problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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1.86 Proof: Any natural number not exceeding $n$! can be represented as the sum of at most $n$ numbers, such that any two of these numbers are not equal, and each number is a divisor of $n!$.
| ［Proof］When $n=1$, the conclusion is obviously true.
Assume that the conclusion holds for $n=k$.
Let $a \leqslant(n+1)!$ and $a=d(n+1)+r$, where $d \leqslant n!, 0 \leqslant r \leqslant n+1$.
By the induction hypothesis,
$$d=d_{1}+d_{2}+\cdots+d_{l}$$
where $d_{i}$ are distinct divisors of $n!$ and $l \leqslant n$. T... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,825 |
6-14 The function $f$ defined on ordered pairs of positive integers satisfies the following three properties:
(1) $f(x, x)=x$;
(2) $f(x, y)=f(y, x)$;
(3) $(x+y) f(x, y)=y f(x, x+y)$.
Try to compute: $f(14,52)$. | [Solution] From (3), we get
$$f(x, x+y)=\frac{x+y}{y} \cdot f(x, y)$$
Thus, when $z>x$, we have
$$f(x, z)=\frac{z}{z-x} f(x, z-x) .$$
Therefore,
$$\begin{aligned}
f(14,52) & =\frac{52}{38} f(14,38) \\
& =\frac{52}{38} \cdot \frac{38}{24} f(14,24) \\
& =\frac{52}{24} \cdot \frac{24}{10} f(14,10) \\
& =\frac{52}{10} f(... | 364 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,826 |
For any positive integer $k, f_{1}(k)$ represents the square of the sum of the digits of $k$, and
$$\begin{array}{l}
\quad f_{n}(k)=f_{1}\left(f_{n-1}(k)\right), n \geqslant 2 . \\
\text { Find: } f_{1988}(11) \text {. }
\end{array}$$ | [Solution] From the given, we have
$$\begin{array}{l}
f_{1}(11)=(1+1)^{2}=4, \\
f_{2}(11)=f_{1}(4)=4^{2}=16, \\
f_{3}(11)=f_{1}(16)=(1+6)^{2}=49, \\
f_{4}(11)=f_{1}(49)=(4+9)^{2}=169, \\
f_{5}(11)=f_{1}(169)=(1+6+9)^{2}=256, \\
f_{6}(11)=f_{1}(256)=(2+5+6)^{2}=169, \\
f_{7}(11)=f_{1}(169)=256 .
\end{array}$$
From this... | 169 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,827 |
6 - 16 Find the value of $10 \operatorname{ctg}(\operatorname{arcctg} 3+\operatorname{arcctg} 7+\operatorname{arcctg} 13+\operatorname{arcctg} 21)$. | [Solution] Let $a_{n}=1+n+n^{2}$, then $a_{1}=3, a_{2}=7, a_{3}=13, a_{4}=21$
We first prove a formula:
$$\operatorname{arcctg}\left(1+n+n^{2}\right)=\operatorname{arctg}(n+1)-\operatorname{arctg} n$$
In fact, we can set
$$\alpha=\operatorname{arctg}(n+1), \beta=\operatorname{arctg} n$$
Thus $0<\beta, 0<\alpha-\beta<... | 15 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,828 |
6・17 Function $f$ is defined on the set of integers and satisfies
$$f(n)=\left\{\begin{array}{l}
n-3, \quad \text { when } n \geqslant 1000 ; \\
f(f(n+5)), \text { when } n<1000 .
\end{array}\right.$$
Find $f(84)$. | [Solution] Since $84+5 \times 184=1004>1000$, we have
$$\begin{aligned}
f(84) & =\underbrace{f \cdots f}_{185 \uparrow}(1004)=\underbrace{f \cdots f}_{184 \uparrow}(1001) \\
& =\underbrace{f \cdots f}_{183 \uparrow}(998)=\underbrace{f \cdots f}_{184 \uparrow}(1003) \\
& =\underbrace{f \cdots f}_{183 \uparrow}(1000)=\un... | 997 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,829 |
6・18 Let $N$ be the set of positive integers, and define the function $f$ on $N$ as follows:
(i) $f(1)=1, f(3)=3$;
(ii) For $n \in N$, we have
$$\begin{array}{l}
f(2 n)=f(n) \\
f(4 n+1)=2 f(2 n+1)-f(n) \\
f(4 n+3)=3 f(2 n+1)-2 f(n)
\end{array}$$
Find all $n$ such that $n \leqslant 1988$ and $f(n)=n$. | [Solution] From conditions (i) and (ii), we can calculate the following table:
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$f(n)$ & 1 & 1 & 3 & 1 & 5 & 3 & 7 & 1 & 9 \\
\hline \hline$n$ & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & \\
\hline$f(n)$ & 5 & 13 & 3 & 11 & 7 & 15 &... | 92 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,830 |
6-19 The set of all positive integers can be divided into two disjoint subsets of positive integers $\{f(1), f(2), \cdots, f(n), \cdots\},\{g(1), g(2), \cdots, g(n), \cdots\}$,
where $\quad f(1)<f(2)<\cdots<f(n)<\cdots$
$$g(1)<g(2)<\cdots<g(n)<\cdots$$
and $\quad g(n)=f(f(n))+1 \quad(n \geqslant 1)$
Find: $f(240)$. | [Solution] First, let's prove the following three equations:
$$\begin{array}{l}
g(n)=f(n)+n \\
f(f(n))=f(n)+n-1 \\
f(f(n)+1)=f(n)+n+1
\end{array}$$
In fact, assume $f(n)=k$, then $g(n)=f(k)+1$. Thus, the two disjoint sets
$$\{f(1), f(2), f(3), \cdots, f(k)\} \text { and }\{g(1), g(2), g(3), \cdots, g(n)\}$$
contain a... | 388 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,831 |
$6 \cdot 21 n$ is a natural number not less than 3, $f(n)$ represents the smallest natural number that is not a divisor of $n$ (for example, $f(12)=5$). If $f(n) \geqslant 3$, then $f(f(n))$ can be made. Similarly, if $f(f(n)) \geqslant 3$, then $f(f(f(n)))$ can be made, and so on. If
$$\underbrace{f(f(\cdots f}_{k \up... | [Solution] For natural number $n$, we discuss by cases.
Case $1$: $n$ is odd.
In this case, $f(n)=2$, so $l_{n}=1$.
Case $2$: $n$ is even.
In this case, we set $n=2^{a}(2 m+1)$, where $a \geqslant 1, m \geqslant 0$.
If all odd numbers $t$ satisfying $1<t<2^{a+1}$ are divisors of $n$, then
$$\begin{array}{l}
f(n)=2^{a+1... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,832 |
6. 22 For a given positive integer $k$, define $f_{1}(k)$ as the square of the sum of the digits of $k$, and let
$$f_{n+1}(k)=f_{1}\left(f_{n}(k)\right),$$
Find: $f_{1991}\left(2^{1990}\right)$.
| [Solution] First, note that for large $k$ values, $f_{1}(k)$ is much smaller than $k$. Since $f_{1}$ is not monotonic, we express this fact in the following form:
If $A \leqslant B$, then the number of digits of $A$ $\leqslant$ the number of digits of $B$ $\leqslant 1+\lg B$,
$$f_{1}(A)<(9 \times(1+\lg B))^{2}<\left(4... | 256 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,833 |
$$\begin{array}{l}
6 \cdot 23 \text { Let } f(0)=f(1)=0 \text {, and } \\
f(v+2)=4^{v+2} f(v+1)-16^{v+1} f(v)+v 2^{v^{2}},(v=0,1,2, \cdots) .
\end{array}$$
Prove: $f(1989), f(1990), f(1991)$ are all divisible by 13 | [Proof] Let $f(v)=g(v) 2^{v^{2}}$, then the given recurrence relation becomes
$$\begin{array}{l}
g(v+2)-2 g(v+1)+g(v)=v 16^{-v-1} \\
\quad v=0,1,2, \cdots .
\end{array}$$
Summing the above equation from 0 to $v-1$, we get
$$g(v+1)-g(v)=\frac{1}{15^{2}}\left[1-(15 v+1) \cdot 16^{-v}\right]$$
Summing again (from 0 to $... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,834 |
1. 88 Let $a, b, c, d$ be natural numbers, and $a b=c d$. Try to determine: can $a+b+c+$ $d$ be a prime number? | [Solution] $a+b+c+d$ cannot be a prime number.
In fact, since $ab=cd$, there exist positive integers $c_{1}, c_{2}, d_{1}, d_{2}$, such that
$$c=c_{1} c_{2}, d=d_{1} d_{2}, a=c_{1} d_{1}, b=c_{2} d_{2} .$$
Thus,
$$\begin{aligned}
a+b+c+d & =c_{1} d_{1}+c_{2} d_{2}+c_{1} c_{2}+d_{1} d_{2} \\
& =\left(c_{1}+d_{2}\right)... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,835 |
$6 \cdot 24 x$ is a real number, $0<x<\pi$, prove: for all natural numbers $n$
$$\sin x+\frac{\sin 3 x}{3}+\frac{\sin 5 x}{5}+\cdots+\frac{\sin (2 n-1) x}{2 n-1}$$
is positive. | [Proof] Let $f(x)=\sin x+\frac{\sin 3 x}{3}+\frac{\sin 5 x}{5}+\cdots+\frac{\sin (2 n-1) x}{2 n-1}$. Using $2 \sin x \sin (2 k-1) x=\cos (2 k-2) x-\cos 2 k x$, we get
$$\begin{aligned}
2 f(x) \sin x= & 1-\cos 2 x+\frac{\cos 2 x-\cos 4 x}{3}+\frac{\cos 4 x-\cos 6 x}{5} \\
& +\cdots+\frac{\cos (2 n-2) x-\cos 2 n x}{2 n-1... | proof | Calculus | proof | Yes | Yes | inequalities | false | 735,836 |
$6 \cdot 25$ A uniform circular disk is suspended from a fine thread tied at the center $O$ of the disk and is in a horizontal position. Weights $p_{1}, p_{2}, p_{3}$ are placed at three different points $A, B, C$ on the edge of the disk, and the balance of the disk is not disturbed. Calculate $\angle A O B, \angle B O... | [Solution] Forces $p_{1}$ and $p_{2}$, parallel to the same direction, have their resultant passing through a point $M$ on the line segment $AB$ (as shown in the figure). Since the disk remains balanced after placing the weight, the resultant of the three forces $p_{1}, p_{2}, p_{3}$ passes through point $O$. Therefore... | \left.\begin{array}{l}
\cos \angle BOC=\frac{p_{1}^{2}-p_{2}^{2}-p_{3}^{2}}{2 p_{2} p_{3}} \\
\cos \angle COA=\frac{p_{2}^{2}-p_{3}^{2}-p_{1}^{2}}{2 p_{3} p_{1}} \\
\cos \angle AOB=\ | Geometry | math-word-problem | Yes | Yes | inequalities | false | 735,837 |
6.26 $N_{0}$ is the set of all non-negative integers, $f(n)$ is a function such that $f: N_{0}$
$\rightarrow N_{0}$, and for each $n \in N_{0}, f(f(n))+f(n)=2 n+3$. Find: $f(1993)$. | [ Sol] First, we prove that if $m, n \in N_{0}, m \neq n$, then $f(m) \neq f(n)$, i.e., $f$ is $1-1$.
