, \{1\}$, and $\{2\}$, so $x_{2}=3$. It is easy to verify that $x_{n}=x_{n-1}+x_{n-2}$ for $n \geqslant 3$. Indeed, if the considered subset contains ... | F_{n+2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,581 |
15.23. In how many ways can the number $n$ be represented as the sum of several addends, each equal to 1 or 2? (Representations differing in the order of the addends are considered different.) | 15.23. Answer: $F_{n+1}$. Let the desired number be $x_{n}$. It is clear that $x_{1}=1$ and $x_{2}=2$. We will prove that $x_{n}=x_{n-1}+x_{n-2}$ for $n \geqslant 3$. Indeed, the number of representations of the number $n$ with the first summand 1 is $x_{n-1}$, and with the first summand $2-x_{n-2}$. | F_{n+1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,582 |
15.24. In how many ways can the number $n$ be represented as a sum of several integer terms $a_{i} \geqslant 2$? (Representations differing in the order of the terms are considered different.) | 15.24. Answer: $F_{n-1}$. Let the desired number be $x_{n}$. It is clear that $x_{1}=0$, $x_{2}=1$, and $x_{3}=1$. We will prove that $x_{n}=x_{n-1}+x_{n-2}$ for $n \geqslant 3$. The first term can be equal to 2; the number of such representations is $x_{n-2}$. The first term can be greater than 2. Then 1 can be subtra... | F_{n-1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,583 |
15.25. In how many ways can the number $n$ be represented as a sum of positive odd summands? (Representations differing in the order of the summands are considered different.)
## 15.5. Special Recurrence Sequences | 15.25. Answer: $F_{n}$. Let the desired number be $x_{n}$. It is clear that $x_{1}=1$ and $x_{2}=1$. We will prove that $x_{n}=x_{n-1}+x_{n-2}$ for $n \geqslant 3$. Indeed, the number of representations with the first term 1 is $x_{n-1}$. If the first term is not less than 3, then 2 can be subtracted from it to obtain ... | F_{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,584 |
15.26. Let $a_{1}=1, a_{2}=0, a_{3}=1$ and
$$
a_{n+3}=\frac{\left(n^{2}+n+1\right)(n+1)}{n} a_{n+2}+\left(n^{2}+n+1\right) a_{n+1}-\frac{n+1}{n} a_{n}
$$
for $n \geqslant 1$. Prove that $a_{n}$ is a square of an integer for any $n \geqslant 1$. | 15.26. Consider the sequence $b_{n}$, for which $b_{1}=1$, $b_{2}=0$ and $b_{n+2}=n b_{n+1}+b_{n}$ for $n \geqslant 1$. Square the equations $b_{n}=b_{n+2}-n b_{n+1}$ and $b_{n+3}=(n+1) b_{n+2}+b_{n+1}$. Then, to eliminate the term $b_{n+2} b_{n+1}$, multiply the first of the obtained equations by $n+1$, the second by ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,585 |
15.27. Let $u_{1}=1, u_{2}=0, u_{3}=1$ and $u_{n}=u_{n-2}+u_{n-3}$ for $n \geqslant 4$. Prove that $u_{2 n}-u_{n-1}^{2}$ and $u_{2 n+1}-u_{n+1}^{2}$ are divisible by $u_{n}$. | 15.27. Let's check that for $0 EXAMPLES AND CONSTRUCTIONS
保留源文本的换行和格式，直接输出翻译结果如下：
15.27. Let's check that for $0 EXAMPLES AND CONSTRUCTIONS | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,586 |
16.1. Is it possible to choose 10 natural numbers such that none of them divides any other, but the square of any number is divisible by each of the numbers? | 16.1. Answer: yes, it is possible. Let $p_{1}, \ldots, p_{10}$ be distinct prime numbers, $N=p_{1} \ldots p_{10}$. Then the numbers $N_{1}=p_{1} N, \ldots, N_{10}=p_{10} N$ have the required properties. Indeed, the number $N_{i}$ is divisible by $p_{i}^{2}$, while the number $N_{j}$, where $j \neq i$, is divisible only... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,587 |
16.2. a) Prove that there are infinitely many pairs of natural numbers $k, l \geqslant 2$, for which $k!l!=n!$ for some natural number $n$.
