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 |
|---|---|---|---|---|---|---|---|---|---|
36.1. Let $f(x)=1+x+x^{2}+x^{3}+\ldots$ and $g(x)=1-x+x^{2}-$ $-x^{3}+\ldots$ Compute the products: a) $f(x) f(x)$; b) $f(x) g(x)$. | 36.1. Answer: a) $1+2 x+3 x^{2}+4 x^{3}+\ldots$; b) $1+x^{2}+x^{4}+x^{6}+\ldots$ | )1+2x+3x^{2}+4x^{3}+\ldots;b)1+x^{2}+x^{4}+x^{6}+\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,197 |
36.2. Let $f(x)=1+x+x^{2}+x^{3}+\ldots$ Compute the formal series $g(x)$ for which $f(x) g(x)=1$. | 36.2. Answer: $g(x)=1-x$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in a mixed format of Russian and English. The mathematical expression and the answer are already in English. The tran... | (x)=1-x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,198 |
36.3. Let \( f(x) = a_{0} + a_{1} x + a_{2} x^{2} + \ldots \) be a formal series, and suppose \( a_{0} \neq 0 \). Prove that there exists a unique formal series \( g(x) = b_{0} + b_{1} x + b_{2} x^{2} + \ldots \) such that \( f(x) g(x) = 1 \).
496 Chapter 36. Formal Series and Generating Functions
## 36.2. Formal Der... | 36.3. The equalities $a_{0} b_{0}=1, a_{1} b_{0}+a_{0} b_{1}=0$, $a_{2} b_{0}+a_{1} b_{1}+a_{0} b_{2}=0, \ldots$ must be satisfied. From these equalities, it follows that $b_{0}=\frac{1}{a_{0}}$, $b_{1}=-\frac{a_{1} b_{0}}{a_{0}}, b_{2}=-\frac{a_{2} b_{0}+a_{1} b_{1}}{a_{0}}$ and so on. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,199 |
36.4. Prove that the formal derivative has the following properties of the ordinary derivative:
a) $D(f(x)+g(x))=D(f(x))+D(g(x))$
b) $D(f(x) g(x))=f(x) D(g(x))+g(x) D(f(x))$;
c) $D\left(f(x)^{n}\right)=n f(x)^{n-1} D(f(x))$ for any natural number $n$;
d) if the formal series $f(x)^{-1}$ exists, then $D\left(f(x)^{-... | 36.4. b) Let $f(x)=\sum_{k=0}^{\infty} a_{k} x^{k}$ and $g(x)=\sum_{k=0}^{\infty} b_{k} x^{k}$. The coefficient of $x^{k-1}$ in the formal series $D(f(x) g(x))$ is $k \cdot \sum_{i+j=k} a_{i} b_{j}$, and the coefficient of $x^{k-1}$ in the formal series $f(x) D(g(x))+g(x) D(f(x))$ is $\sum_{i+j=k} j a_{i} b_{j}+\sum_{i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,200 |
36.5. For the formal series $f=a_{0}+a_{1} x+a_{2} x^{2}+\ldots$, let $S(f)=a_{0}$. Prove that $f=\sum_{n=0}^{\infty} S\left(D^{n}(f)\right) \frac{x^{n}}{n!}$.
## 36.3. Square Root of a Formal Series | 36.5. According to the definition $D^{n}(f)=\sum_{j=n}^{\infty} j(j-1) \ldots(j-n+1) a_{j} x^{j-n}$.
therefore $S\left(D^{n}(f)\right)=n!a_{n}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,201 |
36.6. Prove that for any natural $n$ and any formal series $f(x)=1+a_{1} x+a_{2} x^{2}+\ldots$ there exists a unique formal series $g(x)=1+b_{1} x+b_{2} x^{2}+\ldots$, for which $g(x)^{n}=f(x)$.
The series $g(x)$ we will denote $\sqrt[n]{f(x)}$ or $(f(x))^{1 / n}$. | 36.6. By induction on $n$ it is easy to prove that $\left(1+b_{1} x+b_{2} x^{2}+\ldots\right)^{n}=$ $=1+c_{1} x+c_{2} x^{2}+\ldots$, where $c_{1}=n b_{1}, c_{2}=n b_{2}+p_{n, 2}\left(b_{1}\right), c_{3}=n b_{3}+$ $+p_{n, 3}\left(b_{1}, b_{2}\right), \ldots, c_{k}=n b_{k}+p_{n, k}\left(b_{1}, \ldots, b_{k-1}\right), \ld... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,202 |
36.7. Prove that for any rational number $r$
$$
(1+x)^{r}=1+r x+\frac{r(r-1)}{2!} x^{2}+\ldots+\frac{r(r-1)(r-2) \ldots(r-n+1)}{n!} x^{n}+\ldots
$$
## 36.4. Exponential and Logarithm
We call the formal exponential the formal series
$$
\operatorname{Exp}(x)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots+\frac{x^{n}}{n... | 36.7. According to problem 36.4 d) \( D\left((1+x)^{r}\right) = r(1+x)^{r-1} D(1+x) = r(1+x)^{r-1} \). Therefore, by induction on \( n \), we obtain \( D^{n}\left((1+x)^{r}\right) = r(r-1)(r-2) \ldots (r-n+1)(1+x)^{r-n} \). Now let's use the formula from problem 36.5. If \( f = (1+x)^{r} \), then \( S\left(D^{n}(f)\rig... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,203 |
36.8. Prove that: a) $\operatorname{Exp}((a+b) x)=\operatorname{Exp}(a x) \cdot \operatorname{Exp}(b x)$; b) $\operatorname{Exp}(k x)=(\operatorname{Exp}(x))^{k}$ for any natural number $k$. | 36.8. a) It is required to prove that $\frac{(a+b)^{m}}{m!}=\sum_{k=0}^{m} \frac{a^{k}}{k!} \frac{b^{m-k}}{(m-k)!}$, i.e., $(a+b)^{m}=\sum_{k=0}^{m} \frac{m!}{k!(m-k)!} a^{k} b^{m-k}$. But $\frac{m!}{k!(m-k)!}=C_{m}^{k}$.
b) Follows from a) by induction on $k$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,204 |
36.9. Prove that
$$
\sum_{k=1}^{n}(-1)^{n-k} k^{m} C_{n}^{k}= \begin{cases}0 & \text { for } 0<m<n \\ n! & \text { for } m=n\end{cases}
$$
Let $f=1+a_{1} x+a_{2} x^{2}+\ldots$ be a formal series. The formal logarithm of $f$ is defined as the formal series
$$
\operatorname{Ln}(f)=g-\frac{1}{2} g^{2}+\frac{1}{3} g^{3}... | 36.9. Consider the sum of formal series
$$
\begin{aligned}
\sum_{k=0}^{n}(-1)^{n-k} C_{n}^{k} \operatorname{Exp}(k x)=\sum_{k=0}^{n}(-1)^{n-k} C_{n}^{k} & \left(\sum_{m=0}^{\infty} \frac{k^{m} x^{m}}{m!}\right)= \\
& =\sum_{m=0}^{\infty}\left(\sum_{k=0}^{n}(-1)^{n-k} k^{m} C_{n}^{k}\right) \frac{x^{m}}{m!}
\end{aligne... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,205 |
36.10. Prove that $D(\operatorname{Ln}(f))=f^{-1} D(f)$. | 36.10. It is clear that $D(\operatorname{Ln}(f))=D\left(g-\frac{1}{2} g^{2}+\frac{1}{3} g^{3}-\ldots\right)$. Using the fact that each coefficient of the formal series $\operatorname{Ln}(f)$ is defined by a finite sum, we get
$$
\begin{aligned}
& D\left(g-\frac{1}{2} g^{2}+\frac{1}{3} g^{3}-\ldots\right)=D(g)-\frac{1}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,206 |
36.11. Prove that if $f=1+a_{1} x+a_{2} x^{2}+\ldots$ and $h=1+$ $+b_{1} x+b_{2} x^{2}+\ldots$, then $\operatorname{Ln}(f h)=\operatorname{Ln}(f)+\operatorname{Ln}(h)$. | 36.11. Using the result of problem 36.10 and the properties of the formal derivative, we get $D(\operatorname{Ln}(f h))=(f h)^{-1} D(f h)=$ $=(f h)^{-1}(f D(h)+h D(f))=h^{-1} D(h)+f^{-1} D(f)=D(\operatorname{Ln}(f)+\operatorname{Ln}(h))$.
