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1. 求以下无限简单连分数的值:
(i) $\langle 2,3,1,1,1, \cdots\rangle$;
(ii) $\langle 1,2,3,1,2,3,1,2,3, \cdots\rangle$;
(iii) $\langle 0,2,1,3,1,3,1,3, \cdots\rangle$;
(iv) $\langle-2,2,1,2,1,2,1, \cdots\rangle$. | 1. (i) $(25-\sqrt{5}) / 10$;
(ii) $(4+\sqrt{37}) / 7$;
(iii) $(\sqrt{21}-1) / 10$;
(iv) $-3+\sqrt{2}$. | (i) \frac{25-\sqrt{5}}{10}; (ii) \frac{4+\sqrt{37}}{7}; (iii) \frac{\sqrt{21}-1}{10}; (iv) -3+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,789 |
3. 求以下无理数的无限简单连分数, 前六个渐近分数, 前七个完全商, 以及该无理数和它的前六个渐近分数的差:
(i) $\sqrt{7}$;
(ii) $\sqrt{13}$;
(iii) $\sqrt{29}$;
(iv) $(\sqrt{10}+1) / 3$;
(v) $(5-\sqrt{37}) / 3$. | 3. (i) $\sqrt{7}=\langle 2, \overline{1,1,1,4}\rangle$;
(ii) $\sqrt{13}=\langle 3, \overline{1,1,1,6}\rangle$;
(iii) $\sqrt{29}=\langle 5, \overline{2,1,1,2,10}\rangle$;
(iv) $(\sqrt{10}+1) / 3=\langle\overline{1,2,1}\rangle$;
(v) $(5-\sqrt{37}) / 3=\langle-1,1,1, \overline{1,3,2}\rangle$. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,791 |
6. Prove: If the "smallest non-negative remainder" in the previous problem is changed to "absolute smallest remainder", "smallest positive remainder", or the general remainder in Theorem 2 of § 3, the conclusion still holds. | 6. As above.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,792 |
5. Let $\xi_{0}$ be an irrational number, and its infinite simple continued fraction is $\left\langle a_{0}, a_{1}, a_{2}, \cdots\right\rangle$. Prove:
(i) When $\left.a_{1}\right\rangle 1$, $-\xi_{0}=\left\langle-a_{0}-1,1, a_{1}-1, a_{2}, a_{3}, \cdots\right\rangle$;
(ii) When $a_{1}=1$, $-\xi_{0}=\left\langle-a_{0}-... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,794 |
11. Let $\xi_{0}$ be a real number, $a, b$ be integers, $b \geqslant 1$. Prove: If
$$\left|\xi_{0}-a / b\right|=\min _{0<y<b}\left|\xi_{0}-x / y\right|$$
then $a / b$ must be a convergent or a second convergent of $\xi_{0}$, where $x$ and $y$ take values as in problem 9. Provide an example to show that for a second co... | 11. There must be a unique $n \geqslant 0$ such that $k_{n} \leqslant b < k_{n+1}$. If $b = k_{n}$, then use Theorem 6 (ii); if $k_{n} < b < k_{n+1}$, then use the method of proving Theorem 6 to discuss, considering $\zeta_{0} = \sqrt{2}$, to show that for its second convergent fraction, $(**)$ does not necessarily hol... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,800 |
Theorem 1 Let $n \geqslant 0, h_{n} / k_{n}, h_{n+1} / k_{n+1}$ be two consecutive convergents of an irrational number $\xi_{0}$, then
$$\left|\xi_{0}-h_{n} / k_{n}\right|<1 /\left(2 k_{n}^{2}\right)$$
or
$$\left|\xi_{0}-h_{n+1} / k_{n+1}\right|<1 /\left(2 k_{n+1}^{2}\right)$$
at least one of them holds. | Prove that if neither of the two inequalities holds, then using formula (14) of § 3, we get
$$\begin{aligned}
1 /\left(2 k_{n}^{2}\right)+1 /\left(2 k_{n+1}^{2}\right) & \leqslant\left|\xi_{0}-h_{n} / k_{n}\right|+\left|\xi_{0}-h_{n+1} / k_{n+1}\right| \\
& \leqslant\left|h_{n} / k_{n}-h_{n+1} / k_{n+1}\right|=1 /\left... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,804 |
Lemma 4 Let $x \geqslant 1$ be a real number and $x+x^{-1}<\sqrt{5}$, then it must be true that
$$1 \leqslant x<(\sqrt{5}+1) / 2$$ | Prove that when $x \geqslant 1$, the function $x + x^{-1}$ is an increasing function of $x$. Because, when
$$x_{1} \geqslant 1, x_{2} \geqslant 1, x_{1} \neq x_{2}$$
then,
$$x_{1} + x_{1}^{-1} > x_{2} + x_{2}^{-1}$$
is equivalent to
$$\left(x_{1} x_{2} - 1\right)\left(x_{1} - x_{2}\right) > 0$$
Since $x_{1} x_{2} > ... | 1 \leqslant x < (\sqrt{5} + 1) / 2 | Inequalities | proof | Yes | Yes | number_theory | false | 740,807 |
Theorem 7 Let $\alpha$ be a real number, then for any positive number $x \geqslant 1$, there exist integers $a, b$, satisfying
$$1 \leqslant b \leqslant x, \quad(a, b)=1,$$
such that
$$|\alpha-a / b|<1 /(b x) \text {. }$$ | Prove: The following $[x]+2$ numbers
$$1, \quad j \alpha-[j \alpha], \quad j=0,1, \cdots,[x]$$
are all in the interval $[0,1]$, and thus by the pigeonhole principle, there must be two numbers whose difference does not exceed $([x]+1)^{-1}$. If these two numbers are
$$j_{1} \alpha-\left[j_{1} \alpha\right], \quad j_{2}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,810 |
Theorem 8 Let $\xi_{0}$ be an irrational number, then there exist infinitely many rational fractions $a / b$ such that
$$\left|\xi_{0}-a / b\right|<1 / b^{2}$$