By contradiction.
If $f(m)=f(n)$, then $f(f(m))=f(f(n))$. Thus, by the given condition, we have
$$2 m+3=f(f(m))+f(m)=f(f(n))+f(n)=2 n+3$$
Hence, $m=n$, a contradiction.
Next, find $f(0)$.
Let $f(0)=x, x \geqslant 0$.... | 1994 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,838 |
6. 27 The function $f(k)$ is defined on $N$, taking values in $N$, and is a strictly increasing function (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$), and satisfies the condition $f(f(k))=3 k$. Try t... | [Solution] For any $k \in \mathbb{N}$, by the given condition,
$$\begin{array}{l}
f(f(k))=3 k \\
f(f(f(k)))=f(3 k)
\end{array}$$
Also, by the given condition $f(f(f(k))=3 f(k)$,
so $f(3 k)=3 f(k)$.
If $f(1)=1$, then substituting into (1) gives $f(1)=3$, which is a contradiction.
Therefore, $f(1)=a>1$.
But from (1), $f... | 197 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,839 |
6. 28 Given the value of $\sin \alpha$. Try to find: (a) $\sin \frac{\alpha}{2}$, (b) $\sin \frac{\alpha}{3}$, respectively, how many different values can they have at most? | [Solution] (a) Let $\sin \alpha = x$, the $\alpha$ satisfying this equation must have the form $k \pi + (-1)^k \arcsin x$,
where $k$ is an integer. Thus, the corresponding values of $\frac{\alpha}{2}$ correspond to 4 points on the unit circle:
$$\begin{array}{l}
\frac{1}{2} \arcsin x, \frac{1}{2} \arcsin x + \pi \\
\f... | 4, 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,840 |
6.29 For any real number $x$, the function $f(x)$ has the property
$$f(x)+f(x-1)=x^{2}$$
If $f(19)=94$, then what is the remainder when $f(94)$ is divided by 1000? | [Solution] From the property of $f(x)$, we get
$$f(x)=x^{2}-f(x-1) .$$
Repeatedly applying the above formula, we obtain
$$\begin{aligned}
f(94) & =94^{2}-f(93) \\
& =94^{2}-93^{2}+f(92) \\
& =94^{2}-93^{2}+92^{2}-f(91) \\
& \cdots \\
& =94^{2}-93^{2}+92^{2}-91^{2}+\cdots+20^{2}-f(19) \\
& =187+183+\cdots+43+400-94 \\
... | 561 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,841 |
6.30 For a given positive integer $n$, let $p(n)$ denote the product of the non-zero digits of $n$ (if $n$ is a single digit, then $p(n)$ is equal to that digit). If
$$S=p(1)+p(2)+p(3)+\cdots+p(999)$$
then what is the largest prime factor of $S$? | ［Solution］Since
$$\begin{aligned}
(a+1+2+\cdots+9)^{3}= & a \cdot a \cdot a+a \cdot a \cdot 1+a \cdot a \cdot 2+\cdots \\
& +9 \cdot 9 \cdot 8+9 \cdot 9 \cdot 9
\end{aligned}$$
on the left side, when $a=1$, except for the first term, the rest is exactly equal to
$$\begin{array}{l}
p(1)+p(2)+p(3)+\cdots+p(999) \text {,... | 103 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,842 |
6.31 Try to find the domain of the function: $y=\lg \left(\sin \frac{x}{2}-\frac{1}{2}\right)+\sqrt{\frac{1}{x-2}}$. | [Solution] From the domain of $\lg \left(\sin \frac{x}{2}-\frac{1}{2}\right)$, we get $\sin \frac{x}{2}-\frac{1}{2}>0$, which means $\sin \frac{x}{2}>\frac{1}{2}$.
Therefore, $\quad \frac{\pi}{6}+2 k \pi < \frac{x}{2} < \frac{5 \pi}{6}+2 k \pi$, i.e., $\frac{\pi}{3}+4 k \pi < x < \frac{5 \pi}{3}+4 k \pi$.
And $x-2>0$, ... | 2<x<\frac{5 \pi}{3}, \frac{\pi}{3}+4 k \pi < x < \frac{5 \pi}{3}+4 k \pi \cdot(k=1,2, \cdots) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,843 |
6.34 $a>0$ is a real number. $f$ is a real function defined on all real numbers, and for every real number $x$, we have
$$f(x+a)=\frac{1}{2}+\sqrt{f(x)-f^{2}(x)} .$$
(1) Prove that $f$ is periodic, i.e., there exists a real number $b>0$ such that for every $x$, $f(x+b)=f(x) ;$
(2) For $a=1$, provide a specific non-cons... | [Proof] (1) Clearly $f(x) \geqslant \frac{1}{2}$,
$$\begin{aligned}
{\left[f(x+a)-\frac{1}{2}\right]^{2} } & =f(x)-f^{2}(x) \\
& =\frac{1}{4}-\left[\frac{1}{2}-f(x)\right]^{2}
\end{aligned}$$
It is evident that $\left[f(x+2 a)-\frac{1}{2}\right]^{2}$
$$\begin{array}{l}
=\frac{1}{4}-\left[\frac{1}{2}-f(x+a)\right]^{2}=... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,846 |
6.38 Given that the function $f(x)$ is defined over the entire real line, and when its graph is rotated by $\frac{\pi}{2}$ about the origin, it remains unchanged.
(1) Prove that the equation $f(x)=x$ has exactly one solution.
(2) Provide an example of such a function. | [Proof] (1) Let $f(0)=a$, then by the given condition, we have $f(-a)=0, f(0)=-a$. Therefore, $a=-a$, which implies $a=0, f(0)=0$.
If for $x_{0} \neq 0$, $f\left(x_{0}\right)=x_{0}$, then by the given condition, $f\left(-x_{0}\right)=x_{0}$, $f\left(-x_{0}\right)=-x_{0}$, which contradicts the definition of the functi... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,848 |
6.39 Given for some natural number $n \geqslant 2$, the polynomials $u_{i}(x)=a_{i} x+b_{i}\left(a_{i}\right.$, $b_{i}$ are real numbers, $\left.i=1,2,3\right)$ satisfy the relation
$$u_{1}^{n}(x)+u_{2}^{n}(x)=u_{3}^{n}(x)$$
Prove that these polynomials can be expressed as
$$u_{i}(x)=c_{i}(A \cdot x+B)$$
where $i=1,2... | [Proof] If $a_{1}=a_{2}=0$, then the polynomials $u_{1}$ and $u_{2}$ degenerate into constants. From relation (1), it follows that the polynomial $u_{3}(x)$ also degenerates into a constant, i.e., $a_{3}=0$. In this case, we only need to set $c_{i}=b_{i}(i=1,2,3), A=0, B=1$.
If at least one of the numbers $a_{1}, a_{2... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,849 |
6.40 Given a quadratic function $y=f(x)$ whose graph has a vertex at $(-1,1)$ and intersects the $y$-axis at $(0,2)$.
(1) Find the expression for this quadratic function;
(2) When $x=8, y=$ ?
(3) For any given value of $y$, can we always find a value of $x$? Why? Explain using a graph. | [Solution] (1) Since $(-1,1)$ is the vertex of $y=f(x)$, we can assume $y=a(x+1)^{2}+1$. Given that the intersection with the $y$-axis is $(0,2)$, we have
$$2=a(0+1)^{2}+1 \text {, solving for } a=1 \text {. }$$
Therefore, the quadratic function is
$$y=(x+1)^{2}+1$$
which is $y=x^{2}+2 x+2$.
(2) When $x=8$, $y=9^{2}+... | y=x^{2}+2 x+2, 82, \text{not always possible} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,850 |
6. 41 A train departs from station A, accelerating at 0.5 kilometers/min$^2$ forward, after 2 minutes, it continues to move at a constant speed for another 7 minutes, then decelerates uniformly at 0.5 kilometers/min$^2$ to enter station B, and stops at station B for 2 minutes. Try to draw the speed-time relationship gr... | When $0 \leqslant t \leqslant 2$, $S=\frac{1}{2} a t^{2}=\frac{t^{2}}{4}$; when $2<t \leqslant 9$, $S=\frac{2^{2}}{4}+(t-2)=1+t-2=t-1$; when $9<t \leqslant 11$, $S=(9-1)+(t-9)-\frac{1}{2} a(t-9)^{2}$ $=t-1-\frac{1}{4}(t-9)^{2}$ $=\frac{1}{4}\left(-t^{2}+22 t-85\right) ;$ when $11<t \leqslant 13$, $S=\frac{1}{4}\left(-1... | S=\frac{1}{4}\left(-11^{2}+22 \times 11-85\right)=9 | Calculus | math-word-problem | Yes | Yes | inequalities | false | 735,851 |
6.42 Given the parabola $y=a x^{2}+b x+c$ has a line of symmetry at $x=-2$, it is tangent to a certain line at one point, this line has a slope of 2, and a y-intercept of 1, and the parabola intersects the $y=0$ at two points, the distance between which is $2 \sqrt{2}$. Try to find the equation of this parabola. | [Solution] (1) $y=a x^{2}+b x+c=a\left(x+\frac{b}{2 a}\right)^{2}+c-\frac{b^{2}}{4 a}$, given that $x=-2$ is the axis of symmetry, we get $\frac{b}{2 a}=2$, hence $b=4 a$.