In addition, the formal series $\operatorname{Ln}(f h)$ and $\operatorname{Ln}(f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,207 |
36.12. Prove that $\operatorname{Ln}\left(f^{r}\right)=r \operatorname{Ln}(f)$ for any rational number $r$. | 36.12. First, let's prove that $\operatorname{Ln}\left(f^{-1}\right)=-\operatorname{Ln}(f)$. According to problem 36.11, from the equality $f^{-1} f=1$ it follows that $\operatorname{Ln}\left(f^{-1}\right)+\operatorname{Ln}(f)=$ $=\operatorname{Ln}\left(f^{-1} f\right)=\operatorname{Ln}(1)=0$.
Using problem 36.11 and ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,208 |
36.13. Prove that if $\operatorname{Ln}(f)=\operatorname{Ln}(h)$, then $f=h$. | 36.13. First, let's prove that if $\operatorname{Ln}(f)=0$, then $f=1$. Indeed, $D(\operatorname{Ln}(f))=D(0)=0$, so $f^{-1} D(f)=0$. Therefore, $D(f)=0$ and $f=1$.
If $\operatorname{Ln}(f)=\operatorname{Ln}(h)$, then according to problem 36.11, $\operatorname{Ln}\left(f^{-1} h\right)=\operatorname{Ln}\left(f^{-1}\rig... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,209 |
36.14. Prove that if $g=a_{1} x+a_{2} x^{2}+\ldots$ and $r-$ is any rational number, then
$$
(1+g)^{r}=1+r g+\frac{r(r-1)}{2!} g^{2}+\ldots+\frac{r(r-1)(r-2) \ldots(r-n+1)}{n!} g^{n}+\ldots
$$ | 36.14. We will use the notation
$$
C_{r}^{n}=\frac{r(r-1)(r-2) \ldots(r-n+1)}{n!}
$$
for any rational number $r$ and natural number $n$. Let $h=1+\sum_{n=1}^{\infty} C_{r}^{n} g^{n}$. We need to prove that $(1+g)^{r}=h$. Clearly, $D(h)=D(g) \sum_{n=1}^{\infty} n C_{r}^{n} g^{n-1}$, therefore
$$
\begin{aligned}
(1+g)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,210 |
36.15. Infinite sequences $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are such that $b_{n}=\sum_{i=0}^{n} C_{n}^{i} a_{i}$ for all $n \geqslant 0$. Prove that then $a_{n}=\sum_{i=0}^{n}(-1)^{n-i} C_{n}^{i} b_{i}$ for all $n \geqslant 0$.
## 36.5. Identities for Formal Series | 36.15. Consider the formal series $A(x)=\sum_{n=0}^{\infty} \frac{a_{n} x^{n}}{n!}$ and $B(x)=$ $=\sum_{n=0}^{\infty} \frac{b_{n} x^{n}}{n!}$. By the condition,
$$
\frac{b_{n} x^{n}}{n!}=\sum_{i=0}^{n} C_{n}^{i} \frac{a_{n} x^{n}}{n!}=\sum_{i=0}^{n} \frac{a_{n} x^{n}}{i!(n-i)!}
$$
Thus,
$$
B(x)=\sum_{n=0}^{\infty} \... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,211 |
36.16. Prove that the product $\prod_{m=1}^{\infty}\left(1+x^{m}\right)$ corresponds to a well-defined formal series.
498 Chapter 36. Formal Series and Generating Functions | 36.16. The coefficient of $x^{n}$ depends only on the finite product $\prod_{m=1}^{n}\left(1+x^{m}\right)$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,212 |
36.17. Prove the following identities:
a) $\prod_{m=1}^{\infty}\left(1-x^{2 m-1}\right)^{2}=\prod_{m=1}^{\infty}\left(1-x^{2 m}\right)^{-1} \sum_{k=-\infty}^{\infty}(-1)^{k} x^{k^{2}}$;
b) $\prod_{m=1}^{\infty}\left(1-x^{m}\right)=\prod_{m=1}^{\infty}\left(1+x^{m}\right) \sum_{k=-\infty}^{\infty}(-1)^{k} x^{k^{2}} \q... | 36.17. a) Let
$$
F(y)=\prod_{i=1}^{n}\left(1+x^{2 i-1} y\right)\left(y+x^{2 i-1}\right)=\sum_{i=-n}^{n} A_{i}(x) y^{n+i}
$$
where $A_{i}(x)$ is a polynomial with integer coefficients. A simple check shows that
$$
\begin{gathered}
\left(y+x^{2 n-1}\right) F\left(x^{2} y\right)=x^{2 n-1}\left(1+x^{2 n+1} y\right) F(y)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,213 |
36.18. Let $s(n)$ be the sum of those divisors $d$ of the number $n$ for which $n / d$ is odd $(s(n)=0$ for $n \leqslant 0)$.
a) Let $f=\prod_{n=1}^{\infty} \frac{1-x^{n}}{1+x^{n}}$ and $g=\sum_{n=1}^{\infty} s(n) x^{n-1}$. Prove that $D(f)=-2 f g$.
b) Prove that $s(n)-2 s(n-1)+2 s\left(n-2^{2}\right)-2 s\left(n-3^{2... | 36.18. a) Let $f_{m}=\prod_{n=1}^{m} \frac{1-x^{n}}{1+x^{n}}$. Then
$$
D\left(\operatorname{Ln}\left(f_{m}\right)\right)=D\left(\sum_{n=1}^{m}\left(\operatorname{Ln}\left(1-x^{n}\right)-\operatorname{Ln}\left(1+x^{n}\right)\right)\right)
$$
Chapter 36. Formal Series and Generating Functions
Using the fact that $D(\o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,214 |
36.19. a) Using the result of problem 36.18 b), prove that any prime number $p$ of the form $4k+1$ can be represented as the sum of two squares.
b) Prove that any prime number $p$ of the form $8k+3$ can be represented as the sum of three squares.
## 36.6. Generating Functions
The generating function of the sequence ... | 36.19. a) If the number $n$ is not a square, then the number $s(n)$ is even. Indeed, if $n$ is even and $n / d$ is odd, then $d$ is even; therefore, for even $n$, the number $s(n)$ is the sum of even numbers. If $n$ is odd and not a perfect square, then $s(n)$ can be represented as the sum of even numbers $d+n / d$.
I... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,215 |
36.20. Prove that the generating function of the sequence $a_{k}=C_{n}^{k}, k=0,1, \ldots, n$, is $(1+x)^{n}$. | 36.20. It is clear that $(1+x)^{n}=C_{n}^{0}+C_{n}^{1} x+\ldots+C_{n}^{n} x^{n}=a_{0}+a_{1} x+\ldots$ $\ldots+a_{n} x^{n}$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,216 |
36.21. Let $f(x)=\sum_{k=1}^{\infty} c_{k} x^{k}$ be the generating function for the Catalan numbers, i.e., $c_{1}=1, c_{2}=c_{1} c_{1}$, and $c_{k}=c_{1} c_{k-1}+c_{2} c_{k-2}+\ldots+c_{k-1} c_{1}$ for $k \geqslant 3$.
a) Prove that $(f(x))^{2}=f(x)-x$.
b) Prove that $c_{n}=\frac{(2 n-2)!}{n!(n-1)!}$.
## 36.7. Bern... | 36.21. a) It is clear that $(f(x))^{2}=c_{1} c_{1} x^{2}+\left(c_{1} c_{2}+c_{2} c_{1}\right) x^{3}+\left(c_{1} c_{3}+c_{2} c_{2}+c_{3} c_{1}\right) x^{4}+\ldots=f(x)-x$.
b) The identity $(f(x))^{2}=f(x)-x$ shows that $f(x)=\frac{1}{2} \pm \frac{1}{2} \sqrt{1-4 x}$. Moreover, $f(0)=0$, so $f(x)=\frac{1}{2}-\frac{1}{2}... | c_{n}=\frac{(2n-2)!}{n!(n-1)!} | Algebra | proof | Yes | Yes | olympiads | false | 39,217 |
36.22. Prove that for Bernoulli numbers, when $k>1$, the equality $\sum_{p=0}^{k} B_{p} C_{k}^{p}=B_{k}$ holds, i.e., $\sum_{p=0}^{k-1} B_{p} C_{k}^{p}=0$.
The equality from problem 36.22 can formally be written as $(B+1)^{k}=B_{k}$, where it is implied that instead of $B^{p}$ we write $B_{p}$. | 36.22. From the definition of Bernoulli numbers, it follows that
$$
\left(1+\frac{x}{2!}+\frac{x^{2}}{3!}+\frac{x^{3}}{4!}+\ldots\right) \sum_{n=0}^{\infty} \frac{B_{n}}{n!} x^{n}=1
$$
In the expression on the left side, the coefficient of \(x^{k-1}\) is
$$
\frac{1}{k!} \cdot \frac{B_{0}}{0!}+\frac{1}{(k-1)!} \cdot ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,218 |
36.26. a) Prove that $B_{n}(z+1)-B_{n}(z)=n z^{n-1}$ for $n \geqslant 2$.
b) Prove that for $n \geqslant 1$
$$
1^{n}+2^{n}+3^{n}+\ldots+k^{n}=\frac{1}{n+1}\left(B_{n+1}(k+1)-B_{n+1}(0)\right)
$$ | 36.26. a) By definition
$$
B_{n}(z+1)-B_{n}(z)=\sum_{k=0}^{n} C_{n}^{k} B_{n-k} \sum_{p=0}^{k-1} C_{k}^{p} z^{p}
$$
The expression on the right-hand side does not contain $z^{p}$ for $p>n-1$. Moreover, the coefficient of $z^{n-1}$ is $C_{n}^{n} C_{n}^{n-1}=n$. It remains to prove that for a fixed $p, 0 \leqslant p \l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,222 |
36.28. Prove that
$$
1+\sum_{n=1}^{\infty} p(n) x^{n}=\left(\prod_{n=1}^{\infty}\left(1-x^{n}\right)\right)^{-1}
$$ | 36.28. It is clear that $\left(1-x^{n}\right)^{-1}=1+x^{n}+x^{2 n}+x^{3 n}+\ldots$ Therefore, the coefficient of $x^{m}$ in the formal series $\left(\prod_{n=1}^{\infty}\left(1-x^{n}\right)\right)^{-1}$ is equal to the number of representations of the number $m$ in the form $a_{1}+2 a_{2}+\ldots+k a_{k}$, where $a_{1},... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,224 |
36.29. Prove that
$$
\prod_{n=1}^{\infty}\left(1-x^{n}\right)=1+\sum_{n=1}^{\infty}(-1)^{n}\left(x^{\frac{3 n^{2}-n}{2}}+x^{\frac{3 n^{2}+n}{2}}\right)
$$
(Euler's identity). | 36.29. The infinite product $\prod_{n=1}^{\infty}\left(1-x^{n}\right)$ corresponds to a formal series, in which the coefficient of $x^{m}$ is obtained as follows. Consider all representations of the number $m$ in the form $m=n_{1}+n_{2}+\ldots+n_{k}$, where $n_{1} < n_{2} < \ldots < n_{k}$. Each such representation con... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,225 |
36.30. Prove that
$$
p(n)=\sum_{k=1}^{\infty}(-1)^{k+1}\left(p\left(n-\frac{3 k^{2}-k}{2}\right)+p\left(n-\frac{3 k^{2}+k}{2}\right)\right)
$$
(it is assumed that $p(0)=1$ and $p(m)=0$ for $m<0$). | 36.30. According to problem $36.28\left(1+\sum_{n=1}^{\infty} p(n) x^{n}\right)\left(\prod_{k=1}^{\infty}\left(1-x^{k}\right)\right)=1$. Further, according to problem 36.29
$$
\prod_{n=1}^{\infty}\left(1-x^{n}\right)=1+\sum_{k=1}^{\infty}(-1)^{k}\left(x^{\frac{3 k^{2}-k}{2}}+x^{\frac{3 k^{2}+k}{2}}\right)
$$
We write... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,226 |
36.31. Let $d(n)$ be the number of partitions of the number $n$ into distinct summands, and $l(n)$ be the number of partitions of the number $n$ into odd summands. Prove that $d(n)=l(n)$ (Euler).
## 36.9. Waring's formulas
Let $\sigma_{1}, \ldots, \sigma_{n}$ be the elementary symmetric polynomials in $x_{1}, \ldots,... | 36.31. For $d(n)$ the generating function is equal to
$$
(1+x)\left(1+x^{2}\right)\left(1+x^{3}\right)\left(1+x^{4}\right) \ldots
$$
and for $l(n)$ the generating function is equal to
$$
\frac{1}{1-x} \cdot \frac{1}{1-x^{3}} \cdot \frac{1}{1-x^{5}} \cdot \frac{1}{1-x^{7}} \cdot \cdots
$$
It is clear that
$$
(1+x)\... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,227 |
36.32. Prove that
$$
\frac{(-1)^{k} s_{k}}{k}=\sum \frac{(-1)^{l_{1}+l_{2}+\ldots+l_{n}}\left(l_{1}+l_{2}+\ldots+l_{n}-1\right)!}{l_{1}!\ldots l_{n}!} \sigma_{1}^{l_{1}} \ldots \sigma_{n}^{l_{n}}
$$
where the summation is over all sets of non-negative integers $l_{1}, \ldots, l_{n}$, for which $l_{1}+2 l_{2}+\ldots+n... | 36.32. Consider formal series
$$
\operatorname{Ln}\left(1+\frac{x_{i}}{x}\right)=-\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} \frac{x_{i}^{k}}{x^{k}}, \quad i=1, \ldots, n
$$
(They can be considered as formal series in the variables $x_{i} / x$.) Adding these formal series, we get
$$
\operatorname{Ln}\left(1+\frac{x_{1}}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,228 |
36.33. Prove that
$$
(-1)^{k} \sigma_{k}=\sum \frac{(-1)^{l_{1}+l_{2}+\ldots+l_{n}}}{1^{l_{1}} \cdot 2^{l_{2}} \cdot \ldots \cdot n^{l_{n}} \cdot l_{1}!\ldots l_{n}!} s_{1}^{l_{1}} \ldots s_{n}^{l_{n}}
$$
where the summation is over all sets of non-negative integers \( l_{1}, \ldots, l_{n} \) such that \( l_{1}+2 l_{... | 36.33. Consider the formal series
$$
\operatorname{Ln}\left(1-\frac{x_{1}}{x}\right) \ldots\left(1-\frac{x_{n}}{x}\right)=-\frac{s_{1}}{x}-\frac{s_{2}}{2 x^{2}}-\frac{s_{3}}{3 x^{3}}-\ldots
$$
Taking the formal exponent, we get
$$
\begin{array}{r}
\left(1-\frac{x_{1}}{x}\right) \ldots\left(1-\frac{x_{n}}{x}\right)=\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,229 |
38.2. Suppose that $\frac{\partial f}{\partial x}\left(x_{0}, y_{0}\right) \neq 0$ and $\frac{\partial f}{\partial y}\left(x_{0}, y_{0}\right) \neq 0$. Prove that the tangent at the point $\left(x_{0}, y_{0}\right)$ to the curve given by the equation $f(x, y)=0$ is defined by the equation
$$
\left(x-x_{0}\right) \frac... | 38.2. We will assume that in a sufficiently small neighborhood of the point ( $x_{0}, y_{0}$ ) the curve under consideration is the graph of a function $y=\varphi(x)$, although under the assumptions made about the partial derivatives not being zero, this could be proven.