b) Prove that for each natural number $m$ there are infinitely many sets of natural numbers $a_{1}, \ldots, a_{m} \geqslant 2$, for which $a_{1}!a_{2}!\ldots a_{m}!=n!$ for some $... | 16.2. a) Let $k=l!-1$. Then $k!l!=n!$, where $n=l!$.
b) Let $a_{1}=a_{2}!\ldots a_{m}!-1$. Then $a_{1}!a_{2}!\ldots a_{m}!=n!$, where $n=$ $=a_{2}!\ldots a_{m}!$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,588 |
16.3. Prove that for any natural $n$ there exist $n$ distinct natural numbers, the sum of any two of which is divisible by their difference. | 16.3. For $n=2$ we can take the numbers 1 and 2. Let the numbers $a_{1}, \ldots, a_{n}$ satisfy the required condition. We will show that then the numbers $A, A+a_{1}, \ldots, A+a_{n}$, where $A=a_{1} \ldots a_{n}$, also satisfy the required condition. It is clear that $A+a_{k}+A$ is divisible by $A+a_{k}-A=a_{k}$, sin... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,589 |
16.4. a) Is the set of pairs of natural numbers $a, b$ that have the following property finite or infinite: the prime divisors of the numbers $a$ and $b$ are the same, and the prime divisors of the numbers $a+1$ and $b+1$ are also the same (and at the same time $a \neq b$)?
b) Is the set of pairs of natural numbers $a... | 16.4. a) Answer: infinitely. Let, for example, $a=2^{n}-2$ and $b=2^{n}\left(2^{n}-2\right)$. Then $a+1=2^{n}-1$ and $b+1=2^{2 n}-2 \cdot 2^{n}+1=\left(2^{n}-1\right)^{2}$.
b) Answer: infinitely. Let, for example, $a=2^{k}+1$ and $b=a^{2}-1=$ $=2^{k+1}\left(2^{k-1}+1\right)$. Then the numbers $a$ and $b+1=a^{2}$ have ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,590 |
16.5. Specify pairwise distinct natural numbers $p, q, r, p_{1}, q_{1}, r_{1}$, for which $p^{2}+q^{2}+r^{2}=p_{1}^{2}+q_{1}^{2}+r_{1}^{2}$ and $p^{4}+q^{4}+r^{4}=p_{1}^{4}+q_{1}^{4}+r_{1}^{4}$ (Ramanujan).
## 16.2. Infinite sequences | 16.5. Let's use Ramanujan's identity $f_{2}=f_{4}=0$ from problem 32.23. Setting $a=1, b=2, c=3$ and $d=6$, we obtain the required set of numbers $11,6,5$ and $10,9,1$. | 11,6,510,9,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,591 |
16.6. For any natural $n$ from the sequence $1, 1 / 2, 1 / 3, 1 / 4, \ldots$, can an arithmetic progression of length $n$ be selected? | 16.6. Answer: for any. The numbers $1 / n!, 2 / n!, \ldots, n / n!$ form an arithmetic progression of length $n$ with a difference of $1 / n!$. These numbers have the form $\frac{1}{n!/ k}$, where the numbers $n!/ k$ are integers. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,592 |
16.7. Does there exist an infinite sequence of natural numbers $a_{1}, a_{2}, \ldots$, in which there are no powers of natural numbers and no sum of several different terms of this sequence is a power of a natural number?
## 16.3. Sequences of operations | 16.7. Answer: it exists. Let $a_{1}=2, a_{2}=2^{2} \cdot 3, a_{3}=$ $=2^{2} \cdot 3^{2} \cdot 5, \ldots, a_{n}=2^{2} \cdot 3^{2} \cdot \ldots \cdot p_{n-1}^{2} p_{n}$, where $p_{n}$ is the $n$-th prime number.