(2) $y=a x^{2}+b x+c$ is tangent to the line $l$,
and $l: y=2 x+1$, so the parabola intersects $l$ at one point,
thus we have $a x^{2}+4 a x+c=2 x+... | y=x^{2}+4 x+2 \text{ or } y=\frac{1}{2} x^{2}+2 x+1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,852 |
1. 90 Let $a$ and $b$ be two given natural numbers such that $\frac{a+1}{b}+\frac{b+1}{a}$ is an integer. Prove: the greatest common divisor of $a$ and $b$ does not exceed $\sqrt{a+b}$. | [Solution] Since
$$\begin{array}{l}
\frac{a+1}{b}+\frac{b+1}{a}=\frac{a^{2}+b^{2}+a+b}{a b} \\
(a, b) \mid a, (a, b) \mid b
\end{array}$$
Therefore, if we denote $(a, b)=d$, we have
$$\begin{array}{l}
d^{2} \mid a b \\
d^{2} \mid a^{2}+b^{2} \\
a b \mid\left(a^{2}+b^{2}+a+b\right)
\end{array}$$
Thus, we deduce that
$... | (a, b) \leqslant \sqrt{a+b} | Number Theory | proof | Yes | Yes | inequalities | false | 735,853 |
6. 44 Let $f(x)$ be a continuous function defined on $0 \leqslant x \leqslant 1$, with $f(0)=0, f(1)=1$. Also, for any $x, 0<x<1$, there exists $h$ such that when $0 \leqslant x-h<x+h \leqslant 1$,
$$f(x)=\frac{f(x-h)+f(x+h)}{2} .$$
Prove: For $0 \leqslant x \leqslant 1, f(x)=x$. | [Solution] Let $F(x)=f(x)-x$, then $F(0)=F(1)=0$, and $F$ satisfies the same conditions as $f$. If the maximum value of $F(x)$ is $M>0$, then we take
$$x_{0}=\inf \{x: F(x)=M\}$$
Since $F$ is continuous, $F\left(x_{0}\right)=M>0$, so $x_{0}>0$. And there exists $h, 0 \leqslant x_{0}$ $-h<x_{0}+h \leqslant 1$, such tha... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,854 |
6.46 Let \( M = \{1, 2, 3, \cdots, 20\} \), for any 9-element subset \( S \) of \( M \), the function \( f(S) \) takes an integer value between 1 and 20. Prove that regardless of what \( f \) is, there always exists a 10-element subset \( T \) of \( M \) such that for all \( k \in T \),
\[ f(T - \{k\}) \neq k . \] | [Proof] If a 10-element subset $T$ has the property: for any $k \in T$, there is $f(T-\{k\}) \neq k$, then we call $T$ a "good set". A 10-element subset that is not a "good set" is called a "bad set". That is, if $T$ is a "bad set", it means there is a $k_{0}$ in $T$ such that
$$f\left(T-\left\{k_{0}\right\}\right)=k_{... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,855 |
6.47 Let $X$ be a finite set, and let the rule $f$ assign to each even subset $E$ (a subset consisting of an even number of elements) of $X$ a real number $f(E)$, satisfying the conditions:
(1) There exists an even subset $D$ such that $f(D)>1990$,
(2) For any two disjoint even subsets $A, B$ of $X$, we have
$$f(A \cup... | [Proof] Since $X$ is a finite set, the number of all even subsets of $X$ is finite. Let $P$ be the even subset of $X$ with the least number of elements among those with the maximum value of $f$, and let $Q$ be the complement of $P$ relative to $X$.
It can be proven that such chosen $P, Q$ satisfy the requirements of t... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,856 |
$$6 \cdot 48 \text { Let } N=\{1,2,3, \cdots\} \text {. }$$
Prove whether there exists a function $f: N \rightarrow N$ such that
$$\begin{array}{l}
f(1)=2 \\
f(f(n))=f(n)+{ }_{n}^{n}
\end{array}$$
for all $n \in N$,
$$f(n)<f(n+1)$$
for all $n \in N$. | [Solution] Such a function exists.
We take $\alpha=\frac{\sqrt{5}+1}{2}, \beta=\frac{\sqrt{5}-1}{2}$, clearly $\alpha=1+\beta, \alpha \beta=1$, let $f(n)=[\alpha n+\beta], n \in N$. We prove that $f(n)$ satisfies the conditions $f: N \rightarrow N$, and
$$\begin{array}{l}
f(1)=[\alpha+\beta]=[\sqrt{5}]=2, \\
\begin{ali... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,857 |
6-49 Given the set $G$ of non-constant functions $f$ of the form $f(x)=a x+b$, where $a, b$ are real numbers and $a \neq 0$, and $x$ is a real variable. If $G$ has the following properties:
(1) If $f, g \in G$, then $g_{a} f \in G$, where
$$\left(g_{a} f\right)(x)=g(f(x))$$
(2) If $f \in G$ and $f(x)=a x+b$, then the i... | [Proof] We prove by contradiction that for all $f(x), x_{f}$ in $G$ except $x$ are equal.
Suppose there exist $f, g \in G-\{x\}$ such that $x_{f} \neq x_{g}$.
Let $f(x)=a x+b, g(x)=a^{\prime} x+b^{\prime}$, then from
$$a x_{f}+b=x_{f}, a^{\prime} x_{g}+b^{\prime}=x_{g}$$
we get $a \neq 1, a^{\prime} \neq 1$,
$$x_{f}=\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,858 |
$6 \cdot 50$ Let $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ be real constants, and $x$ be a real variable, such that
$$\begin{aligned}
f(x)= & \cos \left(\alpha_{1}+x\right)+\frac{1}{2} \cos \left(\alpha_{2}+x\right)+\frac{1}{2^{2}} \cos \left(\alpha_{3}+x\right)+\cdots+ \\
& \frac{1}{2^{n-1}} \cos \left(\alpha_{n}+x... | [Proof] First, note that
$$\begin{aligned}
f\left(-\alpha_{1}\right) & =1+\frac{1}{2} \cos \left(\alpha_{2}-\alpha_{1}\right)+\cdots+\frac{1}{2^{n-1}} \cos \left(\alpha_{n}-\alpha_{1}\right) \\
& \geqslant 1-\frac{1}{2}-\cdots-\frac{1}{2^{n-1}}=\frac{1}{2^{n-1}}>0
\end{aligned}$$
Therefore, $f(x)$ is not identically z... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,859 |
6 - 51 Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$ and $g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+c_{0}$ be non-zero polynomials with real coefficients, and suppose there is a real number $\gamma$ such that $g(x)=(x+\gamma) f(x)$; and let $a$ be the largest element of $\left\{|a n|,\left|a_{n-1}\right|, \cdots,\l... | [Proof] Since $g(x)=(x+\gamma) f(x)$, comparing the coefficients on both sides yields
$$\left\{\begin{array}{l}
c_{0}=a_{0} \gamma \\
c_{1}=a_{0}+a_{1} \gamma \\
\cdots \cdots \\
c_{i}=a_{i-1}+a_{i} \gamma \\
\cdots \cdots \\
c_{n}=a_{n-1}+a_{n} \gamma \\
c_{n+1}=a_{n}
\end{array}\right.$$
From (1), we get
$$\left\{\b... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,860 |
$6 \cdot 52$ Let $y_{1}, y_{2}, y_{3}, \cdots \cdots$ be a sequence, where $y_{1}=1$, and for integers $k>0$, we have
$$\begin{array}{l}
y_{2 k}=\left\{\begin{array}{ll}
2 y_{k} & k \text { is even } \\
2 y_{k}+1 & k \text { is odd, }
\end{array}\right. \\
y_{2 k+1}=\left\{\begin{array}{ll}
2 y_{k} & k \text { is odd }... | [Proof] Calculation yields: $y_{1}=1, y_{2}=2 y_{1}+1=3, y_{3}=2 y_{1}=2, y_{4}$ $=2 y_{2}=6, y_{5}=2 y_{2}+1=7, y_{6}=2 y_{3}+1=5, y_{7}=2 y_{3}=4$. That is, $\left\{y_{1}\right\}=\{1\},\left\{y_{2}, y_{3}\right\}=\{2,3\},\left\{y_{4}, y_{5}, y_{6}, y_{7}\right\}=\{4,5,6,7\}$.
Obviously, we only need to prove that th... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,861 |
$$\begin{array}{l}
6 \cdot 53 f, g, h \text { are defined as follows: } \\
f(n)=10 n, n \text { is a positive integer; } \\
g(n)=10 n+4, n \text { is a positive integer; } \\
h(n)=\frac{n}{2}, n \text { is a positive even number. }
\end{array}$$
Prove: Every natural number can be expressed from 4 by a finite number o... | [Proof] By induction. Clearly, $2=h(4), 1=h(2), 5=h(f(1)), 7=$ $h(g(1)), 6=h(h(g(2))), \quad 3=h(6), \quad 10=f(1), 9=$ $h(h(h(h(g(g(1)))))), 8=h(h(h(g(6)))), 4=h(8)$. Therefore, every natural number not greater than 10 can be obtained from 4 through a finite number of $f, g, h$ operations.
Assume the proposition hold... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,862 |
6. 54 Find the value of the constant $c$ such that the function
$$f(x)=\operatorname{arctg} \frac{2-2 x}{1+4 x}+c$$
is an odd function on the interval $\left(-\frac{1}{4}, \frac{1}{4}\right)$. | [Solution] Suppose the constant $c$ exists, and the function $f(x)$ is an odd function, so $f(0) = \operatorname{arctg} 2 + c = 0$. From this, the only possible value for $c$ is $-\operatorname{arctg} 2$.