By the mean value theorem
$$
\begin{aligned}
&... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,230 |
38.4. Find the envelope of the family of lines that cut off from a given right angle a triangle of area $a^{2} / 2$. | 38.4. Let's introduce a coordinate system, directing the axes along the sides of the given right angle. The lines of interest intersect the coordinate axes at points ( $\alpha a, 0$ ) and ( $0, a / \alpha$ ), where $\alpha>0$. The line corresponding to the parameter $\alpha$ is given by the equation $x+\alpha^{2} y=\al... | xy=\frac{^2}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,232 |
38.5. On the sides of an angle with vertex $O$, fixed points $A$ and $B$ are marked. Points $A_{1}$ and $B_{1}$ are chosen on segments $O A$ and $O B$ such that $O B_{1}: B_{1} B = A A_{1}: A_{1} O$. Prove that the envelope of the family of lines $A_{1} B_{1}$ is an arc of a parabola. | 38.5. We can assume that $O$ is the origin, $A=(1,1)$ and $B=(-1,1)$. Then $A_{1}=(1-\alpha, 1-\alpha)$ and $B_{1}=(-\alpha, \alpha)$ for some $\alpha \in[0,1]$. The line passing through points $A_{1}$ and $B_{1}$ is given by the equation $(2 \alpha-1) x+y=2 \alpha(1-\alpha)$. It is easy to verify that if
$\left(x_{0},... | \frac{1+x^{2}}{2} | Geometry | proof | Yes | Yes | olympiads | false | 39,233 |
38.6. Find the envelope of the trajectories of a material point thrown from the origin with a velocity $v_{0}$ in a fixed vertical plane.
\ The curve from problem 38.6 is called the safety parabola. | 38.6. Let's direct the $O y$ axis vertically upwards, and the $O x$ axis along the horizontal component of the velocity $v_{0}$. Then at time $t$, the material point has coordinates $x(t)=v_{0} \cos \alpha \cdot t$, $y(t)=v_{0} \sin \alpha \cdot t-\frac{g t^{2}}{2}$, where $\alpha$ is the angle at which it was launched... | \frac{v_{0}^{2}}{2}-\frac{}{2v_{0}^{2}}x^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 39,234 |
38.8. Prove that an astroid is the trajectory of a marked point on a circle of radius $1 / 4$, which rolls inside a fixed circle of radius 1, with the smaller circle located inside the larger one.
The trajectory of a marked point on a circle of radius $r$, which rolls inside a fixed circle of radius $R$, with the circ... | 38.8. First solution. Let's derive the parametric representation of the trajectory of the marked point in the general case when a circle of radius $r$ rolls inside a circle of radius $R$. This motion of the marked point can be represented as the rotation of the center of the smaller circle along a circle of radius $r_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,236 |
38.9. a) Fix a number $k \neq 0, \pm 1$ and consider the family of lines, each connecting the points $e^{i \varphi}$ and $e^{i k \varphi}$. Prove that the envelope of this family of lines is a hypocycloid or an epicycloid.
b) For each integer $k \neq 0, \pm 1$, find the number of cusps. | 38.9. a) Let $A=e^{i \varphi}, B=e^{i k \varphi}, A^{\prime}=e^{i(\varphi+\alpha)}, B^{\prime}=e^{i k(\varphi+\alpha)}$. Further, let $C$ be the limiting position of the intersection point of the lines $A B$ and $A^{\prime} B^{\prime}$ as $\alpha \rightarrow 0$. Clearly, if $k>0$, then the point $C$ lies on the segment... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,237 |
38.11. Consider a circle $S$ and choose a point $A$ on it. Light rays are emitted from point $A$ and reflect off the circle. Prove that the envelope of the reflected rays is a cardioid (Fig. 38.3).
## 38.3. Curvature
Let $\gamma(t)=(x(t), y(t))$ be a parameterized curve. We will assume that $v(t)=\frac{d \gamma}{d t}... | 38.11. Let's assume that the circle is unitary, and $A=(-1,0)$. Then the ray hitting the point $e^{i \psi}$, after reflection, hits the point $e^{i(2 \psi+\pi)}$. Let $\psi=\varphi+\alpha$. Then $2 \psi+\pi=2 \varphi+2 \alpha+\pi=$ $=2 \varphi+\alpha$ for $\alpha=-\pi$. As a result, we find ourselves in the situation o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,239 |
38.12. Prove that the curvature of a circle of radius $R$ is $1 / R$. | 38.12. A circle of radius $R$ can be parametrically defined by the formulas $x(t)=R \cos \omega t, y(t)=R \sin \omega t$. In this case,
$$
v(t)=(-\omega R \sin \omega t, \omega R \cos \omega t)
$$
Let $\omega=1 / R$. Then $|v(t)|=1$, i.e., $t$ is a natural parameter. In this case, $\frac{d v}{d t}=-R^{-1}(\cos \omega... | 1/R | Geometry | proof | Yes | Yes | olympiads | false | 39,240 |
38.13. Prove that for a curve $\gamma(t)=(x(t), y(t))$ with an arbitrary parameterization $t$ the curvature is computed by the formula
$$
k^{2}=\frac{\left(x^{\prime \prime} y^{\prime}-y^{\prime \prime} x^{\prime}\right)^{2}}{\left(x^{\prime 2}+y^{\prime 2}\right)^{3}}
$$ | 38.13. If $s$ is a natural parameter, then $\frac{d \gamma}{d t}=\frac{d \gamma}{d s} \cdot \frac{d s}{d t}$ and $\left|\frac{d \gamma}{d s}\right|=$ = 1. Therefore, $\left|\frac{d \gamma}{d t}\right|^{2}=\left(\frac{d s}{d t}\right)^{2}$. Hence, $\frac{d \gamma}{d s}=\frac{d \gamma}{d t} \cdot\left|\frac{d \gamma}{d t... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,241 |