holds. | Proof: In Theorem 7, take $\alpha=\xi_{0}$. For $x_{0}=1$, there must exist an integer $a_{0}$ such that
$$0 < y_{0} = \left| \xi_{0} - a_{0} \right| < 1 / x_{0} = 1.$$
For $y_{0} > 0$, by Theorem 7 again, for $x_{1} = y_{0}^{-1}$, there must be
$$1 \leqslant b_{1} < x_{1}, \quad (a_{1}, b_{1}) = 1$$
such that (since $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,811 |
8. Find the possible smallest non-negative remainders, smallest positive remainders, and absolute smallest remainders when $n^{2}, n^{3}, n^{4}, n^{5}$ are divided by $3, 4, 8, 10$. | 8. The table below lists the absolute minimal remainders
\begin{tabular}{|c|c|c|c|c|}
\hline & 3 & 4 & 8 & 10 \\
\hline$n^{2}$ & $\{0,1\}$ & $\{0,1\}$ & $\{0,1,4\}$ & $\{-4,-1,0,1,4,5\}$ \\
$n^{3}$ & $\{-1,0,1\}$ & $\{-1,0,1\}$ & $\{-3,-1,0,1,3\}$ & $\{-4,-3,-2,-1,0,1,2,3,4,5\}$ \\
$n^{4}$ & $\{0,1\}$ & $\{0,1\}$ & $\{... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,814 |
3. Let $\xi_{0}$ be an irrational number. For a given $x \geqslant 1$, how can we use convergents to find $a / b$ such that Theorem 7 (taking $\alpha=\xi_{0}$) holds? For example, with $\xi_{0}=\sqrt{7}, \sqrt{13}, \sqrt{23}, x=10^{2}, 10^{3}, 10^{4}$, find specific $a / b$. | 3. For $\alpha=\xi_{0}$, there must exist a unique $n$ such that $k_{n}<x \leqslant k_{n+1}$. Taking $a / b$ as $h_{n} / k_{n}$ satisfies the requirement. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,815 |
4. Prove: (i) For any given real number $c>2$, there must exist an irrational number $\xi$ such that there are only finitely many rational numbers $h / k (k \geqslant 1)$ satisfying $|\xi-h / k|<1 / k^{c}$;
(ii) For any real number $c$, there must exist an irrational number $\xi$ such that there are infinitely many rat... | 4. (i) First, fix $n_{0} \geqslant 1$, and appropriately choose $a_{0}, \cdots, a_{n_{0}}$ such that $k_{n_{0}-2}^{c-2} \geqslant 3$. Then sequentially select positive integers $a_{n+1}\left(n \geqslant n_{0}\right)$ satisfying $a_{n+1}+2 \leqslant k_{n}^{c-2}$. $\xi_{0}=\left\langle a_{0}, a_{1}, \cdots, a_{n}, \cdots... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,816 |
5. Construct a table of rational fractions as follows: First, write down the fractions $0 / 1,1 / 1$ in the first row (from left to right). After the fractions in the $(n-1)$-th row have been written, write the fractions in the $n$-th row as follows: First, rewrite all the fractions from the $(n-1)$-th row in the $n$-t... | 5. (i) Prove by induction that any two adjacent fractions (from left to right) in the $(n+1)$-th row must be one of the following three cases: (a) $a / b, a^{\prime} / b^{\prime}$; (b) $a / b, (a + a^{\prime}) / (b + b^{\prime})$; (c) $(a + a^{\prime}) / (b + b^{\prime}), a^{\prime} / b^{\prime}$, where $a / b, a^{\pri... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,817 |
11. Let $a, b, c, d$ be integers, $b>0, d>0, a d-b c=1$; and let $n=\max (b, d)$, $a / b, c / d$ both belong to the $n$-th Farey sequence. Prove:
(i) $a / b, c / d$ must be consecutive fractions in the $n$-th Farey sequence;
(ii) they are not necessarily consecutive in the $(n+1)$-th Farey sequence. | 11. (i) Without loss of generality, let $b=n$. If they are not adjacent, then there must be $n$-th order Farey fractions $x_{1} / y_{1}, x_{2} / y_{2}$, such that $a / b \geqslant x_{1} / y_{1}>x_{2} / y_{2}>c / d$, and $x_{1} / y_{1}, x_{2} / y_{2}, c / d$ are three adjacent fractions in the $n$-th order Farey sequenc... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,823 |
12. If $a / b, a^{\prime} / b^{\prime}, a^{\prime \prime} / b^{\prime \prime}$ are three consecutive fractions in the $n$-th order Farey sequence, prove: $a^{\prime} / b^{\prime}=\left(a+a^{\prime \prime}\right) /\left(b+b^{\prime \prime}\right)$. | 12. Use question 5 (iv).
... | null | Number Theory | proof | Yes | Yes | number_theory | false | 740,824 |
Theorem $1 \alpha$ is a quadratic irrational if and only if there exists a non-square integer $d$ and rational numbers $r, s, s \neq 0$, such that
$$\alpha=r+s \sqrt{d}$$
Furthermore, $\alpha$ is a real quadratic irrational if and only if $d>0$. | Necessity: Suppose $\alpha$ satisfies the quadratic equation (1). $\alpha$ must be one of the two numbers given by equation (2), so we can take $d=b^{2}-4 a c, r=-b /(2 a), s=1 /(2 a)$ or $-1 /(2 a)$, which gives equation (4). If $\alpha$ is a real number, it must have $d>0$.