We now prove that in the interval $\left(-\frac{1}{4}, \frac{1}{4}\right)$, the function
$$f(x) = \operatorname{ar... | -\operatorname{arctg} 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,863 |
6. 56 Given a function $f$ defined on the set of positive integers satisfies: $f(1)=1$, $f(2)=2, \quad f(n+2)=f(n+2-f(n+1))+f(n+1-f(n)) \quad(n \geqslant 1)$
(1) Prove: $0 \leqslant f(n+1)-f(n) \leqslant 1$, and when $f(n)$ is odd, $f(n+1)=f(n)+1$;
(2) Try to find: all values of $n$ that satisfy $f(n)=2^{10}+1$, and pr... | [Solution] (1) We first prove a lemma: For any natural number $n, f(n+1) - f(n) \in \{0,1\}$
We use mathematical induction.
When $n=1$, $f(2)-f(1)=1$, the lemma holds.
Assume when $n=1$, the stronger conclusion to be proved holds.
In fact, when $m=2$, the equation becomes $f(n)=3$, which has a unique solution $n=2^2$... | 2^{11} | Number Theory | proof | Yes | Yes | inequalities | false | 735,864 |
6. 57 Prove: There does not exist a function $f$ from the set of non-negative integers to the set of non-negative integers such that for every $n$, $f(f(n))=n+1987$.
保留了源文本的换行和格式。 | [Proof] Suppose such a function exists, then for every $n \in N$ (the set of non-negative integers), we have
$$f(n+1987)=f(f(f(n)))=f(n)+1987$$
By applying induction, for each $n, t \in N$, we have
$$f(n+1987 t)=f(n)+1987 t .$$
On the other hand, consider $\gamma \in N, \gamma \leqslant 1986$, then
$$f(\gamma)=1987 k... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,865 |
6. 59 Let $S_{r}=x^{r}+y^{r}+z^{r}$, where $x, y, z$ are real numbers. It is known that when $S_{1}=0$, for
$$(m, n)=(2,3),(3,2),(2,5) \text { or }(5,2),$$
we have
$$\frac{S_{m+n}}{m+n}=\frac{S_{m}}{m} \cdot \frac{S_{n}}{n}$$
Determine all other positive integer pairs $(m, n)$ that satisfy (1). | [Solution] Clearly, when $x=k+1, y=-k, z=-1$, it can make $S_{1}=x+y+z=0$.
$$\begin{array}{l}
\text { If } \quad \frac{S_{m+n}}{m+n}=\frac{S_{m}}{m} \cdot \frac{S_{n}}{n} \\
\text { then } \frac{x^{m+n}+y^{m+n}+z^{m+n}}{m+n}=\frac{x^{m}+y^{m}+z^{m}}{m} \cdot \frac{x^{n}+y^{n}+z^{n}}{n} \text {. } \\
\text { Substitutin... | (m, n)=(2,3),(3,2),(2,5),(5,2) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,867 |
$6 \cdot 60$. Let $N$ be the set of natural numbers, $k \in N$, and suppose there is a function $f: N \rightarrow N$ that is strictly increasing, and for each $n \in N$, $f(f(n))=k n$.
Prove that for every $n \in N$,
$$\frac{2 k}{k+1} n \leqslant f(n) \leqslant \frac{k+1}{2} n$$ | [Proof] Since $f: N \rightarrow N$ is strictly increasing, we have
$$\begin{array}{l}
f(n) \geq n \\
f(n+m) \geq f(n)+m
\end{array}$$
From (1), we can assume $f(n)=n+m$ (where $m$ is a non-negative integer), thus by the given we have
$$\begin{aligned}
k n & =f(f(n))=f(n+m) \geqslant f(n)+m \\
& =f(n)+f(n)-n
\end{align... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,868 |
6. 61 Let $f$ be a function mapping the set of real numbers $\mathbb{R}$ to itself, and for any $x \in \mathbb{R}$, we have $|f(x)| \leqslant 1$, and
$$f\left(x+\frac{13}{42}\right)+f(x)=f\left(x+\frac{1}{6}\right)+f\left(x+\frac{1}{7}\right).$$
Prove: $f$ is a periodic function, i.e., there exists a non-zero real num... | [Proof] From the given,
$$\begin{aligned}
f\left(x+\frac{13}{42}\right)+f(x) & =f\left(x+\frac{7}{42}\right)+f\left(x+\frac{6}{42}\right) \\
f\left(x+\frac{7}{42}\right)-f(x) & =f\left(x+\frac{13}{42}\right)-f\left(x+\frac{6}{42}\right) \\
& =f\left(x+\frac{19}{42}\right)-f\left(x+\frac{12}{42}\right) \\
& =\cdots \cdo... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,869 |
6.63. $N$ is the set of all positive integers, $f: N \rightarrow N$ is a function that satisfies the inequality, for any $x \in N$,
$$f(x)+f(x+2) \leqslant 2 f(x+1)$$
Prove: There exists a line in the plane that contains infinitely many points $(n, f(n))$. | [Proof] Let
$$d(x)=f(x+1)-f(x), x \in N$$
By the given condition, $\square$
$$d(x+1) \leqslant d(x)$$
Therefore, the sequence $\{d(x) \mid x \in N\}$ is monotonically decreasing.
We will now prove: $d(x) \geqslant 0, x \in N$.
By contradiction.
If there exists $k$ such that $d(k) < 0$, then for any $n > k$, by the mo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,870 |
1・92 $x, y, p, n, k$ are all natural numbers, and satisfy
$$x^{n}+y^{n}=p^{k}$$
Prove: If $n$ is an odd number greater than 1, and $p$ is an odd prime, then $n$ can be expressed as a power of $p$ with a natural number as the exponent. | ［Proof］Let $m$ be the greatest common divisor of $x, y$, then
$$x=m x_{1}, y=m y_{1}, x_{1}, y_{1} \in \mathbb{N}, (x_{1}, y_{1})=1$$
From the given, we have
$$m^{n}\left(x_{1}^{n}+y_{1}^{n}\right)=p^{k}$$
Thus, there exists a non-negative integer $\alpha$, such that
$$\begin{array}{l}
m=p^{\alpha} \\
x_{1}^{n}+y_{1}... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,871 |
6・64 Given integers $n \geqslant k \geqslant 0$, define the number $c(n, k)$:
$c(n, 0)=c(n, n)=1$ when $n \geqslant 0$;
$c(n+1, k)=2^{k} c(n, k)+c(n, k-1)$ when $n \geqslant k \geqslant 1$.
Prove: $c(n, k)=c(n, n-k)$ for all integers $n, k$ satisfying $n \geqslant k \geqslant 0$. | [Proof] For $m \geqslant 1$ let
$$\begin{array}{l}
f(m)=\left(2^{1}-1\right)\left(2^{2}-1\right) \cdots\left(2^{m}-1\right) \\
f(0)=1
\end{array}$$
Let
$$a(n, k)=\frac{f(n)}{f(k) f(n-k)}$$
Then for $n \geqslant 0$, we have
$$a(n, 0)=a(n, n)=1$$
Also, because
$$f(m) \cdot\left(2^{m+1}-1\right)=f(m+1), \text{ for } m ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,872 |
6. $65 X=\{0, a, b, c\}$ and $M(X)=\{f \mid f: X \rightarrow X\}$ is the set of all functions from $X$ to itself. Define the addition operation $\oplus$ on $X$ as follows:
\begin{tabular}{|c|c|c|c|c|}
\hline$\oplus$ & 0 & $a$ & $b$ & $c$ \\
\hline 0 & 0 & $a$ & $b$ & $c$ \\
\hline$a$ & $a$ & 0 & $c$ & $b$ \\
\hline$b$ ... | [Solution] (1) From the addition operation table, for $x \in \{a, b, c\}$, we have
$$\begin{array}{l}
0 \oplus x = x \oplus 0 = x \\
x \oplus x = 0 \\
b \oplus a = a \oplus b = c, \quad c \oplus a = a \oplus c = b \\
c \oplus b = b \oplus c = a .
\end{array}$$
Therefore, we have
$$x \oplus y = y \oplus x, \forall x, y... | 64 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,873 |
6・66 $a, b, c$ are positive real numbers, $\alpha$ is a real number, and suppose
$$\begin{array}{l}
f(\alpha)=a b c\left(a^{\alpha}+b^{\alpha}+c^{\alpha}\right) \\
g(\alpha)=a^{\alpha+2}(b+c-a)+b^{\alpha+2}(a-b+c)+c^{\alpha+2}(a+b-c)
\end{array}$$
Determine the relationship in size between $f(\alpha)$ and $g(\alpha)$. | 【Solution】 From the given,
$$\begin{aligned}
f(\alpha)-g(\alpha)= & a b c\left(a^{\alpha}+b^{\alpha}+c^{\alpha}\right)+\left(a^{\alpha+3}+b^{\alpha+3}+c^{\alpha+3}\right) \\
& -a^{\alpha+2}(b+c)-b^{\alpha+2}(c+a)-c^{\alpha+2}(a+b) \\
= & {\left[a^{\alpha+1} b c+a^{\alpha+3}-a^{\alpha+2}(b+c)\right] } \\
& +\left[b^{\al... | f(\alpha) \geqslant g(\alpha), \alpha \in R | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,874 |
6.67 Let $a_{1}, a_{2}, \cdots, a_{n}$ denote any permutation of the integers $1,2, \cdots, n$. Let $f(n)$ be the number of permutations that satisfy the conditions:
(1) $a_{1}=1$,
(2) $\left|a_{i}-a_{i+1}\right| \leqslant 2, i=1,2, \cdots, n-1$
Determine whether $f(1996)$ is divisible by 3. | 【Solution】Obviously, $f(1)=f(2)=1, f(3)=2$.
Let $n \geqslant 4$. Since $a_{1}=1$, then $a_{2}=2$ or 3.