38.14. Calculate the curvature of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at each point.
## 38.4. Osculating Circle
Consider two curves on a plane given by the equations $y=f(x)$ and $y=g(x)$. These curves intersect if $f\left(x_{0}\right)=g\left(x_{0}\right)$. The curves touch (have a contact of orde... | 38.14. Consider the following parameterization of an ellipse: $x(t) = a \cos t, y(t) = b \sin t$. Then
$$
\begin{gathered}
x'' y' - y'' x' = -ab(\cos^2 t + \sin^2 t) = -ab, \\
x'^2 + y'^2 = a^2 \sin^2 t + b^2 \cos^2 t
\end{gathered}
$$
Therefore, according to problem 38.13,
$$
k^2 = \frac{a^2 b^2}{(a^2 \sin^2 t + b^... | k^2=\frac{^2b^2}{(^2\sin^2+b^2\cos^2)^3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 39,242 |
38.15. Prove that the center of curvature is the limiting position of the intersection point of nearby normals.
## 38.5. Focal Points. Evolute
Let $\gamma(s)=(x(s), y(s))-$ be a curve with natural parameterization. Fix a point $q=\left(x_{0}, y_{0}\right)$ in the plane and consider the function $F(s)=|\gamma(s)-q|^{2... | 38.15. We can assume that the curve is given by the equation $y=f(x)$, and $f^{\prime}(0)=0$. We are interested in the center of curvature for the point $(0, f(0))$. The normal to the curve at the point $(\varepsilon, f(\varepsilon))$ is given by the equation $(x-\varepsilon)+f^{\prime}(\varepsilon)(y-f(\varepsilon))=0... | (0,f(0)+\frac{1}{f^{\\}(0)}) | Calculus | proof | Yes | Yes | olympiads | false | 39,243 |
38.16. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ find the focal point corresponding to the point $\left(a \cos t_{0}, b \sin t_{0}\right)$; find the radius of curvature of the ellipse at this point. | 38.16. For a fixed point ( $x_{0}, y_{0}$ ), consider the function $F(t)=\left(x_{0}-a \cos t\right)^{2}+\left(y_{0}-b \sin t\right)^{2}$ on the ellipse. We are interested in the point $\left(x_{0}, y_{0}\right)$ for which $F^{\prime}\left(t_{0}\right)=0$ and $F^{\prime \prime}\left(t_{0}\right)=0$, i.e.,
$$
\begin{al... | R^{2}=\frac{(^{2}\sin^{2}t_{0}+b^{2}\cos^{2}t_{0})^{3}}{^{2}b^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,244 |
38.17. Find the equation of the evolute of an ellipse.
I A cycloid is a curve described by a fixed point on a circle rolling along a fixed straight line.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 38.17. The expressions for the foci obtained in the solution of problem 38.16 show that they lie on the curve
$$
(a x)^{2 / 3}+(b y)^{2 / 3}=\left(a^{2}-b^{2}\right)^{2 / 3}
$$ | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,245 | |
38.18. Prove that the evolute of a cycloid is the same cycloid (obtained from the original curve by a parallel translation).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 38.18. Let the fixed line be the $O x$ axis, and at the initial moment, the fixed point coincides with the origin $O$. We will also assume that the circle has a radius of 1 and rolls in the upper half-plane. Suppose that after some time, the fixed point has moved to point $X$; at this moment, the rolling circle $S_{1}$... | proof | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,246 |
39.1. Given a non-empty finite set. Prove that the number of its subsets containing an even number of elements is equal to the number of subsets containing an odd number of elements.
## 39.2. Operations on Sets
In set theory, the following notations are used:
$x \in X: x$ is an element of set $X$ ( $x$ belongs to se... | 39.1. We fix one element $x$ in the given set. Subsets of this set can be paired as follows: we take an arbitrary subset that does not contain $x$, and as the second set of the pair, we take the same subset and add the element $x$ to it. In each pair, one set consists of an even number of elements, and the other - an o... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,247 |
39.2. a) Prove that $A \cup(B \cap C)=(A \cup B) \cap(A \cup C)$.
b) Prove that $A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$. | 39.2. a) If $x \in A \cup (B \cap C)$, then $x \in A$ or $x \in B \cap C$. If $x \in A$, then $x \in A \cup B$ and $x \in A \cup C$, so $x \in (A \cup B) \cap (A \cup C)$. If $x \in B \cap C$, then $x \in B$ and $x \in C$, so $x \in A \cup B$ and $x \in A \cup C$, which means $x \in (A \cup B) \cap (A \cup C)$.
If $x ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,248 |
39.3. Prove that $\overline{A \cup B}=\bar{A} \cap \bar{B}$ and $\overline{A \cap B}=\bar{A} \cup \bar{B}$ (De Morgan's laws).
- Identities similar to those proven in problems 39.2 and 39.3 are conveniently proven using so-called Venn diagrams, in which the considered sets are depicted as intersecting circles. In the ... | 39.3. $x \in \overline{A \cup B} \Leftrightarrow x \notin A \cup B \Leftrightarrow x \notin A$ and $x \notin B \Leftrightarrow x \in \bar{A} \cap \bar{B}$. $x \in \overline{A \cap B} \Leftrightarrow x \notin A \cap B \Leftrightarrow x \notin A$ or $x \notin B \Leftrightarrow x \in \bar{A} \cup \bar{B}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,249 |
39.5. Prove that the segments $[0,1]$ and $[0, a]$ are equinumerous for any positive number $a$. | 39.5. A one-to-one mapping $[0,1] \rightarrow[0, a]$ is given by the formula $x \mapsto a x$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,251 |
39.6. Prove that the interval $(0,1)$ and the ray $(1,+\infty)$ are equinumerous. | 39.6. A one-to-one mapping $(0,1) \rightarrow(1,+\infty)$ is given by the formula $x \mapsto 1 / x$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,252 |
39.7. Prove that the interval $(-1,1)$ is equinumerous to the set of all real numbers. | 39.7. The function $f(x)=\operatorname{tg}\left(\frac{\pi x}{2}\right)$ defines a one-to-one mapping of the interval $(-1,1)$ onto the set of all real numbers. | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,253 |
39.8. Prove that the set of all infinite sequences $a_{1}, a_{2}, \ldots$, where $a_{i}=0$ or 1 for all $i$, is equinumerous to the set of all subsets of the natural numbers. | 39.8. Let us associate with the sequence $a_{1}, a_{2}, \ldots$ the set of those natural numbers $n$ for which $a_{n}=1$. As a result, we obtain a one-to-one correspondence between sequences of the specified type and sets of natural numbers. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,254 |