Sufficiency: Suppose $\alpha$ is given by ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,826 |
Theorem 3 Let the integer $d$ not be a square number, then the sum, difference, product, and quotient of numbers of the form $r+s \sqrt{d}(r, s$ being rational numbers) are still numbers of this form. | Let $\alpha_{1}=r_{1}+s_{1} \sqrt{d}, \alpha_{2}=r_{2}+s_{2} \sqrt{d}, r_{1}, r_{2}, s_{1}, s_{2}$ be rational numbers. We have
$$\begin{array}{l}
\alpha_{1} \pm \alpha_{2}=\left(r_{1} \pm r_{2}\right)+\left(s_{1} \pm s_{2}\right) \sqrt{d} \\
\alpha_{1} \alpha_{2}=\left(r_{1} r_{2}+d s_{1} s_{2}\right)+\left(r_{1} s_{2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,828 |
Theorem 4 (i) $\xi_{0}$ is a purely periodic continued fraction if and only if there exists $k \geqslant 1$, such that
$$\xi_{0}=\xi_{k};$$
(ii) If $\xi_{0}$ is a purely periodic continued fraction with period $l$, then the necessary and sufficient condition for equation (15) to hold is $l \mid k$;
(iii) If $\xi_{0}$ i... | Proof: By definition, $\xi_{0}$ is a purely periodic continued fraction if and only if equation (13) holds. By Corollary 3 of §3, equation (13) is equivalent to equation (15). This proves (i). When the period of the purely periodic continued fraction $\xi_{0}$ is $l$, if $k \geqslant 1$ makes equation (13) hold, then $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,829 |
Example 1 Find the value of $\xi_{0}=\langle-1,1,4, \overline{3,1,1,1,3,7}\rangle$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve for $\xi_{0}$, the maximal pure cyclic part is $\xi_{3}=\langle\overline{3,1,1,1,3,7}\rangle . \xi_{3}$ satisfies
$$\xi_{3}=\left\langle 3,1,1,1,3,7, \xi_{3}\right\rangle$$
First, calculate $\langle 3,1,1,1,3\rangle$ and $\langle 3,1,1,1,3,7\rangle$. We have
$$\begin{aligned}
\langle 3,1,1,1,3\rangle & =\langle ... | \frac{3-\sqrt{57}}{24} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,830 |
Theorem 5 (i) The value of a pure periodic continued fraction $\xi_{0}$ is necessarily a real quadratic irrational number, $\xi_{0}>1$, and its conjugate $\xi_{0}^{\prime}$ satisfies $-1<\xi_{0}^{\prime}<0$;
(ii) The value of a periodic continued fraction is a real quadratic irrational number. | (i) Let the period of the pure periodic continued fraction $\xi_{0}$ be $l$, then we have
$$\xi_{0}=\left\langle a_{0}, \cdots, a_{l-1}, \xi_{0}\right\rangle$$
From $a_{0}=a_{l} \geqslant 1$ we know $\xi_{0}>1$. By formula (17) in §3, we have
$$\xi_{0}=\frac{h_{l-1} \xi_{0}+h_{l-2}}{k_{l-1} \xi_{0}+k_{l-2}}$$
Here $h... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,831 |
Theorem 6 Let $\xi_{0}$ be a real quadratic irrational number, then
(i) the infinite simple continued fraction representation of $\xi_{0}$ must be a periodic continued fraction.
(ii) Let $\xi_{0}^{\prime}$ be the conjugate of $\xi_{0}$. If $\xi_{0}>1,-1<\xi_{0}^{\prime}<0$, then the infinite simple continued fraction r... | To prove the theorem, we need to introduce a representation form of quadratic irrational numbers (see formula (16)), which also plays a crucial role in specifically finding the solutions of the Pell equation (see §6). By Theorem 1, a real quadratic irrational number \(\xi_{0}\) can certainly be represented as (why)
\[
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,832 |
Example 2 Find the periodic continued fraction of $\xi_{0}=(\sqrt{14}+1) / 2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | We solve according to the method of Theorem 6, which requires finding the smallest $k>h \geqslant 0$ such that $\xi_{h} = \xi_{k}$, i.e., equation (24) holds. To satisfy condition (16), $\xi_{0}$ should be expressed as
$$\begin{array}{c}
\xi_{0}=(\sqrt{56}+2) / 4, \quad d=56 \\
c_{0}=2, \quad q_{0}=4, \quad a_{0}=\left... | \xi_{0}=(\sqrt{14}+1) / 2=\langle 2, \overline{2,1,2,3}\rangle | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,833 |
Example 3 Find the periodic continued fraction of $\xi_{0}=\sqrt{73}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | To make condition (16) hold, $\xi_{0}$ should be expressed as
$$\xi_{0}=(\sqrt{73}+0) / 1, \quad q_{0}=1, c_{0}=0, d=73$$
$a_{0}=[\sqrt{73}]=8$. Using equation (22), we obtain
$$\begin{array}{c}
c_{1}=8 \cdot 1-0=8, \quad q_{1}=\left(73-8^{2}\right) / 1=9 \\
\xi_{1}=(\sqrt{73}+8) / 9, \quad a_{1}=1
\end{array}$$
Below... | \xi_{0}=\sqrt{73}=\langle 8, \overline{1,1,5,5,1,1,16}\rangle | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,834 |
Theorem 7 Let $\xi_{0}$ be given by (16),
$$\xi_{0}=\left\langle a_{0}, a_{1}, a_{2}, \cdots\right\rangle$$
and let $h_{n}, k_{n}$ be given by (2) in $\S 2$, and $c_{n}, q_{n}$ by (21). Then, we have
$$\begin{aligned}
(-1)^{n+1} c_{n}= & \left(q_{0} h_{n-1} h_{n-2}-c_{0}\left(h_{n-1} k_{n-2}+h_{n-2} k_{n-1}\right)\rig... | Proof: From equation (17) in §3, we have
$$\xi_{0}=\left(h_{n-1} \xi_{n}+h_{n-2}\right) /\left(k_{n-1} \xi_{n}+k_{n-2}\right), \quad n \geqslant 0$$
Substituting the expressions for $\xi_{0}$ and $\xi_{n}$ from equation (21) into the above equation, we get
By comparing coefficients, we obtain
$$\begin{array}{c}
\left(... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,835 |
Theorem 8 Let $d>1$ be a non-square number, $\xi_{0}=\sqrt{d}+[\sqrt{d}]$; and let $\xi_{j}, a_{j}$ be the same as in Theorem 4 of §3, and $c_{j}, q_{j}$ be determined by the representation (21) of $\xi_{j}$ given in Theorem 6. Then
(i) $q_{j}=1$ if and only if $l \mid j$, where $l$ is the period of the pure periodic c... | Let $\xi_{0}$ have its conjugate number $\xi_{0}^{\prime}$. We have
$$\xi_{0}>1, \quad-11, \quad-11, \quad-1c_{j}>\sqrt{d}$$
This is impossible. Therefore, for any $j \geqslant 0, q_{j} \neq-1$. Proof completed. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,837 |
1. Let $\left\langle a_{0}, a_{1}, a_{2}, \cdots\right\rangle$ be a periodic continued fraction with period $l$, and let $h_{n} / k_{n}$ be its convergents, $\xi_{n}=\left\langle a_{n}, a_{n+1}, \cdots\right\rangle$.