The number $a_{k}+a_{k+l}+\ldots+a_{k+r}$ is divisible by $p_{k}$ and not divisible by $p_{k}^{2}$, hence it ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,593 |
16.8. Cards with numbers $1,2,3, \ldots, 32$ are stacked in order. It is allowed to take any number of cards from the top and insert them between the remaining cards, without changing the order of the cards in each of the two parts, and otherwise arbitrarily.
a) Prove that the cards can be arranged in any order with 5... | 16.8. a) After renumbering the cards, we can assume that the cards are stacked in a random order, and they need to be sorted in ascending order. Let's call a disorder a pair of adjacent cards with numbers $i$ and $i+1$, on which the numbers $a_{i}>a_{i+1}$ are written. We will divide the cards into blocks where there a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,594 |
16.9. Provide an example of a rational function $R(x)$ (i.e., the ratio of two polynomials), which is different from a constant and satisfies $R(x)=R(1 / x)$ and $R(x)=R(1-x)$. | 16.9. By replacing $x$ with $1-x$ and $1 / x$, from each number one can obtain a total of six different numbers: $x, 1 / x, 1-x, \frac{1}{1-x}, \frac{x}{x-1}$, $\frac{x-1}{x}$. Therefore, the function
$$
R(x)=x^{2}+\frac{1}{x^{2}}+(1-x)^{2}+\frac{1}{(1-x)^{2}}+\frac{x^{2}}{(1-x)^{2}}+\frac{(x-1)^{2}}{x^{2}}
$$
has th... | R(x)=x^{2}+\frac{1}{x^{2}}+(1-x)^{2}+\frac{1}{(1-x)^{2}}+\frac{x^{2}}{(1-x)^{2}}+\frac{(x-1)^{2}}{x^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,595 |
16.10. Does there exist a polynomial $f(x, y)$ of two real variables that is everywhere non-zero but takes values arbitrarily close to zero? | 16.10. Yes, it exists. Let, for example, \( f(x, y) = x^2 + (xy - 1)^2 \). If \( f(x, y) = 0 \), then \( x = 0 \) and \( xy - 1 = 0 \), which cannot be. On the other hand, for any \( \varepsilon > 0 \), we can set \( x = \sqrt{\varepsilon} \) and \( xy - 1 = \sqrt{\varepsilon} \), i.e., \( y = 1 + \frac{1}{\sqrt{\varep... | 2\varepsilon | Algebra | proof | Yes | Yes | olympiads | false | 38,596 |
16.11. Does there exist a polynomial $f(x)$ with integer coefficients that takes the same values for some different real $x$, but takes different values for all different rational $x$?
## 16.5. Different examples and constructions | 16.11. Answer: it exists. Let, for example, $f(x)=x^{3}-2 x$. This polynomial equals zero at $x=0, \pm \sqrt{2}$. Suppose that $x=m / n$ and $y=p / q$ are irreducible fractions ($n>0$ and $q>0$), and $f(x)=f(y)$. Then
$$
q^{3} m\left(m^{2}-2 n^{2}\right)=n^{3} p\left(p^{2}-2 q^{2}\right)
$$
The numbers $p$ and $q$ ar... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,597 |
16.12. Can $2 n$ people be paired in $2 n-1$ ways so that each person is paired with every other person exactly once | 16.12. A n s w e r: Yes, it is possible. Place one person at the center of the circle, and arrange all the others along the circumference at equal intervals. The pairing will be constructed as follows. Choose one person standing on the circumference and draw a diameter through them. One pair consists of the chosen pers... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,598 |
16.13. a) There is a piece of chain consisting of 60 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, ..., 60 g can be formed (an unbuckled link also weighs 1 g)?