39.9. Prove that the segment $[0,1]$ is equinumerous to the half-interval $[0,1)$.
## 39.4. Countable Sets
A set is called countable if it is equinumerous to the set of natural numbers. | 39.9. Let's choose an infinite sequence of pairwise distinct numbers $a_{1}, a_{2}, \ldots$, strictly between 0 and 1. We will construct a mapping $f:[0,1] \rightarrow[0,1)$ as follows: $f(1)=a_{1}$, $f\left(a_{i}\right)=a_{i+1}$ and $f(x)=x$, if $x$ is different from 1 and from $a_{i}$ for all $i$. It is clear that th... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,255 |
39.10. Prove that the set of integers is countable. | 39.10. In the sequence $0,1,-1,2,-2,3,-3, \ldots$ each integer occurs exactly once. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,256 |
39.11. Prove that the set of rational numbers is countable. | 39.11. For each natural $k$, we write the sequence $\frac{-k+1}{1}, \frac{-k+2}{2}, \ldots, \frac{-1}{k-1}, \frac{0}{k}, \frac{1}{k-1}, \ldots, \frac{k-1}{1}$. We write such sequences for $k=1,2,3, \ldots$ After that, we will go through the obtained sequence and cross out all the numbers that have already appeared to u... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,257 |
39.12. Prove that any subset of a countable set is either finite or countable. | 39.12. Let us number the elements of the given countable set, and let the elements of the given subset have numbers $i_{1}<i_{2}<$ $<i_{3}<\ldots$ The set of these numbers is either finite (in which case the given subset is finite), or infinite (in which case the given subset is countable). | proof | Other | proof | Yes | Yes | olympiads | false | 39,258 |
39.13. Prove that the union of a countable set of countable sets is countable. | 39.13. Let the $k$-th countable set consist of elements $a_{k 1}$, $a_{k 2}$, $a_{k 3}$, $\ldots$ Then the sequence of elements $a_{11} ; a_{12}, a_{21} ; a_{13}$, $a_{22}, a_{31} ; a_{14}, a_{23}, a_{32}, a_{41} ; \ldots$ contains all elements of these sets. It is only necessary to remove the repeating elements from i... | proof | Other | proof | Yes | Yes | olympiads | false | 39,259 |
39.14. Prove that for each natural number $n$ the set of all sequences of $n$ natural numbers is countable. | 39.14. The set of ordered pairs of natural numbers can be represented as the union of a countable set of countable sets (the first number of the pair can be considered as the number of the set). Therefore, this set is countable (problem 39.13). Let's enumerate the pairs of natural numbers. A triplet of natural numbers ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,260 |
39.16. Prove that the set of all algebraic numbers is countable. | 39.16. The set of all polynomials with integer coefficients of a given degree $n$ is countable. Each of them has no more than $n$ distinct roots, so the set of roots of polynomials of degree $n$ with integer coefficients is countable (polynomials of the form $p x^{n}-q x^{n-1}$ show that it is infinite). Therefore, the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,262 |
39.17. Prove that the set of pairwise non-intersecting segments on a line is finite or countable. | 39.17. Each of the segments contains a rational point, and for non-intersecting segments, these points are different. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,263 |
39.18. Prove that the set of pairwise non-intersecting circles on the plane is finite or countable. | 39.18. Each of the circles contains a point, both coordinates of which are rational, and for non-intersecting circles these points are different. | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,264 |
39.19. Prove that if the set $X$ is infinite, and the set $Y$ is finite or countable, then the set $X \cup Y$ is equinumerous to $X$.
## 39.5. Power of the Continuum
A set is said to have the power of the continuum if it is equinumerous to the set of real numbers. | 39.19. We can consider that $X$ and $Y$ do not intersect. Indeed, instead of $Y$, we can take the set $Y^{\prime}$, consisting of those elements of $Y$ that do not belong to $X$. The set $Y^{\prime}$ is a subset of $Y$, so it is finite or countable.
Let $X_{1}$ be a countable subset of $X$, and $X_{2}$ be the compleme... | proof | Other | proof | Yes | Yes | olympiads | false | 39,265 |
39.21. a) Prove that the set of all infinite sequences $a_{1}, a_{2}, \ldots$, where $a_{i}=0$ or 1, has the power of the continuum.
b) Prove that the set of all subsets of the set of natural numbers has the power of the continuum. | 39.21. a) It is sufficient to prove that the set of points of the segment $[0,1]$ is equipotent to the set of all infinite sequences $a_{1}, a_{2}, \ldots$, where $a_{i}=0$ or 1. We can associate such a sequence with the number $\frac{a_{1}}{2}+\frac{a_{2}}{2^{2}}+\frac{a_{3}}{2^{3}}+\ldots$ This representation of a nu... | proof | Other | proof | Yes | Yes | olympiads | false | 39,267 |
39.22. Prove that the set of all points in a square has the power of the continuum.
## 39.6. Properties of Cardinality | 39.22. A point in the square can be represented as a pair of points from the segment. According to problem 39.21, the set of points of the segment is equinumerous with the set of infinite sequences $a_{1}, a_{2}, \ldots$, where $a_{i}=0$ or 1. Therefore, it is sufficient to prove that the set of ordered pairs of such s... | proof | Other | proof | Yes | Yes | olympiads | false | 39,268 |
39.23. Prove that no set \( X \) is equipotent to the set \( P(X) \) of all its subsets (Cantor). | 39.23. Suppose we have managed to establish a one-to-one correspondence $\varphi: X \rightarrow P(X)$. Let $Y$ be the set of those elements $x \in X$ for which $x \notin \varphi(x)$. We will prove that no element of the set $X$ corresponds to this subset. Indeed, suppose that $Y=\varphi(y)$. Then
$$
y \in Y \Leftright... | proof | Other | proof | Yes | Yes | olympiads | false | 39,269 |
1. Write a million: a) using three hundreds and operation signs; b) using six tens and operation signs. | 1. a) $100 \cdot 100 \cdot 100=1000000$; б) $10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10=1000000$. | 1000000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,271 |
2. Write down and read all seven-digit numbers, the sum of the digits in each of which is equal to 2. How many such numbers are there? | 2. $2000000,1100000,1010000,1001000,1000100,1000010,1000001$. Bcero 7 numbers. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,272 |
3. The number 686 is written on a piece of paper. How can you get a number that is 303 greater than the given one, without making any writings or calculations?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 3. By turning the sheet of paper halfway around a full rotation, we get the number 989, which is greater than 686 by $303(989-686=303)$. | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,273 | |
4. Tanya wrote a certain two-digit number on a piece of paper; Svetlana, sitting opposite, perceived the written number as another, smaller by 75. What number did Tanya write?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 4. Svetа saw a number that results from rotating a piece of paper by half a full turn, so the number could only contain the digits $0,1,8,6$ and 9. The only number that satisfies this condition is 91. Svetа saw 16, so $91-16=75$. | Logic and Puzzles | other | Yes | Yes | olympiads | false | 39,274 | |
6. In all three equalities XIII $=$ VII - VI, VII $=\mathbf{V}-\mathbf{I}, \mathrm{XI}+\mathbf{V}=\mathbf{V}$, composed of sticks, errors have been made. Move one stick in each of them so that the equalities become correct. | 6. $\mathrm{XIII}-\mathrm{VII}=\mathrm{VI} ; \mathrm{VI}=\mathrm{V}+\mathrm{I} ; \mathrm{XI}-\mathrm{V}=\mathrm{VI}$. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,276 |
7. How many two-digit numbers are there in which: a) there is at least one digit 3; b) the number of tens is less than the number of units? | 7. a) The digit 3 starts 10 numbers (from 30 to 39 inclusive), and ends in this digit in a total of 9 numbers, but the number 33 has already been counted, so there are 8 such numbers left. Therefore, the total number of such numbers is $10+8=18$.