(i) Prove: $\xi_{n+1}+k_{n-1} / k_{n}=\left\langle\xi_{n+1}, a_{n}, \cdots, a_{1}\right\rangle$;
(ii) P... | 1. (i) Use Problem 6 from Exercise 1;
(ii) Using (i), Problem 13 from Exercise 3, and for any fixed $j, 0 \leqslant j \leqslant l-1$, when $n \equiv m_{0}+j(\bmod l), n \rightarrow \infty$, the limit of $\xi_{n+1}+k_{n-1} / k_{n}$ is $\lambda_{m_{0}+j}$.
(iii) The two limit points of $\langle 2,5,1, \overline{1,2}\rang... | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,840 |
2. 求以下二次无理数的循环连分数表示式, 它的纯循环部分及周期: $\square$
(i) $(5+\sqrt{37}) / 3$;
(ii) $\sqrt{43}$;
(iii) $(6+\sqrt{43}) / 7$;
(iv) $\sqrt{80}+8$;
(v) $(3+\sqrt{7}) / 2$;
(vi) $\sqrt{26 / 5}$. | 2. (i) $\langle\overline{3,1,2}\rangle$;
(ii) $\langle 6, \overline{1,1,3,1,5,1,3,1,1,12}\rangle$;
(iii) $\langle 1,1,3,1,5,1,3,1,1, \sqrt{43}+6\rangle=\langle\overline{1,1,3,1,5,1,3,1,1,12}\rangle$;
(iv) $\langle\overline{16,1}\rangle$; (v) $\langle 2, \overline{1,4,1,1}\rangle$; (vi) $\langle 2, \overline{3,1,1,3,4}\... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,841 |
5. Let $\xi_{0}=\left\langle\overline{a_{0}, a_{1}, \cdots, a_{n}}\right\rangle, \xi_{0}^{\prime}$ be the conjugate of $\xi_{0}$. Prove:
$$-1 / \xi_{0}^{\prime}=\left\langle\overline{a_{n}, a_{n-1}, \cdots, a_{0}}\right\rangle .$$ | 5. Let $\eta_{0}=-1 / \xi_{0}^{\prime}$.
$\xi_{0}=\left(h_{n} \xi_{0}+h_{n-1}\right) /\left(k_{n} \xi_{0}+k_{n-1}\right), \eta_{0}=\left(h_{n} \eta_{0}+k_{n}\right) /\left(h_{n-1} \eta_{0}+k_{n-1}\right)$. Use Exercise 1, Question 6. | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,844 |
11. Under the notation of the previous question, find $j$ that satisfies
(i) $0 \bmod 3 \cap 0 \bmod 5=j \bmod 15$;
(ii) $1 \bmod 3 \cap 1 \bmod 5=j \bmod 15$;
(iii) $-1 \bmod 3 \cap -2 \bmod 5=j \bmod 15$. | 11. (i) $j=0$;
(ii) $j=1$;
(iii) $j=8$. | 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,847 |
9. Let $a$ be an odd number. Prove:
(i) When $a>1$, $\sqrt{a^{2}+4}=\langle a, \overline{(a-1) / 2,1,1,(a-1) / 2,2 a}\rangle$;
(ii) When $a>3$,
$$\sqrt{a^{2}-4}=\langle a-1, \overline{1,(a-3) / 2,2,(a-3) / 2,1}, \overline{2 a-2}\rangle$$
Provide specific examples to illustrate the application of (i) and (ii). | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,849 |
11. Let $l$ be a positive integer. Prove: there exist infinitely many $\sqrt{d}$, whose periodic continued fraction has a period of $l$. | 11. Let $c_{1}=2, c_{2}=5$, and $c_{s}=2 c_{s-1}+c_{s-2}, s \geqslant 3$. When taking $d=\left(u c_{s}+1\right)^{2}+2 u c_{s-1}+1$, the period of $\sqrt{d}$ is $s+1$. Here $u$ is any positive integer. $\sqrt{d}=\left\langle\left(u c_{s}+1\right), \overline{2, \cdots, 2,2\left(u c_{s}+1\right)}\right\rangle$, with $s$ t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,851 |
Prove:
$$\left\langle a_{1}, \cdots, a_{l-1}\right\rangle=\left\langle a_{l-1}, \cdots, a_{1}\right\rangle, \quad \text { i.e. } a_{j}=a_{l-j}, 1 \leqslant j \leqslant l / 2 \text {. }$$ | 12. Utilize Question 5, take $\xi_{0}=[\sqrt{d}]+\sqrt{d} .-1 / \xi_{0}=1 /(\sqrt{d}-[\sqrt{d}])=$ $\left\langle\overline{a_{1}}, \cdots, a_{l-1}, 2[\sqrt{d}]\right\rangle$. But from Question 5, we know that $-1 / \xi_{0}^{\prime}=\left\langle\overline{a_{l-1}, \cdots, a_{1}, 2[\sqrt{d}]}\right\rangle$. | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,852 |
13. Let $\xi_{0}$ be a quadratic irrational number, with a periodic continued fraction of period $l$, and $h_{n} / k_{n}$ be its convergents. Prove:
(i) When $\xi_{0}$ is a purely periodic continued fraction, there exist integers $a, b, c, d, a d-b c=(-1)^{l}$, such that
$$\binom{h_{n+l}}{k_{n+l}}=\left(\begin{array}{l... | 13. (i) From equation (5) in §2, we know that the system of equations $a h_{0}+b k_{0}=h_{l}, a h_{1}+b k_{1}=h_{l+1}$ can determine the integers $a, b$; the system of equations $c h_{0}+d k_{0}=k_{l}, c h_{1}+d k_{1}=k_{l+1}$ can determine the integers $c, d$. These $a, b, c, d$ are the ones we seek. Similarly, prove ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,853 |
14. Let the continued fraction of $\sqrt{d}$ be as in Question 13, and let $h_{n} / k_{n}$ be its convergents. Prove: for positive integer $m$, when $1 \leqslant j \leqslant m l$,