If $a_{2}=2$, the number of permutations of $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ is equal to the number of permutations of $a_{2}-1, a_{3}-1, \cdots, a_{n}-1$. The latter is the permutation of $1,2, \cdots, n-1$ and s... | 1 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,875 |
6・68 $f(n)$ is defined on the set of positive integers, and satisfies
$$f(1)=2, f(n+1)=(f(n))^{2}-f(n)+1, n=1,2,3, \cdots$$
Prove that for all integers $n>1$,
$$1-\frac{1}{2^{2^{n-1}}}<\frac{1}{f(1)}+\frac{1}{f(2)}+\cdots+\frac{1}{f(n)}<1-\frac{1}{2^{2^{n}}}$$ | [Proof] If $f(n) \geqslant 2$, then from
$$f(n+1)=f(n)(f(n)-1)+1$$
we can get
$$f(n+1) \geqslant 2$$
Noting that $f(1)=2$, therefore, for all integers $n \geqslant 1$, we have
$$f(n) \geqslant 2 \text {. }$$
Transform the given equation to
$$f(n+1)-1=f(n)(f(n)-1),$$
it is clear that both sides of the equation are p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,876 |
6・70 Proof: Any function defined on the entire real line can be expressed as the sum of two functions, each of whose graphs has an axis of symmetry. | [Proof] Let $f(x)$ be any given function. We prove that $f(x)$ can be expressed as $f_{1}(x)+f_{2}(x)$, where $f_{1}(x)$ is symmetric about $x=0$, and $f_{2}(x)$ is symmetric about $x=a, a>0$.
In the interval $[-a, a]$, let $f_{1}(x) \equiv 0, f_{2}(x)=f(x)$.
In the interval $[a, 3a]$, let
$$f_{2}(x)=f_{2}(2a-x), f_{1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,877 |
6・72 Let $U$ be a finite set, and $f, g$ be bijections from $U$ to itself. Let
$$\begin{array}{l}
S=\{w \in U: f(f(w))=g(g(w))\} \\
T=\{w \in U: f(g(w))=g(f(w))\}
\end{array}$$
and assume $U=S \cup T$. Prove that for $w \in U, f(w) \in S$ if and only if $g(w) \in S$. | [Proof] Since the status of $f$ and $g$ is the same, it is sufficient to prove that $f(w) \in S \Rightarrow g(w) \in S$.
By contradiction.
If there exists $w \in U, f(w) \in S$ and $g(w) \bar{\in} S$.
Then, by $g(w) \in U$ and $S \cup T=U$, we know $g(w) \in T$.
Therefore, by the definitions of $S$ and $T$, we have
$$\... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,879 |
$6 \cdot 73$ For $x \neq 0, f(x)=\frac{x^{2}+1}{2 x}$, define $\left\{\begin{array}{l}f^{(0)}(x)=x, \\ f^{(n)}(x)=f\left(f^{(n-1)}(x)\right),\end{array}\right.$ for all positive integers $n$ and $x \neq 0$. Prove that for all non-negative integers $n$ and $x \neq-1,0,1$, we have
$$\frac{f^{(n)}(x)}{f^{(n+1)}(x)}=1+\fra... | [Proof] Use mathematical induction.
When $n=0$, for $x \neq-1,0,1$, we have
$$\frac{f^{(0)}(x)}{f^{(1)}(x)}=\frac{x}{f(x)}=\frac{2 x^{2}}{x^{2}+1},$$
and
$$\begin{aligned}
1+\frac{1}{f\left(\frac{x+1}{x-1}\right)} & =1+\frac{2\left(\frac{x+1}{x-1}\right)}{\left(\frac{x+1}{x-1}\right)^{2}+1} \\
& =1+\frac{2(x+1)(x-1)}{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,880 |
1.93 Let $p$ be the product of $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$. If $p - x_{k}$ is an odd number, $k=1,2, \cdots, n$. Prove: $x_{1}, x_{2}, \cdots, x_{n}$ are all irrational numbers. | [Proof] Suppose $\frac{a}{b}-\frac{c}{d}$ is an integer, and $a, b, c, d$ are all integers, $(a, b)=1$, $(c, d)=1, b>0, d>0$, then from
$$\frac{a}{b}-\frac{c}{d}=\frac{a d-b c}{b d}$$
we know
$$b:(a d-b c)$$
thus
$$b|a d, b| d$$
Similarly, we get $d \mid b$. Therefore, $b=d$.
Assume $x_{1}, x_{2}, \cdots, x_{n}$ are... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,881 |
6・74 Let $R$ be the triangular region in the plane with vertices $A(4,1), B(-1,-6), C(-3,2)$ (including the interior and boundary of the triangle, as shown in the figure). Try to find the maximum and minimum values of the function $4 x-3 y$ as $(x, y)$ varies over $R$ (you need to prove your conclusion). | [Solution] Let $\lambda=4 x-3 y$. It is evident that when $\lambda$ is fixed and $(x, y)$ varies, we obtain a straight line on the plane. By letting $\lambda$ vary, we get a series of parallel lines on the plane. On each of these lines, the value of $4 x-3 y$ is the same. When the line passes through point $A$, $\lambd... | 14 \text{ and } -18 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,882 |
6.75 Given that $x, y, z$ are positive numbers, and satisfy the equation
$$x y z(x+y+z)=1$$
Find the minimum value of the expression $(x+y)(y+z)$. | [Solution] From the given conditions and the arithmetic-geometric mean inequality, we have
$$\begin{aligned}
(x+y)(y+z) & =(x+y+z) y+x z \\
& =\frac{1}{x y z} \cdot y+x z \\
& =\frac{1}{x z}+x z \\
& \geqslant 2 .
\end{aligned}$$
On the other hand, when $x=1, y=\sqrt{2}-1, z=1$, the given equation is satisfied, and
$$... | 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,883 |
6・76 Given $\left|x_{i}\right|<1, i=1,2, \cdots, n$. Also,
$$\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|=19+\left|x_{1}+x_{2}+\cdots+x_{n}\right|$$
What is the minimum value of the integer $n$? | ［Solution］From the given, we have
$$\sum_{i=1}^{n}\left|x_{i}\right|-\left|\sum_{i=1}^{n} x_{i}\right| \leqslant \sum_{i=1}^{n}\left|x_{i}\right|19$$,
thus we have $\quad n \geqslant 20$.
On the other hand, when $n=20$, we take
$$x_{i}=\left\{\begin{array}{l}
\frac{19}{20}, 1 \leqslant i \leqslant 10 \\
-\frac{19}{20},... | 20 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,884 |
6.77 Find the maximum value of $k$ such that $3^{11}$ can be expressed as the sum of $k$ consecutive positive integers. | [Solution] Let
$$3^{11}=(n+1)+(n+2)+\cdots+(n+k),$$
where $n$ is a non-negative integer, and $k$ is a natural number, i.e.,
$$\begin{array}{l}
3^{11}=n k+\frac{k(k+1)}{2} \\
2 \cdot 3^{11}=k(2 n+k+1)
\end{array}$$
Clearly, $k<2 n+k+1$. To make $k$ as large as possible, while $n$ is a non-negative integer, the best po... | 486 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,885 |
$\begin{array}{c}6 \cdot 78 \text { Let } f(x)=|x-p|+|x-15|+|x-p-15| \text {, where } \\ 0<p<15 \text {. Find the minimum value of } f(x) \text { for } x \text { in the interval } p \leqslant x \leqslant 15. \end{array}$ | [Solution] From the given, we know $0 < p < 15$.
Therefore, $f(x) = (x - p) + (15 - x) + (p + 15 - x)$, which simplifies to $f(x) = 30 - x$.
The minimum value of this function on the interval $[p, 15]$ is $f(15) = 15$. | 15 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,886 |
6. 79 Find the minimum value of $f(x)=\frac{9 x^{2} \sin ^{2} x+4}{x \sin x}(0<x<\pi)$. | [Solution] Let $t=x \sin x$. By $00$. Then
$$f(x)=\frac{9 t^{2}+4}{t} \geqslant \frac{2 \times 3 t \times 2}{t}=12 .$$
Therefore, the minimum value of $f(x)$ is 12. | 12 | Calculus | math-word-problem | Yes | Yes | inequalities | false | 735,887 |
$6 \cdot 80$ Let two complex numbers $x, y$, the sum of their squares is 7, and the sum of their cubes is 10, then the largest real value that $x+y$ can take is what? | [ Sol] From the given conditions
$$\left\{\begin{array}{l}
x^{2}+y^{2}=7 \\
x^{3}+y^{3}=10
\end{array}\right.$$
Let $x+y=t$. Then we have
$$\begin{array}{l}
(x+y)^{3}=\left(x^{3}+y^{3}\right)+3 x y(x+y) \\
(x+y)^{3}=\left(x^{3}+y^{3}\right)+3(x+y) \cdot \frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2}
\end{array}$$
Thus,... | 4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,888 |
6. 81 Function $f$ is defined on the real number domain, and satisfies the following conditions: for any real number $x$,
$$f(2+x)=f(2-x), f(7+x)=f(7-x) .$$
If $x=0$ is a root of $f(x)=0$, then $f(x)=0$ should have at least how many roots in the interval $-1000 \leqslant x \leqslant 1000$? | ［Solution］From the given conditions, we have
$$\begin{array}{l}
f(x)=f(2+x-2)=f[2-(x-2)]=f(4-x), \\
f(4-x)=f[7-(x+3)]=f[7+(x+3)]=f(10+x)
\end{array}$$
From (1) and (2), we get $f(x)=f(10+x)$.