b) From 10 to 20, there are 8 such numbers (12, 13, ..., 19), from 21 to... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,277 |
8. How many natural numbers not exceeding 1000 are there such that each subsequent digit is greater than the previous one? | 8. Among two-digit numbers, there are 36 (see problem 7). Let's count how many such numbers there are among three-digit numbers. From 123 to $129-7$ numbers, from 134 to $139 \cdots 6$ and so on; finally, one number 189. Thus, among three-digit numbers, where the hundreds digit is 1, there are $7+6+\ldots+2+1=28$ such ... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,278 |
13. In the number 513879406, cross out 4 digits so that the remaining digits in the same order form: a) the largest number; b) the smallest number. | 13. a) $89406 ;$ б) 13406 . | 89406 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,282 |
14. All numbers from 1 to 40 are written in a row. Without changing the order of writing, erase 60 digits so that the remaining digits express: a) the smallest number; b) the largest number. | 14. a) In writing numbers from 1 to 40 inclusive, 71 digits are used. To make the remaining 11 digits express the smallest number, one must erase 60 larger digits, and in the remaining number, the smallest digits should be on the left. First, there are three zeros (remaining from the numbers 10, 20, 30). Then come 1, 2... | 12333330 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,283 |
17. Students were given the task to write several different three-digit numbers that do not contain the digit 7 in their notation. How many such numbers can be written in total? | 17. First, let's calculate the number of three-digit numbers in which there is at least one digit 7. Firstly, we notice that from 700 to 800, there are 100 such numbers; secondly, within each other hundred, there are 19 such numbers. (For example, from 100 to 200, the numbers with the tens digit 7 will be 10. These are... | 648 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,286 |
18. Can a two-digit number equal the sum of its digits? | 18. Suppose there exists a number $10 x+y$, then $10 x+y=$ $=x+y, 9 x=0$, from which $x=0$, which contradicts our assumption. Therefore, such a number does not exist. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,287 |
19. The lock code is a seven-digit number, the first three digits of which are the same, and the last four digits are also the same. The sum of all the digits of this number is a two-digit number, the first digit of which matches the first digit of the code, and the last digit matches the last digit of the code. Find t... | 19. Let the cipher be $\overline{a a b b b b b}$, then by the condition $3 a+4 b=10 a+b$, from which $7 a=3 b$. Since $a$ and $b$ are digits, the equality is possible only when $a=3$ and $b=7$. Therefore, the cipher is 3337777. (The notation $\bar{a} a b b b b$ represents a seven-digit number, where $b$ is the unit of ... | 3337777 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,288 |
20. Andrei was asked to name the apartment number that his family received in the new building. He replied that this number is 17 times greater than the number in the units place of the apartment number. What is the apartment number? | 20. If the desired number is $10 a+b$, then $10 a+b=17 b$, from which $5 a=8 b$. The equality is possible only when $a=8$ and $b=5$. Answer: 85. | 85 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,289 |
21. Observant Yura noticed that if in a two-digit number expressing the distance in kilometers they traveled today, a zero is inserted between the tens and units digits, the resulting number is 9 times greater than the original. What distance did they travel? | 21. Let the distance be expressed by the number $10 a+b$. If we insert a zero between the tens and units digits, we get $100 a+b$. According to the condition, we have $100 a+b=9(10 a+b)$, so $10 a=8 b$, which means $5 a=4 b, a=4, b=5$. Answer: 45 km. | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,290 |
22. If to any two-digit number we append on the right the number written with the same digits but in reverse order, we get a four-digit number that is divisible by 11 without a remainder. Prove it. | 22. Let the two-digit number be $\overline{a b}$, then the obtained four-digit number is $\bar{a} \bar{b} \bar{b} a$. Since $\overline{a b b a}=1000 a+100 b+10 b+a=1001 a+110 b$ and $1001: 11$, as well as $110: 11$, then $\overline{a b b a}: 11$.
Note. The notation $1001: 11$ means that 1001 is divisible by 11. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,291 |
23. The distance between two cities in kilometers is expressed as a two-digit number, where the left digit is equal to the difference between this number and the number written with the same digits but in reverse order. What is this distance? | 23. Let the desired number be $10 a+b$. Then
$$
a=(10 a+b)-(10 b+a), a=9 a-9 b, 9 b=8 a
$$
from which $a=9, b=8$. Answer: 98 km. | 98 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,292 |
24. Find the number whose sum of digits equals the difference between 328 and the sought number. | 24. Obviously, the desired number is a three-digit number. Let it be $100a + 10b + c$, then $328 - 100a - 10b - c = a + b + c$, which simplifies to $328 = 101a + 11b + 2c$, from which $a = 3$. Therefore, $11b + 2c = 328 - 303 = 25$, from which $b = 1$ and $c = 7$. Thus, the desired number is 317. | 317 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,293 |
25. Does there exist a three-digit number that is reduced three times by moving its initial digit to the end of the number? | 25. Suppose there is a number $100 a+10 b+c$ that satisfies the condition of the problem, then $100 a+10 b+c=3(100 b+10 c+a), 97 a-290 b-29 c=0$, from which $c=\frac{97 a}{29}-10 b$. Since $a$ is a digit, different from 0, then $1 \leqslant a \leqslant 9$. At the same time, the number $\frac{97 a}{29}$ cannot be an int... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,294 |
27. Find the digit that has the property that if it is appended to the end of any natural number, the resulting number is equal to the sum of three terms, the first of which is the original number, the second is the number represented by the sought digit, and the third is the product of the first two terms. | 27. Let the natural number be $a$, and the appended digit be $x$. Then according to the condition $a \cdot 10 + x = a + x + a x$, from which $9 a = a x$, and therefore, $x = 9$. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,296 |
28. Write down a four-digit number where each subsequent digit is 1 greater than the previous one, then write the number with the same digits but in reverse order and subtract the smaller number from the larger one. Repeat this several times with different numbers and compare the results. Solve the problem in general t... | 28. Let the digits of the number $a, a+1, a+2$ and $a+3$, then one number is $1000 a + 100(a+1) + 10(a+2) + (a+3)$, and the number written with the same digits but in reverse order is $1000(a+3) + 100(a+2) + 10(a+1) + a$. Let's find the difference between these numbers: $1000 a + 3000 + 100 a + 200 + 10 a + 10 + a - 10... | 3087 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,297 |
29. All natural numbers from 1 to 100 inclusive are divided into 2 groups - even and odd. Determine in which of these groups the sum of all digits used to write the numbers is greater and by how much. | 29. Odd numbers $1,3,5,7, \ldots, 99$; even numbers $2,4,6,8, \ldots, 100$. Let's compare the sums of the single-digit numbers in these groups: $1+3+5+7+9=25,2+$ $+4+6+8=20,25-20=5$. Consider the numbers with the tens digit 1. The sum of the digits will be greater in the group that has a greater sum of the units digits... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,298 |
30. The number of students studying in the V-VI grades of the school is expressed as a three-digit number. From the digits of this number (without repeating them), 6 different two-digit numbers can be formed, the sum of which is twice the number of students in the V-VI grades. How many students are there in these grade... | 30. Let the desired number be $100a + 10b + c$, then we can form the following two-digit numbers: $10a + b, 10b + a, 10a + c, 10c + a, 10b + c, 10c + b$. According to the condition, we have
$(10a + b) + (10b + a) + (10a + c) + (10c + a) + (10b + c) + (10c + b) = 2(100a + 10b + c), 22a + 22b + 22c = 2(100a + 10b + c), ... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,299 |
31. The number of poplars planted by the students is expressed as a three-digit number, and the number of lindens as a two-digit number. In total, they planted 144 trees. If the extreme digits in these two numbers are swapped and the resulting numbers are added, the sum is 603. How many poplars and how many lindens wer... | 31. Since the sum of a three-digit and a two-digit number is 144, the hundreds digit of the three-digit number is 1. The sum of this digit and the tens digit of the two-digit number is 3, as the sum of the numbers written in reverse order is 603; therefore, the tens digit of the two-digit number is $2(3-1=2)$. The unit... | 115 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,300 |