$$\binom{h_{m l-1}}{k_{m l-1}}=k_{j-1}\binom{h_{m l-j}}{k_{m l-j}}+k_{j-2}\binom{h_{m l-j-1}}{k_{m l-j-1}}$$
and
$$\binom{d k_{m l-1}}{h_{m... | 14. Prove formulas $(*)$ and $(* *)$ by induction (on $j$). When proving $(* *)$, use the following relationship: Let $\xi_{0}=[\sqrt{d}]+\sqrt{d}=\left\langle\overline{a_{0}, a_{1}, \cdots, a_{l-1}}\right\rangle$. From $\xi_{m l+1}=\xi_{1}$, we get
$$\xi_{0}=\left(h_{m l}+h_{m l-1}\left(\xi_{0}-a_{0}\right)\right) /\l... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,854 |
15. Let the irrational number $\alpha=\left\langle\overline{a_{0}, a_{1}}\right\rangle, a_{0}=c a_{1}, \beta$ be its conjugate; and let $h_{n} / k_{n}$ be the convergents of $\alpha$. Prove:
$$\begin{array}{ll}
h_{n}=c^{-[(n+1) / 2]} \frac{\alpha^{n+2}-\beta^{n+2}}{\alpha-\beta}, & n \geqslant-2 \\
k_{n}=c^{-[(n+1) / 2... | 15. $\alpha$ is the positive root of the quadratic equation $x^{2}-a_{0} x-c=0$, i.e.,
$$\alpha=\left(a_{0}+\sqrt{a_{0}^{2}+4 c}\right) / 2$$
Let $u_{n}=\left(\alpha^{n}-\beta^{n}\right) /(\alpha-\beta), p_{n}=c^{-[(n+1) / 2]} u_{n+2}, q_{n}=c^{-[(n+1) / 2]} u_{n+1}$. Use induction to prove the desired conclusion: Fir... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,855 |
16. Let $u_{1}, u_{2}, \cdots, u_{n}, \cdots$ be the Fibonacci sequence, i.e., $u_{1}=u_{2}=1, u_{n+2}=$ $u_{n+1}+u_{n}, n \geqslant 1$. Prove:
$$u_{n}=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right\}, \quad n \geqslant 1 .$$ | 16. A special case of the previous problem, taking $\alpha=(\sqrt{5}+1) / 2$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note at the end is not part of the translation but is provided to clarify the instruction. The actual translation is above. | null | Algebra | proof | Yes | Yes | number_theory | false | 740,856 |
Theorem 1 Let $\xi_{0}=\sqrt{d}$, with a periodic continued fraction of length $l$, and convergents $h_{n} / k_{n}$. Then
(i) When $l$ is even, the indeterminate equation (2) has no solution, and all positive solutions of the indeterminate equation (1) are
$$x=h_{j l-1}, \quad y=k_{j l-1}, \quad j=1,2,3, \cdots$$
(ii) ... | Proof: By Theorem 8 of §3, if $x, y$ is a positive solution of the indeterminate equation (1) or (2), then there must be some $n \geqslant 0$ such that $x=h_{n}, y=k_{n}$. On the other hand, by formula (38) of §5, we have
$$h_{n}^{2}-d k_{n}^{2}=(-1)^{n+1} q_{n+1}$$
By Corollary 9 of §5 (where $\tilde{\xi}_{0}, \tilde... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,857 |
Lemma 2 If $a, b, c$ are all integers and $a|b, b| c$, then $a \mid c$.
| Proof. Since $a \mid b$, by Definition 1 there is an integer $d$ such that $b = a d$. Also, since $b \mid c$, there is an integer $e$ such that $c = b e$. From $c = b e$ and $b = a d$, we have $c = a d e$. Since $d$ and $e$ are integers, $d e$ is also an integer. By Definition 1 and $c = a d e$, we have $a \mid c$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,859 |
Lemma 8 Assume $a$ and $b$ are positive integers, and $a>b$,
$$a=b q+r, \quad 0<r<b$$
where $\boldsymbol{a}$ and $r$ are positive integers, then the greatest common divisor of $\boldsymbol{a}$ and $b$ is equal to the greatest common divisor of $b$ and $r$, i.e.,
$$(a, b)=(b, r)$$ | Proof: Let $(a, b)=d$. By Definition 5, there exist two integers $m, n$ such that $a=d m, b=d n$. From $r=a-b q=(m-q n) d$, it follows that $d \mid r$. Since $d \mid r, d \mid b$, it follows that $d \mid (b, r)$, i.e., $(b, r) \geqslant d$. Suppose $(b, r)=D>d$, then $D \mid b, D \mid r$. From $a=b q+r$, it follows tha... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,860 |
Theorem 1 Let the binary linear indeterminate equation be
$$a x+b y=c$$
(where $a, b, c$ are all positive integers and $(a, b)=1$) have a set of integer solutions $x=$ $x_{0}, y=y_{0}$, then all integer solutions of (8) can be expressed as
$$x=x_{0}-b t, \quad y=y_{0}+a t,$$
where $t=0, \pm 1, \pm 2, \pm 3, \cdots$. | Since $x_{0}, y_{0}$ are integer solutions of equation (8), they of course satisfy $a x_{0}+b y_{0} = c$. Therefore,
$$a\left(x_{0}-b t\right)+b\left(y_{0}+a t\right)=a x_{0}+b y_{0}=c_{0}$$
This shows that equation (12) is a solution to equation (8).