b) The same question for a chain consisting of 150 l... | 16.13. a) Answer: 3 links. Let's determine the largest $n$ for which it is sufficient to break $k$ links of an $n$-link chain so that all integer weights from 1 to $n$ can be formed from the resulting parts. If $k$ links are broken, any number of links from 1 to $k$ can be formed from them. But we cannot form $k+1$ lin... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,599 |
16.14. Is it possible to arrange 10 bus routes in a city and set up stops on them such that for any 8 routes chosen, there will be a stop not lying on any of them, while any 9 routes pass through all stops. | 16.14. Answer: yes, it is possible. Let's draw 10 pairwise intersecting lines. Suppose the routes pass through these lines, and the stops are the points of intersection of the lines. Any 9 routes pass through all stops, since each stop lying on the remaining line is passed by one of the 9 lines corresponding to these r... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,600 |
16.15. Given $n$ integers $a_{1}=1, a_{2}, a_{3}, \ldots, a_{n}$, and $a_{i} \leqslant a_{i+1} \leqslant 2 a_{i}(i=1,2, \ldots, n-1)$, and the sum of all numbers is even. Can these numbers be divided into two groups such that the sums of the numbers in these groups are equal? | 16.15. Answer: Yes, it is possible. We will assign the number $a_{n}$ to one group and the number $a_{n-1}$ to the other. Then we will sequentially assign the numbers $a_{n-2}, a_{n-3}, \ldots, a_{1}$ to the group with the smaller sum of numbers (if the sums are equal, the number can be assigned to either group). Let $... | S_{1}=S_{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,601 |
16.16. A radio tube has seven contacts arranged in a circle and plugged into a socket with seven holes. Can the contacts of the tube and the holes of the socket be numbered in such a way that, when the tube is plugged in, at least one contact will always be in its correct position (i.e., in the hole with the same numbe... | 16.16. Answer: yes, it is possible. Number the contacts of the lamp in order, and the holes of the plug in the reverse order. Then the contact with number $k$ falls into the hole with number $a-k$, where $a$ is a fixed number. (More precisely, we are talking not about the numbers themselves, but about their remainders ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,602 |
16.17. a) There are 555 weights weighing 1 g, 2 g, 3 g, 4 g, ..., 555 g. Divide them into three equal-weight piles.
b) There are 81 weights weighing $1^{2}$ g, $2^{2}$ g, $3^{2}$ g, ..., $81^{2}$ g. Divide them into three equal-weight piles. | 16.17. a) Nine weights with masses $n, n+1, \ldots, n+8$ can be divided into three equal-weight piles: 1) $n, n+4, n+8$; 2) $n+1, n+5, n+6$; 3) $n+2, n+3, n+7$. This allows the weights with masses $1,2, \ldots, 549=61 \cdot 9$ to be divided into three equal-weight piles. The remaining six weights with masses $550,551, ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,603 |
16.18. Consider all possible ten-digit numbers written using the digits 2 and 1. Divide them into two classes such that the sum of any two numbers from each class results in a number that contains at least two threes in its representation. | 16.18. Let's assign to the first class all numbers in which an even number of twos appears, and to the second class all numbers in which an odd number of twos appears. Two numbers of the same class either contain the same number of twos, or in one number there are at least two more twos than in the other. If two number... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,604 |
16.19. A right angle is divided into an infinite number of squares with a side length of 1. Can a natural number be written in each of these squares so that in each row of squares, parallel to one or the other side of the right angle, any natural number appears exactly once?
See also problem 20.9. | 16.19. Answer: Yes, it is possible. Let $A$ be a square with side $2^{n}$, which is divided into squares with side 1, and a natural number is written in each of them. We associate with it a square
| $A+2^{n}$ | $A$ |
| :---: | :---: |
| $A$ | $A+2^{n}$ |
with side $2^{n+1}$. If we start with a square with side 1, in ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,605 |
17.1. Prove that the decimal representation of a rational number is a periodic fraction.
Note. For the converse statement, see problem 9.3. | 17.1. It is sufficient to consider rational numbers of the form $p / q$, where $1 \leqslant p<q$. In this case, the digits of the decimal representation $p / q=0, a_{1} a_{2} a_{3} \ldots$ are found as follows: $a_{k}=\left[\frac{10^{k} p}{q}\right]-10\left[\frac{10^{k-1} p}{q}\right]$. Let $10^{k-1} p=M q+r_{k}$, wher... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,606 |
17.2. Prove that if natural numbers $a$ and $b$ are coprime, then $a^{n}-1$ is divisible by $b$ for some natural $n$. | 17.2. Among the numbers $a, a^{2}, \ldots, a^{b+1}$, two of them give the same remainder when divided by $b$. Let these numbers be $a^{k}$ and $a^{l}$, where $k>l$. Then $a^{k}-a^{l}=a^{l}\left(a^{k-l}-1\right)$ is divisible by $b$. The numbers $a^{l}$ and $b$ are coprime, so $a^{k-l}-1$ is divisible by $b$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,607 |
17.3. Let $p<q$ be natural numbers, and $q$ is not divisible by 2 and 5. According to problem 17.2, one can choose $n$ such that $10^{n}-1$ is divisible by $q$. Prove that $p / q$ is a purely periodic fraction, the length of whose period divides $n$.