Thus, from $f(0)=0$, we have
$$f(0)=f( \pm 10)=f( \pm 20)=\cdots=f( \pm 1000)=0$$
This gives us 201 roots of $f(x)=0$ in the i... | 401 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,889 |
6. 82 Find the smallest real number $A$, such that for every quadratic polynomial $f(x)$ satisfying the condition
$$|f(x)| \leqslant 1 \quad(0 \leqslant x \leqslant 1)$$
the inequality $f^{\prime}(0) \leqslant A$ holds. | [Solution] Let the quadratic trinomial be $f(x)=a x^{2}+b x+c$. When $0 \leqslant x \leqslant 1$, it satisfies the inequality
$$|f(x)| \leqslant 1.$$
In particular, there should be the inequalities
$$|f(0)| \leqslant 1,\left|f\left(\frac{1}{2}\right)\right| \leqslant 1,|f(1)| \leqslant 1.$$
Since $f(0)=c, f\left(\fra... | 8 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,890 |
6・83 Given distinct real numbers $a_{1}, a_{2}, \cdots, a_{n}$, find the minimum value of the function defined by
$$y=\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right|$$
where $x$ is a real number. | [Solution] First, note that when $a<b$,
$$|x-a|+|x-b|=\left\{\begin{array}{l}
a+b-2 x, \text { when } x \leqslant a, \\
-a+b, \text { when } a \leqslant x \leqslant b, \\
2 x-a-b, \text { when } x \geqslant b.
\end{array}\right.$$
Therefore, on each point of the interval $a \leqslant x \leqslant b$, the sum $|x-a|+|x-... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,891 |
1-10 Starting from 1, write natural numbers in sequence, ask what is the number at the one millionth position?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | [Solution] Since there are 9 one-digit numbers, 90 two-digit numbers, 900 three-digit numbers, $\cdots$, and $900000 \times 6>1000000$. Therefore, the required digit is not in any seven-digit number. Also, because $9 + 90 \times 2 + 900 \times 3 + 9000 \times 4 + 90000 \times 5 = 488889$, the required digit must be in ... | null | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,892 |
1.94 $n$ is a positive odd number, prove: $n(n-1)^{(n-1)^{n}+1}+n$ is a multiple of $((n-1)^{n} +1)^{2}$. Does this property hold for even $n$? Prove your conclusion. | [Proof] When $n$ is a positive odd number,
$$\begin{aligned}
(n-1)^{n}+1 & =n^{n}-c_{n}^{1} \cdot n^{n-1}+\cdots-c_{n}^{n-2} n^{2}+c_{n}^{n-1} \cdot n \\
& =n^{2} q,
\end{aligned}$$
where $q$ is a positive odd number and is coprime with $n$. Thus,
$$(n-1)^{n^{2} q}=n^{n^{2} \varphi}-c_{n^{1} q}^{12} \cdot n^{n^{2} q-1... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,893 |
6. 84 For a finite set $A$, there exists a function $f: N \rightarrow A$ with the following property: if $|i-j|$ is a prime number, then $f(i) \neq f(j), N=\{1,2, \cdots\}$, find the minimum number of elements in the finite set $A$.
| [Solution] Starting from 1, the smallest natural number whose difference with it is a prime number is 3; the smallest natural number whose difference with the previous two is a prime number is 6; the smallest natural number whose difference with the first three is a prime number is 8. This means that for the set $M=\{1... | 4 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,894 |
6. 85 Find the maximum value of the following expression:
$$|\cdots|\left|x_{1}-x_{2}\right|-x_{3}\left|-\cdots-x_{1990}\right|,$$
where $x_{1}, x_{2}, \cdots, x_{1990}$ are distinct natural numbers from 1 to 1990. | [Solution] When $x \geqslant 0, y \geqslant 0$, it is obvious that $|x-y| \leqslant \max \{x, y\}$ and $\max \{\max \{x, y\}\}, z=\max \{x, y, z\}$. Therefore, when $n=1,2, \cdots, 1990$, we have
$$|\cdots|\left|x_{1}-x_{2}\right|-x_{3}\left|-\cdots-x_{n}\right| \leqslant \max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}... | 1989 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,895 |
6.88 The function $F(x)=\left|\cos ^{2} x+2 \sin x \cos x-\sin ^{2} x+A x+B\right|$ has a maximum value $M$ on $0 \leqslant x \leqslant \frac{3}{2} \pi$ that depends on the parameters $A$ and $B$. For what values of $A$ and $B$ is $M$ minimized? Prove your conclusion. | [Solution] (1) $F(x)=\left|\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)+A x+B\right|$ When $A=B=0$, $F(x)$ becomes $f(x)=\sqrt{2}\left|\sin \left(2 x+\frac{\pi}{4}\right)\right|$. In the interval $\left[0, \frac{3}{2} \pi\right]$, there are three points $x_{1}=\frac{\pi}{8}, x_{2}=\frac{5 \pi}{8}, x_{3}=\frac{9 \pi}{8}... | A=B=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,897 |
6. 89 Determine the maximum value of $m^{2}+n^{2}$. Where $m, n$ are integers, and $m, n \in\{1$, $2, \cdots, 1981\},\left(n^{2}-m n-m^{2}\right)^{2}=1$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6. 89 Determine the max... | ［Solution］First, consider the case where $m=n$.
Since $\left(n^{2}-m n-m^{2}\right)^{2}=1$, we have $(m n)^{2}=1$, and since $m, n \in\{1,2,3, \cdots, 1981\}$, it must be that $m=n=1$. Next, consider the case where $m \neq n$.
Clearly, $n>m$. Let $n=m+u_{k}$.
Then $\left[\left(m+u_{k}\right)^{2}-m\left(m+u_{k}\right)-... | 3524578 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,898 |
$6 \cdot 90$ Let the numbers $x_{1}, \cdots, x_{1991}$ satisfy the condition
$$\left|x_{1}-x_{2}\right|+\cdots+\left|x_{1990}-x_{1991}\right|=1991,$$
and $y_{k}=\frac{1}{k}\left(x_{1}+\cdots+x_{k}\right), k=1, \cdots, 1991$. Try to find the maximum value that the following expression can achieve.
$$\left|y_{1}-y_{2}\r... | [Solution] For $k=1, \cdots, 1990$, we have the relation
$$\begin{aligned}
& \left|y_{k}-y_{k+1}\right| \\
= & \left|\frac{1}{k}\left(x_{1}+\cdots+x_{k}\right)-\frac{1}{k+1}\left(x_{1}+\cdots+x_{k+1}\right)\right| \\
= & \left|\frac{1}{k(k+1)}\left(x_{1}+\cdots+x_{k}-k x_{k+1}\right)\right| \\
\leqslant & \frac{1}{k(k+... | 1990 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,899 |
$6 \cdot 91$ Consider the function $f(x)$ on $[0,1]$ satisfying the conditions:
(1) $f(x) \geqslant 0$, for any $x \in[0,1]$;
(2) $f(1)=1$;
(3) $f(x)+f(y) \leqslant f(x+y), x, y, x+y \in[0,1]$
Try to find the smallest constant $c$, such that for any function $f(x)$ satisfying (1) (3), we have $f(x) \leqslant c x, x \i... | [Solution] The minimum value of $c$ is 2.
In fact, using the function
$$f(x)=\left\{\begin{array}{l}
0, \quad 0 \leqslant x \leqslant \frac{1}{2} \\
1, \quad \frac{1}{2} < x \leqslant 1
\end{array}\right.$$
If $x > \frac{1}{2}$, then $f(x) \leqslant 1 < 2x$.
If $0 < x \leqslant \frac{1}{2}$, then we can choose $n$ such... | 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,900 |
6.92 Let $n$ be a given positive integer, and the sum
$$\begin{aligned}
\sum_{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|= & \left|x_{1}-x_{2}\right|+\left|x_{1}-x_{3}\right|+\cdots+\left|x_{1}-x_{n}\right|+ \\
& \left|x_{2}-x_{3}\right|+\left|x_{2}-x_{4}\right|+\cdots+\left|x_{2}-x_{n}\right|+ \\
& \cdots+\le... | [Solution] Without loss of generality, let $0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n} \leqslant 1$. Let
$$S=\sum_{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|=\sum_{1 \leqslant i<j \leqslant n}\left(x_{j}-x_{i}\right)$$
This sum has $C_{n}^{2}$ terms. Each $x_{k}$ appears in $n-1$ of t... | \left[\frac{n^{2}}{4}\right] | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,901 |
$6 \cdot 93$ The bus route consists of a circular loop with a circumference of 10 miles and a straight line from the main station to point $Q$ on the loop, which is 1 mile long (as shown in the figure). Two buses serve on this route. Each loop takes 20 minutes. Bus 1 leaves the main station, first travels along the str... | [Solution] From the given information, the car's speed is $\frac{3}{5}$ miles per minute, and $w(x) = w(12-x)$, so it is sufficient to consider the case where $0 < x \leq 6$.
If $0 < x \leq 1$, the person will obviously take the car heading to the main station, with a maximum waiting time of 10 minutes and a travel ti... | 25 | Other | math-word-problem | Yes | Yes | inequalities | false | 735,902 |
$1 \cdot 95$ Let $m$ and $n$ be known positive integers, $m$ has $d$ digits when represented in decimal, where $d \leqslant n$. Find the sum of all digits in the decimal representation of $\left(10^{n}-1\right) m$. | [Solution] Let $f(x)$ denote the sum of the digits of the positive integer $x$, and let $m=\overline{a_{1} a_{2} \cdots a_{d} 00 \cdots 0}, a_{d} \neq 0$.