Let $x^{\prime}, y^{\prime}$ be any integer solution of equation (8... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,861 |
Example 3 Find all integer solutions to $111 x-321 y=75$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Since
$$111=3 \times 37, \quad 321=3 \times 107$$
$$75=3 \times 25$$ and $$111 x-321 y=75$$, we get
$$37 x-107 y=25$$
Thus, the solutions to $$111 x-321 y=75$$ are exactly the same as those to equation (13). Now, let's solve
$$37 s+107 t=1$$
Since
$$107=2 \times 37+33, \quad 37=33+4, \quad 33=8 \times 4+1$$
We have
... | x=-8+107 t, \quad y=-3+37 t | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,862 |
Example 5 At the end of the fifth century, the ancient Chinese mathematician Zhang Qiujian proposed an indeterminate equation problem in his compiled "Suanjing" — a famous "Hundred Chickens Problem" in the history of world mathematics:
A rooster is worth five coins, a hen is worth three coins, and three chicks are wor... | Let $x, y, z$ represent the number of roosters, hens, and chicks, respectively. We then have the following equations:
$$\begin{array}{l}
5 x+3 y+\frac{z}{3}=100 \\
x+y+z=100
\end{array}$$
From (16) and (17), we have
$$15 x+9 y+z-x-y-z=14 x+8 y=200$$
That is,
$$7 x+4 y=100$$
To solve this problem, we need to find the... | \begin{array}{l}
x=12, y=4, z=84 \\
x=8, y=11, z=81 \\
x=4, y=18, z=78 \\
x=0, y=25, z=75
\end{array} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,864 |
Theorem 2 All positive integer solutions of the indeterminate equation (21) satisfying the conditions
$$x>0, \quad y>0, \quad z>0, \quad(x, y)=1, \quad 2 \mid x$$
can be expressed by the following formulas:
$$x=2 a b, \quad y=a^{2}-b^{2}, \quad z=a^{2}+b^{2}$$
where $a$ and $b$ are positive integers, and $a>b,(a, b)=... | Given that $a$ and $b$ are positive integers and $a > b$, from equation (28) we have $x = 2ab > 0$, $y = a^2 - b^2 > 0$, $z = a^2 + b^2 > 0$, which means $x, y, z$ satisfy the conditions $x > 0$, $y > 0$, $z > 0$, and $2 \mid x$ in equation (27). From equation (28), we have
$$\begin{aligned}
x^2 + y^2 & = 4a^2b^2 + (a^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,865 |
Example 6 Find all integer solutions of the following indeterminate equation:
$$4 x-9 y+5 z=8$$ | Let $t$ be an integer and have
$$4 x-9 y=t$$
From (29) and (30), we have
$$t+5 z=8$$
By $x=-2 t, y=-t$ being a set of integer solutions to (30), and by Theorem 1, we get that $x=-2 t+9 u, y=-t+4 u, u=0, \pm 1, \pm 2, \cdots$ are all integer solutions to (30). By Theorem 1 and $t=3, z=1$ being a set of integer solutio... | \begin{array}{l}
x=-6+10 v+9 u \\
y=-3+5 v+4 u \\
z=1+v
\end{array} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,866 |
Example 7 Find all integer solutions to the indeterminate equation $x y=x^{2}+6$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, it seems you might have wanted the translation of the problem statement itself, which i... | Solve: From $x y=x^{2}+6$ we get $x(x-y)=-6$, so $x \mid(-6)$. When $x=1$, $y=7$; when $x=-1$, $y=-7$. When $x=2$, $y=5$; when $x=-2$, $y=-5$. When $x=3$, $y=5$; when $x=-3$, $y=-5$. When $x=6$, $y=7$; when $x=-6$, $y=-7$. Therefore, all integer solutions to $x y=x^{2}+6$ are
$$\left.\left.\left.\left.\left.\left.\begi... | \begin{array}{l}
x=1, y=7 \\
x=-1, y=-7 \\
x=2, y=5 \\
x=-2, y=-5 \\
x=3, y=5 \\
x=-3, y=-5 \\
x=6, y=7 \\
x=-6, y=-7
\end{array} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,867 |
Lemma 3 The indeterminate equation
$$x^{4}+y^{4}=z^{4}$$
has no positive integer solutions. | To prove that
$$x^{4}+y^{4}=u^{2}$$
has no positive integer solutions, we assume that equation (39) has positive integer solutions. Among all the positive integer solutions that satisfy (39), there must be a solution where \( \boldsymbol{u} \) is the smallest, i.e., there exists a smallest positive integer \( u_{1} \)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,868 |
1. Determine whether the following equations have integer solutions:
(i) $x^{3}+3 x^{2}+4 x+2=0$.
(ii) $x^{9}+x^{5}-x^{4}-2 x^{2}+3 x-37=0$.
(iii) $x^{7}+3 x^{5}+3 x+1005973=0$. | 1.
(i) Solution: If the equation has integer solutions, they must be factors of the constant term. The constant term 2 has four factors: $1, -1, 2$, and -2. Since the coefficients of each term in the equation are positive, there cannot be positive solutions, because substituting a positive value for $x$ into the equati... | not found | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,869 |
2. Find the integer solutions of the following indeterminate equations:
(i) $7 x+15 y=0$.
(ii) $9 x-11 y=1$.
(iii) $17 x+40 y=280$.
(iv) $133 x-105 y=217$.
(v) $49 x-56 y+14 z=35$ | 2.
(i) Solution: Since $(7,15)=1$, only the constant term is zero, so the integer solutions of the equation are: $x=-15i, y=7t, t=0, \pm 1, \pm 2, \cdots$.