$$
* * *
$$ | 17.3. Let $10^{n}-1=r q$. Then $\frac{p}{q}=\frac{r p}{r q}=\frac{r p}{10^{n}-1}=r p \cdot 10^{-n}+$ $+r p \cdot 10^{-2 n}+r p \cdot 10^{-3 n}+\ldots$ In this case, $r p<r q=10^{n-1}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,608 |
17.4. Prove that from the integers $a_{1}, a_{2}, \ldots, a_{n}$ one can select one or several consecutive numbers such that their sum is divisible by $n$. | 17.4. Suppose that none of the numbers $S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\ldots, S_{n}=a_{1}+\ldots+a_{n}$ is divisible by $n$. Then, when divided by $n$, these numbers can give no more than $n-1$ different remainders: $1,2, \ldots, n-1$. Therefore, two numbers $S_{p}$ and $S_{q}$, where $p>q$, give the same remainder... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,609 |
17.5. From the numbers $1,2, \ldots, n$, more than $\frac{n+1}{2}$ different numbers are chosen. Prove that one of the chosen numbers divides another. | 17.5. Match each selected number with its greatest odd divisor. The number of odd numbers not exceeding $n$ is $n / 2$ for even $n$ and $(n+1) / 2$ for odd $n$. Therefore, the greatest odd divisor of two of the selected numbers is the same. The larger of these two numbers is divisible by the smaller. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,610 |
17.6. Prime numbers $a_{1}, a_{2}, \ldots, a_{p}$ form an increasing arithmetic progression and $a_{1}>p$. Prove that if $p$ is a prime number, then the common difference of the progression is divisible by $p$. | 17.6. Consider the remainders of the numbers $a_{1}, \ldots, a_{p}$ when divided by $p$. The numbers $a_{1}, \ldots, a_{p}$ are prime and all are strictly greater than $p$, so none of them is divisible by $p$. Thus, we have $p$ remainders, all different from $p$. Therefore, there are two numbers $a_{i}$ and $a_{j}$ tha... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,611 |
17.7. Let $n>1$ be a natural number, $e-$ the smallest integer exceeding $\sqrt{n}$. Prove that for any natural number $a$, coprime with $n$, one can choose natural numbers $x$ and $y$, not exceeding $e-1$, such that $a y \equiv \pm x(\bmod n)$.
See also problem 15.10.
## 17.2. Various problems | 17.7. Consider $e^{2}$ numbers $a y+x$, where $x, y=0,1,2, \ldots, e-1$. Since $e^{2}>n$, among these numbers there will be numbers $a y_{1}+x_{1}$ and $a y_{2}+x_{2}$, giving the same remainder when divided by $n$. Consequently, $a\left(y_{1}-y_{2}\right) \equiv x_{2}-x_{1}(\bmod n)$. The numbers $\left|y_{1}-y_{2}\ri... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,612 |
17.8. Prove that among any $n \geqslant 2$ people, there will be two people who have the same number of acquaintances among these $n$ people. | 17.8. We can associate with each of the $n$ people a number equal to the number of their acquaintances among these $n$ people. This number can take one of $n$ values: $0,1, \ldots, n-1$. But if someone has no acquaintances at all, then no one can be acquainted with everyone. Conversely, if someone is acquainted with ev... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,613 |
17.9. Among the numbers $1,2, \ldots, 100$, 55 numbers are chosen. Prove that among them there will be two numbers whose difference is: a) 9; b) 10; c) 12; d) 13. e) Show that among them there may not be two numbers whose difference is 11. | 17.9. a) Let's divide the numbers from 1 to 100 into 6 sets: numbers from 1 to 18, from 19 to 36, from 37 to 54, from 55 to 72, from 73 to 90, and from 91 to 100. If no more than 9 numbers are chosen from each of these sets, then a total of no more than \(6 \cdot 9 = 54\) numbers are chosen, so in at least one of these... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,614 |
17.10. Prove that any sequence of $m n+1$ pairwise distinct numbers contains either an increasing subsequence of $m+1$ numbers or a decreasing subsequence of $n+1$ numbers.