Write $\left(10^{n}-1\right) m=m \cdot 10^{n}-m$ as a subtraction in vertical form, and note that $d \leqslant n$, we have $a_{1} a_{2} \cdots a_{d} 00 \cdots 000 \... | 9n | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,903 |
6・94 Determine the largest real number $z$, such that
$$x+y+z=5, \quad xy+yz+zx=3,$$
and $x, y$ are also real numbers. | [Solution] From $x+y+z=5$, we get
$$(x+y)^{2}=(5-z)^{2},$$
From $x y+y z+z x=3$, we get
$$x y=3-z(5-z) .$$
Thus, we have
$$\begin{aligned}
(x-y)^{2} & =(x+y)^{2}-4 x y=(5-z)^{2}-4[3-z(5-z)] \\
& =-3 z^{2}+10 z+13=(13-3 z)(1+z),
\end{aligned}$$
Therefore,
$$\begin{array}{l}
(13-3 z)(1+z) \geqslant 0 \\
-1 \leqslant z... | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,904 |
6.96 Let $a$ and $d$ be non-negative numbers, $b$ and $c$ be positive numbers, and $b+c \geqslant a+d$. Find the minimum value of the following expression:
$$\frac{b}{c+d}+\frac{c}{a+b} .$$ | [Solution] If we simultaneously swap $a$ with $d$ and $b$ with $c$, the conditions and conclusions of the problem remain unchanged. Therefore, we can assume without loss of generality that $a+b \geqslant c+d$. Thus, we have
$$\begin{aligned}
& \frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+... | \sqrt{2}-\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,905 |
6・97 Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, try to find the minimum value of the following expression
$$S=\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} .$$ | ［Solution］ Squaring the expression, we get
$$S^{2}=\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}}+2\left(x^{2}+y^{2}+z^{2}\right)$$
By the arithmetic mean and geometric mean inequality, we have
$$\begin{array}{l}
\frac{1}{2}\left(\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}\right)... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,906 |
6.98 I. Sketch the graphs of the following functions:
(1) $y=\sqrt{4 x^{2}+12 x+9}$;
(2) $y=\sqrt{(2 x+3)^{2}}$;
(3) $y=\frac{\left|4 x^{2}+12 x+9\right|}{2 x+3}$.
II. Consider the function $f(x)=\frac{x}{1+x^{2}}$.
(1) If $-1<x_{1}<x_{2}<1$, compare $f\left(x_{1}\right)$ and $f\left(x_{2}\right)$;
(2) Prove that $f(1)... | [Solution] I. From the given, we have (1) $y=|2 x+3|$,
$$\text { (2) } y=|2 x+3|$$
(3) Since $4 x^{2}+12 x+9=(2 x+3)^{2} \geqslant 0$, we have $\left|4 x^{2}+12 x+9\right|=4 x^{2}+12 x+9=(2 x+3)^{2}$.
Therefore, the original function is $y=2 x+3$, but $x \neq -\frac{3}{2}$.
The graph is as follows:
II. From the given, ... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,907 |
6.99 $k$, $l$ are natural numbers, find the number with the smallest absolute value in the form of $36^{k}-5^{l}$, and prove that the number found is indeed the smallest.
In the form of $36^{k}-5^{l}$, find the number with the smallest absolute value, and prove that the number found is indeed the smallest. | [Solution] The last digit of $36^{k}$ is $6$, and the last digit of $5^{l}$ is $5$. Therefore, when $36^{k}>5^{l}$, the last digit of $\left|36^{k}-5^{l}\right|$ is 1; when $36^{k}<5^{l}$, the last digit of $\left|36^{k}-5^{l}\right|$ is 9.
If $\left|36^{k}-5^{l}\right|=1$, i.e., $\square$
$$36^{k}-5^{l}=1$$
Then we ... | 11 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,908 |
6. $100 n$ positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ sum up to 1. Let $S$ be the largest of the following numbers:
$$\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \cdots, \frac{x_{n}}{1+x_{1}+x_{2}+\cdots+x_{n}} .$$
Find the smallest possible value of $S$. For what values of $x_{1}, x_{2}, \cdots, x_{n}$ is... | [Solution] Let $y_{0}=1, y_{k}=1+x_{1}+x_{2}+\cdots+x_{k}(1 \leqslant k \leqslant n)$.
Then $\square$
$$x_{k}=y_{k}-y_{k-1}$$
From the given information,
$$\frac{x_{k}}{1+x_{1}+x_{2}+\cdots+x_{k}} \leqslant S$$
which is $\frac{y_{k}-y_{k-1}}{y_{k}} \leqslant S$,
thus we have $1-S \leqslant \frac{y_{k-1}}{y_{k}} \quad... | 1-2^{-\frac{1}{n}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,909 |
6-101. Let $x, y$ be positive numbers, and $S$ be the smallest number among $x, y+\frac{1}{x},-\frac{1}{y}$. Find the possible maximum value of $S$, and the values of $x, y$ that achieve this maximum value. | [Solution] We consider, when $S$ takes the maximum value, it can simultaneously satisfy the three inequalities
$$\begin{array}{l}
x \geqslant S \\
y+\frac{1}{x} \geqslant S \\
\frac{1}{y} \geqslant S
\end{array}$$
and at least one of them is an equality.
Since when $x=y=1$, $S=1$, we can assume $S>0$.
From (3) we get ... | \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,910 |
$6 \cdot 103$ Find the smallest positive integer $n$ (where $n>1$) such that the quadratic mean of the first $n$ natural numbers is an integer. Here, the quadratic mean of $n$ numbers $a_{1}, a_{2}, \cdots, a_{n}$ is given by
$$\left(\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}\right)^{\frac{1}{2}} .$$ | ［Solution］Let $\left(\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}\right)^{\frac{1}{2}}=m, m \in N$.
Then $\square$
$$\begin{array}{l}
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=m^{2} \\
\frac{(n+1)(2 n+1)}{6}=m^{2} \\
(n+1)(2 n+1)=6 m^{2}
\end{array}$$
Since $6 m^{2}$ is even and $2 n+1$ is odd, $n+1$ must be even, hence $n$ is odd. Th... | 337 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,912 |
1.96 Does there exist natural numbers $x, y$ such that $x^{2}+y$ and $y^{2}+x$ are both squares of integers. | [Solution] Without loss of generality, let $x \geqslant y$, then
$$x^{2}<x^{2}+y \leqslant x^{2}+x<(x+1)^{2}$$
That is, $x^{2}+y$ is not a perfect square of an integer, hence such $x, y$ do not exist. | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,913 |
6. 104 Let $a, b$ be natural numbers, $1 \leqslant a \leqslant b, M=\left[\frac{a+b}{2}\right]$. The function $f: Z \rightarrow Z$ is defined as
$$\begin{array}{l}
f(n)=\left\{\begin{array}{l}
n+a, \text { if } n<M \\
n-b, \text { if } n \geqslant M
\end{array}\right. \\
f^{1}(n)=f(n), f^{+1}(n)=f\left(f^{i}(n)\right),... | [Proof] It is clear that we only need to consider the case where $(a, b)=1$.
Let the set
$$S=\{n \mid M-b \leqslant n \leqslant M+a-1, n \in \mathbb{Z}\}.$$
Then $f(s) \subseteq S$, and $0 \in S$.
Suppose $k \geqslant 1$ and $f^{k}(0)=0$. Since $f(m)=m+a$ or $m-b$, $k$ can be written as $k=r+s$ and $r a-s b=0$. Since ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,914 |
6. 105 Let $S$ be the unit circle in the complex plane (i.e., the set of complex numbers with modulus 1), and $f$ be a mapping from $S$ to $S$. For any $z \in S$, define
$$\begin{array}{l}
f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \cdots, \\
f^{(k)}(z)=\underbrace{f(f(\cdots(f}_{k \uparrow f}(z) \cdots))),
\end{array}$$
If... | [Solution] We first prove:
If $z_{0} \in S$ is a fixed point of $f^{(n)}$, i.e., $f^{(n)}\left(z_{0}\right)=z_{0}$, and $z_{0}$ is an $m$-periodic point of $f$, then it must be that $m \mid n$.
In fact, let $n=p m+q, 0 \leqslant q<m$, then we have
$$\begin{aligned}
z_{0} & =f^{(n)}\left(z_{0}\right) \\
& =f^{(p m+q)}\... | m^{1989} - m^{117} - m^{153} - m^{663} + m^{9} + m^{39} + m^{51} - m^{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,915 |
6-106 For the function $y=\lg (2 \sin x)$, answer the following questions, and then draw its graph.
(1) What is its domain?
(2) For what values of $x$ is $y=0$?
(3) For what values of $x$ does the function have a maximum value? What is the maximum value?
(4) Is this function periodic? If so, find its smallest positive ... | [Solution] (1) Since negative numbers and zero do not have logarithms, we have $2 \sin x>0$, i.e., $\sin x>0$,
the domain is $2 k \pi<x<(2 k+1) \pi$ (where $k$ is an integer).
(2) From $\lg (2 \sin x)=0$,
we have $2 \sin x=1, \sin x=\frac{1}{2}$,
thus $x=2 k \pi+\frac{\pi}{6}$ or $x=2 k \pi+\frac{5 \pi}{6}$ (where $k$... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,916 |
6. 107 Given vibrations $y_{1}=\sin x, y_{2}=\sin ^{2} x$; study the properties of their combined motion $y=y_{1}+y_{2}=\sin x+\sin ^{2} x$ (the range of $y$, maximum value, minimum value, zeros, variation, symmetry, periodicity, and plot the graph). | [Solution] (1) Since $y=\sin x+\sin ^{2} x \leqslant 2$, and when $x=\left(2 n+\frac{1}{2}\right) \pi$ (where $n$ is an integer), $y=2$.
Also, since $y=\left(\sin x+\frac{1}{2}\right)^{2}-\frac{1}{4} \geqslant-\frac{1}{4}$,
and when $x=\left(2 n-\frac{1}{6}\right) \pi$ or $\left(2 n+\frac{7}{6}\right) \pi$ (where $n$ ... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,917 |
$6 \cdot 109$ Given the family of curves
$$2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0$$
where $\theta$ is a parameter. Find the maximum value of the length of the chord intercepted by the line $y=2 x$ on this family of curves. | [Solution] Obviously, the family of curves always passes through the origin, and the line $y=2 x$ also passes through the origin. Therefore, the length of the chord intercepted by the family of curves on $y=2 x$ depends on the coordinates of the other intersection point of the family of curves with $y=2 x$.
Substituti... | 8 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,919 |
6・110 For positive integers $a, n$, define
$$F_{n}(a)=q+r$$
where $q, r$ are non-negative integers, $a=q^{n+r}$, and $0 \leqslant r<n$.