(ii) Solution: Since $11=9+2, 9=2 \times 4+1$, we get $1=9-2 \times 4=9-4 \times (11-9)=9 \times 5-11 \times 4$, so $x=5$, $y=4$ is a set of integer solutions of th... | \begin{array}{l}
(i) \quad x=-15t, \quad y=7t, \quad t=0, \pm 1, \pm 2, \cdots \\
(ii) \quad x=5+11t, \quad y=4+9t, \quad t=0, \pm 1, \pm 2, \cdots \\
(iii) \quad x=-40t, \ | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,870 |
Example 4 Find the greatest common divisor of 6731 and 2809 | Solve:
\begin{aligned}
6731 & =2809 \times 2+1113 \\
2809 & =1113 \times 2+583 \\
1113 & =583 \times 1+530 \\
583 & =530+53 \\
530 & =53 \times 10+0
\end{aligned}
So $(6731,2809)=53$. For convenience in writing, the series of calculations above can be abbreviated as follows:
2 \begin{tabular}{|r|r|r}
6731 & 2809 & 2 \... | 53 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,871 |
4. Find the positive integer solutions for
$$\left\{\begin{array}{l}
5 x+7 y+2 z=24 \\
3 x-y-4 z=4
\end{array}\right.$$ | 4. Solution: This is a system of three linear equations. We can use the elimination method to eliminate one of the unknowns, thus obtaining a system of two linear equations. By multiplying the first equation by 2 and adding it to the second equation, we can eliminate \( z \) and get
$$13 x+13 y=52$$
which simplifies t... | x=3, y=1, z=1 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,873 |
5. Take a total of ten coins of one cent, two cents, and five cents, to pay eighteen cents. How many different ways are there to do this?
保留源文本的换行和格式，直接输出翻译结果如下：
5. Take a total of ten coins of one cent, two cents, and five cents, to pay eighteen cents. How many different ways are there to do this? | 5. Solution: Let $x, y, z$ represent the number of 1-cent, 2-cent, and 5-cent coins, respectively. Therefore, we have the following equations:
$$\begin{array}{l}
x+2 y+5 z=18 \\
x+y+z=10
\end{array}$$
Subtracting the second equation from the first, we get $y+4 z=8$.
We need to find the non-negative integer solutions t... | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 740,874 |
6. There is 7 zhang 5 chi of cloth, to be cut into material for adult and children's clothing. An adult's garment requires 7 chi 2 cun of cloth, and a child's garment requires 3 chi. How many pieces of each should be cut to use up all the cloth?
Note: In traditional Chinese units, 1 zhang = 10 chi, and 1 chi = 10 cun. | 6. Solution: Let $x, y$ represent the number of adult and children's clothes cut, respectively, then we have the equations
i.e.,
$$\begin{array}{l}
7.2 x+3 y=75 \\
12 x+5 y=125
\end{array}$$
We now seek its non-negative integer solutions. First, solve
$$12 u+5 v=1$$
Since $12=2 \times 5+2, 5=2 \times 2+1$, we get
$$... | null | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,875 |
7. A two-digit number is three times the product of its tens and units digits. What is this two-digit number?
The above text has been translated into English, retaining the original text's line breaks and format. | 7. Solution: Let the tens digit be $x$, and the units digit be $y$. According to the problem, we have
$$10 x+y=3 x y$$
Dividing both sides of the equation by $x$ gives
$$10+\frac{y}{x}=3 y$$
Since 10 and $3 y$ are both integers, $\frac{y}{x}$ must also be an integer. Let $z=\frac{y}{x}$, then the original equation be... | 24 \text{ and } 15 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,876 |
8. Prove: If $a$ and $b$ are two coprime positive integers, then there exist two integers $x, y$ such that
$$a x + b y = 1$$ | 8. Proof: The proof is divided into three steps.
(i) Suppose $R_{1}$ and $R_{2}$ are two integers that can be written in the form $a x + b y$, where $a$ and $b$ are fixed positive integers, and $x$ and $y$ are integers. Then $k_{1} R_{1} + k_{2} R_{2} \left(k_{1}, k_{2}\right.$ are integers) can also be written in the ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,877 |
9. Prove that all positive integer solutions of the indeterminate equation
$$x^{2}+y^{2}=z^{2}$$
satisfying the conditions $(x, y)=1,2 \mid x$ can be expressed as
$$x=2 a b, \quad y=a^{2}-b^{2}, \quad z=a^{2}+b^{2}$$
where $a$ and $b$ are positive integers, $a>b,(a, b)=1,2 \nmid(a+b)$. | 9. Proof: (i) First prove that $\frac{z+y}{2}$ and $\frac{z-y}{2}$ are both integers, and
$$\left(\frac{z+y}{2}, \frac{z-y}{2}\right)=1$$
Since $x$ is even, $x^{2}$ is even, and by $(x, y)=1$ it follows that $y$ must be odd. Thus, $y^{2}$ is also odd, and therefore $x^{2}+y^{2}$ is odd. Since $x, y, z$ satisfy
$$x^{2}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,878 |
10. Prove: For a right-angled triangle with integer side lengths, when the difference between the hypotenuse and one of the legs is 1, its three side lengths can be expressed as: \(2b+1, 2b^2+2b, 2b^2+2b+1\), where \(b\) is any positive integer. | 10. Proof: Let $x, y$ be the lengths of the legs, and $z$ be the length of the hypotenuse, and $z-x=1$. By the Pythagorean theorem, we have
$$x^{2}+y^{2}=z^{2}$$