220 Chapter 17. The Pigeonhole Principle. The Extreme Rule | 17.10. We associate with the member $a_{k}$ of the given sequence two numbers $x_{k}$ and $y_{k}$, where $x_{k}$ is the greatest length of an increasing subsequence starting with $a_{k}$, and $y_{k}$ is the greatest length of a decreasing subsequence starting with $a_{k}$. Suppose that $x_{k} \leqslant m$ and $y_{k} \l... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,615 |
17.11. The numbers $[a],[2 a], \ldots,[N a]$ are distinct from each other, and the numbers $[1 / a],[2 / a], \ldots,[M / a]$ are also distinct from each other. Find all such $a$.
## 17.3. Approximations of irrational numbers by rational numbers | 17.11. Answer: $\frac{N-1}{N} \leqslant|a| \leqslant \frac{M}{M-1}$.
The numbers $[x]$ and $[y]$ are different if and only if the numbers $[-x]$ and $[-y]$ are different. Therefore, it is sufficient to consider the case when $a>0$. If $a<\frac{N-1}{N}$, then among the numbers $[a],[2 a], \ldots,[N a]$ there are coinci... | \frac{N-1}{N}\leqslant||\leqslant\frac{M}{M-1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,616 |
17.13. Let $\alpha$ be a positive number. Prove that for any number $C>1$ one can choose a natural number $x$ and an integer $y$ such that $x<C$ and $|x \alpha-y| \leqslant 1 / C$. | 17.13. First, assume that the number $C$ is an integer. Consider the number 1 and the numbers $k \alpha-[k \alpha]$ for $k=0,1, \ldots, C-1$. These $C+1$ numbers lie on the interval $[0,1]$, so the distance between some two of them does not exceed $1 / C$. If these turn out to be the numbers $k_{1} \alpha-\left[k_{1} \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,618 |
17.14. a) Let $\alpha$ be an irrational number. Prove that for any numbers $a<b$, one can choose integers $m$ and $n$ such that $a<m \alpha-n<b$.
b) Prove that the numbers $m$ and $n$ can be chosen to be natural numbers.
## 17.4. The Rule of the Extreme
- The Rule of the Extreme consists in considering the element f... | 17.14. a) Let $\Delta=b-a$. For each integer $m_{1}$, one can choose an integer $n_{1}$ such that $0 \leqslant m_{1} \alpha - n_{1} \leqslant 1$. Divide the interval $[0,1]$ into equal segments, each of which is less than $\Delta$. Let the number of these segments be $k$. Then among the numbers $m_{1} \alpha - n_{1}, \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,619 |
17.15. On the squares of an infinite chessboard, natural numbers are written such that each number is equal to the arithmetic mean of the four neighboring numbers - the one above, below, to the left, and to the right. Prove that all numbers on the board are equal to each other. | 17.15. Among the written numbers, we can choose the smallest one. Indeed, first, we take an arbitrary written number $n$. If there is a number that is smaller than the chosen number, then on the next step, we choose it, and so on. This process is finite because we can choose no more than $n$ different numbers.
Let $m$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,620 |
17.16. In each cell of an infinite sheet of graph paper, some real number is written. Prove that there is a cell in which the number written does not exceed the numbers written in at least four of the eight adjacent cells. | 17.16. Consider a square consisting of 16 cells, and cut out the 4 corner cells. Among the remaining 12 cells, choose the one with the smallest number (if there are several such cells, choose any one of them). The selected cell has the required property. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,621 |
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