32. Two identical two-digit numbers were written on the board. To one of them, 100 was added to the left, and to the other, 1 was added to the right, as a result of which the first number became 37 times larger than the second. What numbers were written on the board? | 32. When multiplying the second number, ending with the digit 1, by 37, we get a number ending with the digit 7. This means the original numbers ended with the digit 7, i.e., $\overline{a 7}$. According to the condition, $\overline{100 a 7}=\overline{a 71} \cdot 37$, $\overline{a 71} \cdot 37=(100 a+71) \cdot 37=3700 a... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,301 |
33. How many six-digit numbers are there in which all digits are odd? | 33. The total number of different six-digit numbers is 900000. (How to determine this?) Of these, half end with an odd digit, i.e., 450000. Of these numbers, those with the second digit from the right being odd are $450000: 2=225000$, of which those with the third digit from the right being odd are $225000: 2=112500$, ... | 15625 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,302 |
34. Using each of the nine digits $1,2,3, \ldots, 9$ exactly once, write down three numbers such that their sum is the greatest possible, and the difference between the largest and smallest numbers is as small as possible. Find this sum and difference. | 34. To make the sum the largest, the hundreds should be 9, 8, and 7, the tens should be 6, 5, 4, and the units should be 3, 2, and 1. The larger number has 9 hundreds, and the smaller one has 7. To make their difference the smallest, the larger number should have as few tens and units as possible, while the smaller num... | 2556,178 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,303 |
36. The number of students in grades V-VI of one of the schools is expressed by a three-digit number. If you find the product of the digits of this number, and then the product of the digits of the resulting number, then all these three numbers can be written as:
$$
\triangle \bigcirc O ; \triangle \square ; \square
$... | 36. By comparing the numbers $\Delta \square$ and $\square$, we get that $\triangle=1$, because only by multiplying the number $\square$ by 1 do we get the same number $\square$. By comparing 100 and $1 \square$, we get that $\bigcirc=4$, because only by multiplying one by 4 and then again by 4 do we get a number with ... | 144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,305 |
37. In the hundred-digit number $12345678901234567890 \ldots 1234567890$, all digits in odd positions were erased. In the resulting fifty-digit number, digits in odd positions were erased again. The erasing continued until nothing was left. What was the last digit to be erased? | 37. The first time, 50 odd digits were crossed out, leaving 50 even ones. The second time, 25 digits were crossed out, including 5 twos, 5 fours, 5 sixes, 5 eights, and 5 zeros. 25 remained in the following sequence: $4,8,2,6,0,4$, $8,2,6,0, \ldots, 2,6,0$. (The group of digits 4, 8, 2, 6,0 is written in sequence 5 tim... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,306 |
38. "How many students are there in our city?" the children asked. "And you will find out yourself if you know that the number representing the number of students is the largest of all such numbers, where any two adjacent digits always form a number that is divisible by 23," the teacher replied. Using this information,... | 38. Let the desired number be $\overline{x_{1} x_{2} x_{3} \ldots x_{n}}$. It is known that $\overline{x_{i} x_{i+1}}: 23$. For $\overline{x_{i} x_{i+1}}$, the possible values are: $00,23,46,69,92$. Considering these 5 possible options, we get 46923. This number will be the largest of the possible such numbers, since n... | 46923 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,307 |
39. Find all three-digit numbers $x$, in the notation of which the digits do not repeat, such that the difference between this number and the number written with the same digits but in reverse order is also a three-digit number consisting of the same digits as the number $x$. | 39. Let $x=100 a+10 b+c$. Then $(100 a+10 b+c)-(100 c+10 b+a)=$ $=99(a-c)$. The difference of digits $(a-c)$ can take values from 2 to 8. By assigning such values ( $a-c$ ), we get $198,297,396,495,594,693,792$. Let's consider all possible cases.
1) The desired number can consist of the digits 1,9 and 8. Then we have
... | 954459 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,308 |
40. Write the number 1980 in words and count the total number of letters in these words. Write the number expressing the count of letters in words and count the letters in them again, then repeat this operation 10 more times and write down all 13 numbers in sequence.
a) Identify the pattern in writing this sequence st... | 40. "One thousand nine hundred eighty" has 26 letters, "twenty-six" has 13 letters, "thirteen" has 10 letters, "ten" has 6 letters, "six" has 5 letters, "five" has 4 letters, "four" has 6 letters, "six" has 5 letters, "five" has 4 letters. We have:
$1980, 26, 13, 10, 6, 5, 4, 6, 5, 4, 6, 5, 4$.
a) Starting from the f... | 6,5,4 | Other | math-word-problem | Yes | Yes | olympiads | false | 39,309 |
1. For a gift to his son, the father bought 4 books. All books except the first cost 84 k., except the second - 80 k., except the third - 76 k., and except the fourth - 72 k. What is the cost of each book? | 1. If we add four numbers $84, 80, 76, 72$, we get the tripled cost of all books, so the cost of the books is: $(84+80+76+72): \dot{3}=$ $=312: 3=104$ (k.). Therefore, the first book costs $104-84=20$ (k.), the second $104-80=24$ (k.), the third $104-76=28$ (k.), and the fourth $104-72=32$ (k.). | 20,24,28,32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,310 |
3. In a school from I to VIII grade, there is only one class per grade. In grades I-IV, there are a total of 130 students. In grade V, there are 7 more students than in grade II, in grade VI, there are 5 fewer students than in grade I, in grade VII, there are 10 more students than in grade IV, and in grade VIII, there ... | 3. Since there are 8 more students in grades V-VIII than in grades I-IV $(7-5+10-4=8)$, there are a total of 138 students in grades V-VIII ( $130+8=138$ ), and a total of 268 students in grades I-VIII $(130+138=268)$. | 268 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,312 |
4. Vitya and Vova collected 27 kg of waste paper together. If the number of kilograms of waste paper collected by Vitya were increased by 5 times, and that collected by Vova by 3 times, they would have 111 kg together. How many kilograms did each boy collect? | 4. If the number of kilograms of waste paper collected by the two boys is tripled, we would get $81 (27 \cdot 3=81)$. Then $111-81=30$ (kg) is the doubled number of kilograms of waste paper collected by Vitya, which means Vitya collected 15 kg $(30: 2=15)$, then Vova - 12 kg $(27-15=12)$. | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,313 |
5. Over three days, juice was sold in the cafeteria. On the first day, 1 large and 4 medium cans were sold, on the second day - 2 large and 6 liter cans, and on the third day - 1 large, 3 medium, and 3 liter cans. How many liters of juice were sold over 3 days, if the same amount of juice was sold each day? | 5. Comparing what was sold on the first and third days (I - 1 large jar and 4 medium; III - 1 large jar, 3 medium, and 3 liter jars), we get that the medium jar has a capacity of 3 liters. Therefore, for:
I day - 1 large jar and $3 \cdot 4=12$ (L);
II day - 2 large jars and 6 L. From this, the capacity of the large j... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,314 |
6. The sum of the minuend, subtrahend, and difference is 1024. Find the minuend, subtrahend, and difference, if the difference is 88 less than the subtrahend. | 6. We use such a schematic notation.
Since the minuend + subtrahend + difference $=1024$ and minuend $=$ subtrahend + difference, then minuend $\cdot 2=1024$, from which the minuend is 512. If the difference is $x$, then the subtrahend is $x+88$, and then $x+x+88=$ $=512$, from which $x=212$. Therefore, the difference... | 512,300,212 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,315 |
8. How many different products divisible by ten can be formed from the numbers $2,3,5,7,9 ?$ | 8. For the product of numbers to be divisible by 10, the factors must include the numbers 2 and 5. Taking this into account, the following products satisfy the condition of the problem:
1) $2 \cdot 5$
2) $2 \cdot 5 \cdot 3$
3) $2 \cdot 5 \cdot 7$
4) $2 \cdot 5 \cdot 9$;
5) $2 \cdot 5 \cdot 3 \cdot 7$
6) $2 \cdot 5 \cdo... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,317 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.