Find the largest positive integer $A$ such that there exist positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$, for any positive integer $a \leqslant A$, we have
$$F_{n_{6}}... | [Solution] Let the maximum positive integer $B$ that satisfies the condition "there exist positive integers $n_{1}, n_{2}, \cdots, n_{k}$, such that for any positive integer $a \leqslant B$,
$$F_{n_{k}}\left(F_{n_{k-1}}\left(\cdots\left(F_{n_{1}}(a)\right) \cdots\right)\right)=1 "$$
be denoted as $x_{k}$,
Clearly, the... | 53590 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,920 |
6. The axial cross-section of a large round wine glass is the graph of the function $y=x^{4}$. A cherry, a small sphere with radius $r$, is placed into the wine glass. What is the maximum radius $r$ for the cherry to touch the lowest point of the bottom of the glass (in other words, what is the maximum $r$ such that a ... | [Solution] The original problem is equivalent to: the circle passing through the origin
$$(y-r)^{2}+x^{2}=r^{2}$$
is completely within the region $y \geqslant x^{4}$, find the maximum value of $r$.
Therefore, when $0 \leqslant x \leqslant r$, we have
$$r-\sqrt{r^{2}-x^{2}} \geqslant x^{4}$$
Let $x=r \sin \theta$. (1)... | \frac{3}{4} \sqrt[3]{2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 735,922 |
6・113 Let the sequence $a(n), n=1,2,3, \cdots$, be defined as follows:
$a(1)=0$, and for $n>1$,
$$a(n)=a\left(\left[\frac{n}{2}\right]\right)+(-1)^{\frac{n(n+1)}{2}} .$$
(a) Find the maximum and minimum values of $a(n)$ for $n<1996$, and provide all values of $n$ at which these extrema are attained.
(b) For $n<1996$, h... | [Solution] For $n>1$,
$$a(n)=a\left(\left[\frac{n}{2}\right]\right)+(-1)^{\frac{n(n+1)}{2}},$$
we can derive that for $n \geqslant 1$,
$$\begin{aligned}
a(2 n) & =a(n)+(-1)^{n(2 n+1)} \\
& =a(n)+(-1)^{n} \\
a(2 n+1) & =a(n)+(-1)^{(2 n+1)(n+1)} \\
& =a(n)+(-1)^{n+1} .
\end{aligned}$$
Assume that for $n \geqslant 2$, t... | 346 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,923 |
6-114 Let $P(x, y)$ be a point on $|5 x+y|+|5 x-y|=20$, find the maximum and minimum values of $x^{2}-x y+y^{2}$. | [Solution] Squaring both sides of $|5 x+y|+|5 x-y|=20$, and simplifying, we get $25 x^{2}+y^{2}+\left|25 x^{2}-y^{2}\right|=200$.
If $|5 x| \geqslant|y|$, then we have
$$50 x^{2}=200, x= \pm 2,|y| \leqslant 10$$
If $|5 x| \leqslant|y|$, then we have
$$2 y^{2}=200, y= \pm 10,|5 x| \leqslant 10,|x| \leqslant 2$$
Theref... | Q_{\text {maximum }}=124, Q_{\text {minimum }}=3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,924 |
6. 115 Let $r_{1}, r_{2}, \cdots, r_{m}$ be $m$ given positive rational numbers, such that $\sum_{k=1}^{m} r_{k}=1$. For every positive integer $n$, define the function $f$ as
$$f(n)=n-\sum_{k=1}^{m}\left[r_{k} n\right]$$
Find the minimum and maximum values of $f(n)$. | ［Solution］From the given information,
$$f(n)=\sum_{k=1}^{m}\left(r_{k} n-\left[r_{k} n\right]\right)$$
Since $0 \leqslant r_{k} n-\left[r_{k} n\right]<1$, for all $1 \leqslant k \leqslant m$,
we have $\quad 0 \leqslant f(n) \leqslant m-1$.
Next, we prove that the bounds on both sides of the above inequality can be ach... | 0 \text{ and } m-1 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,925 |
$6 \cdot 116$ Let $A=\{1,2,3, \cdots, 17\}$, for the mapping $f: A \rightarrow A$, denote
$$f^{[1]}(x)=f(x), f^{[k+1]}(x)=f\left(f^{[k]}(x)\right), k \in N .$$
Suppose the one-to-one mapping $f$ from $A$ to $A$ satisfies the condition: there exists a natural number $M$ such that:
(1) When $m<M, 1 \leqslant i \leqslant... | [Solution] The maximum possible value of $M$ is $M_{0}=8$.
First, we prove that $M_{0} \geqslant 8$.
In fact, we can define the mapping $f(i) \equiv 3 i-2(\bmod 17)$, where $i \in A, f(i) \in A$.
If $f(i) \equiv f(j)(\bmod 17)$, then
$$\begin{aligned}
3 i-2 & \equiv 3 j-2(\bmod 17) \\
i & \equiv j(\bmod 17)
\end{align... | 8 | Combinatorics | proof | Yes | Yes | inequalities | false | 735,926 |
6. $118 \quad x_{1}, x_{2}, \cdots, x_{1993}$ satisfy
$$\begin{array}{l}
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+\left|x_{1992}-x_{1993}\right|=1993, \\
y_{k}=\frac{x_{1}+x_{2}+\cdots+x_{k}}{k},(k=1,2, \cdots, 1993)
\end{array}$$
Then, what is the maximum possible value of $\left|y_{1}-y_{2}\right|+\l... | [Solution]
$$\begin{array}{l}
\left|y_{k}-y_{k+1}\right|=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\frac{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}{k+1}\right| \\
=\frac{1}{k(k+1)} \left| (x_{1}-x_{k+1})+(x_{2}-x_{k+1})+\cdots+(x_{k}-x_{k+1}) \right| \\
\leqslant \frac{1}{k(k+1)}\left(\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\... | 1992 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,927 |
6・119 Every natural number $n$ greater than 2 can be expressed as the sum of several pairwise distinct natural numbers. Let the maximum number of terms be $A(n)$. Find $A(n)$ (expressed in terms of $n$). | [Solution] Let $n=a_{1}+a_{2}+\cdots+a_{A(n)}, a_{1}<a_{2}<\cdots<a_{A(n)}$ be natural numbers.
Obviously, $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{A(n)} \geqslant A(n)$, therefore, $n \geqslant 1+2+\cdots+$ $A(n)=\frac{1}{2} A(n)(A(n)+1)$,
thus, the natural number $A(n) \leqslant\left[\frac{\sqrt{8 n+1}-1}{2... | A(n)=\left[\frac{\sqrt{8 n+1}-1}{2}\right] | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,928 |
6. 121 Let $a_{1}, a_{2}, \cdots, a_{10}$ be any 10 distinct natural numbers, and their sum is 1995. Try to find the minimum value of
$$a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{9} a_{10}+a_{10} a_{1}$$ | [Solution] Write 10 natural numbers in a clockwise direction on a circle. Thus, the expression in the problem is the sum of the products of every two adjacent numbers, which we will simply call the "sum value".
Let the sum of 10 distinct natural numbers be $N$, then $N \geqslant 55$. Denote the minimum "sum value" of ... | 6064 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,930 |
6-123 Let the complex number $z=x+i y(x, y \in R)$ satisfy $|z-i| \leqslant 1$. Find the maximum and minimum values of $A$ $=x\left(|z-i|^{2}-1\right)$ and the corresponding $z$ values. | [Solution] Let $z-i=x+(y-1)i=r(\cos \theta+i \sin \theta)$.
Then $\left\{\begin{array}{l}|z-i|=r \leqslant 1,0 \leqslant r \leqslant 1, \\ x=r \cos \theta\end{array}\right.$
Thus,
$$\begin{array}{l}
A=r \cos \theta\left(r^{2}-1\right)=r\left(r^{2}-1\right) \cos \theta \\
A^{2}=r^{2}\left(r^{2}-1\right)^{2} \cdot \cos ... | A_{\text{max}} = \frac{2 \sqrt{3}}{9}, \, A_{\text{min}} = -\frac{2 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,932 |
1.98 Given two positive prime numbers $p, q$. An integer $n$ is called a "good number" if it can be expressed in the form $n = px + qy$, where $x, y$ are non-negative integers. Conversely, it is called a "bad number".
(1) Prove: There exists an integer $c$ such that among the integers $n$ and $c-n$, one is always a goo... | [Solution] (1) Since $p$ and $q$ are coprime natural numbers, every integer $z$ can be expressed as $z = px + qy$.
Any such representation can be obtained from a fixed $z = pa + qb$ using the general formula $z = p(a - qt) + q(b + pt)$, where $t$ is an integer, and there exists a unique representation such that $0 \le... | \frac{(p-1)(q-1)}{2} | Number Theory | proof | Yes | Yes | inequalities | false | 735,933 |
6-124 Let $f(n)$ be a function defined on all positive integers and taking positive integer values. For all positive integers $m, n$, we have
$$f(f(m)+f(n))=m+n .$$
Find all possible values of $f(1988)$. | [Solution] Let $f(1)=t$. If $f(m)=n$, then
$$f(2 n)=f(f(m)+f(m))=2 m$$
In particular, when $m=1, n=t$,
$$f(2 t)=2$$
We prove that $t=1$. Otherwise, let
$$t=b+1, b \in \mathbb{N}$$
Suppose $f(b)=c, c \in \mathbb{N}$, then $f(2 c)=2 b$, and
$$2 c+2 t=f(f(2 c)+f(2 t))=f(2 b+2)=f(2 t)=2$$
Thus, $t+c=1$, which is a cont... | 1988 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,934 |
6. 126 Find all functions $f: Z^{*} \rightarrow R$ that satisfy
$f(n+m)+f(n-m) \equiv f(3 n), n, m \in Z^{*}\left(Z^{*}\right.$ is the set of non-negative integers), $n \geqslant m$. | [Solution] In the identity concerning $f(x)$, let $m=0$, we get $2 f(n) \equiv f(3 n)$, $n \in Z^{*}$. When $n=m=0$, we have $f(0)=0$. Next, in the identity, let $n=m$, we get $f(2 n)+f(0) \equiv f(3 n)$, that is, $f(2 n) \equiv f(3 n)$. Therefore, on one hand, for any $m \in Z^{*}$, we have
$$f(4 m)=f(6 m)=f(9 m)$$
O... | f(n) \equiv 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,935 |
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