Since $z-x=1$, it follows that $(x, z)=1$. Therefore, by equation (6), it must be that $(x, y)=1$. According to the discussion in this chapter, the solutions... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,879 |
11. Prove that the indeterminate equation
$$x^{4}-4 y^{4}=z^{2}$$
has no positive integer solutions. | 11. Proof: Squaring both sides of the equation, we get
$$\begin{aligned}
z^{4}= & \left(x^{4}-4 y^{4}\right)^{2} \\
& =x^{8}-8 x^{4} y^{4}+16 y^{8} \\
& =\left(x^{8}+8 x^{4} y^{4}+16 y^{8}\right)-16 x^{4} y^{4} \\
& =\left(x^{4}+4 y^{4}\right)^{2}-(2 x y)^{4}
\end{aligned}$$
Thus,
$$(2 x y)^{4}+z^{4}=\left(x^{4}+4 y^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,880 |
Example 5 Find the greatest common divisor of 735000, 421160, and 238948. | $$\begin{aligned}
735000 & =238948 \times 3+18156 \\
238948 & =18156 \times 13+2920 \\
18156 & =2920 \times 6+636 \\
2920 & =636 \times 4+376 \\
636 & =376 \times 1+260 \\
376 & =260 \times 1+116 \\
260 & =116 \times 2+28 \\
116 & =28 \times 4+4 \\
28 & =7 \times 4+0
\end{aligned}$$
So the greatest common divisor of 7... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,882 |
Lemma 1 When $a$ is an integer and $m$ is a positive integer, then we have
$$a \equiv a(\bmod m)$$ | Prove that since $a-a=m \times 0$, it follows that $m \mid(a-a)$, hence $a \equiv$ $a(\bmod m)$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Proof: Since $a-a=m \times 0$, it follows that $m \mid (a-a)$, hence $a \equiv a \pmo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,884 |
Lemma 2 If $a, b$ are integers and $m$ is a positive integer, then when
$$a \equiv b \quad(\bmod m)$$
holds, we have
$$b \equiv a(\bmod m) .$$ | Proof: Since $a \equiv b(\bmod m)$, we get $a-b=m t$, where $t$ is an integer. Also, $b-a=m(-t)$, and since $-t$ is also an integer, we have $m \mid (b-a)$, thus $b \equiv a(\bmod m)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,885 |
Lemma 3 If $a, b, c$ are all integers and $m$ is a positive integer, then when
$$\begin{array}{ll}
a \equiv b & (\bmod m) \\
b \equiv c & (\bmod m)
\end{array}$$
both hold, we have
$$a \equiv c \cdot(\bmod m)$$ | Given that $a \equiv b(\bmod m)$, we get
$$a-b=m t$$
where $t$ is an integer. From $b \equiv c(\bmod m)$, we get
$$b-c=m s$$
where $s$ is an integer. Adding equations (1) and (2) yields
$$a-c=a-b+b-c=m t+m s=m(t+s)$$
Since $t+s$ is also an integer, we have
$$a \equiv c(\bmod m)$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,886 |
If $a, b, c, d$ are all integers, and $m$ is a positive integer, then when
$$\begin{array}{ll}
a \equiv b & (\bmod m) \\
c \equiv d & (\bmod m)
\end{array}$$
both hold, we have
$$\begin{array}{ll}
a+c \equiv b+d & (\bmod m) \\
a-c \equiv b-d & (\bmod m)
\end{array}$$ | Given that $a \equiv b(\bmod m)$, we get
$$a-b=m s$$
where $s$ is an integer. Since $c \equiv d(\bmod m)$, we get
$$c-d=m t,$$
where $t$ is an integer. From equations (3) and (4), we have
$$\begin{array}{l}
(a+c)-(b+d)=a-b+c-d=m s+m t \\
\quad=m(s+t)
\end{array}$$
Since $s+t$ is an integer, we have $a+c \equiv b+d(\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,887 |
Lemma 5 If $a, b, c$ are all integers, and $m$ is a positive integer, then when
$$a \equiv b \quad(\bmod m)$$
holds, we have
$$a c \equiv b c(\bmod m)$$ | Given that $a \equiv b(\bmod m)$, we get
$$a-b=m s,$$
where $s$ is an integer. Multiplying both sides of (5) by $c$, we obtain
$$a c-b c=(a-b) \cdot c=m c s.$$
Since $c s$ is an integer, we have $a c \equiv b c(\bmod m)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,888 |
Lemma 6 If $a, b, c, d$ are all integers, and $m$ is a positive integer, then when
$$\begin{array}{ll}
a \equiv b & (\bmod m) \\
c \equiv d & (\bmod m)
\end{array}$$
both hold, we have
$$a c \equiv b d \quad(\bmod m)$$ | From $a \equiv b(\bmod m)$ and Lemma 5, we get
$$a c \equiv b c \quad(\bmod m)$$
From $c \equiv d(\bmod m)$ and Lemma 5, we get
$$b c \equiv b d \quad(\bmod m)$$
From equations (6), (7) and Lemma 3, we have $a c \equiv b d \quad(\bmod m)$. | a c \equiv b d \quad(\bmod m) | Number Theory | proof | Yes | Yes | number_theory | false | 740,889 |
Lemma 7 If $a, b$ are integers, and $m$ and $n$ are positive integers, then when
$$a \equiv b(\bmod m)$$
holds, we have
$$a^{n} \equiv b^{n} \quad(\bmod m)$$ | Proof: When $n=1$, the lemma is clearly true. Now assume $n \geqslant 2$. In Lemma 6, take $c=a, d=b$. By Lemma 6, we have
$$a^{2} \equiv b^{2} \quad(\bmod m)$$
Thus, the lemma holds when $n=2$. Now assume $n \geqslant 3$. In Lemma 6, take $c=$ $a^{2}, d=b^{2}$. Then by (8) and Lemma 6, we have
$$a^{3} \equiv b^{3} \q... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,890 |
Lemma 8 If $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ are all integers, and $m$ and $n$ are positive integers, then when
$$\begin{array}{r}
a_{1} \equiv b_{1} \quad(\bmod m) \\
a_{2} \equiv b_{2} \quad(\bmod m) \\
a_{n} \equiv b_{n} \quad(\bmod m)
\end{array}$$
all hold, we have
$$a_{1}+a_{2}+\cdots+a_... | Given $a_{1} \equiv b_{1}(\bmod m), a_{2} \equiv b_{2}(\bmod m)$ and Lemma 4, we have
$$a_{1}+a_{2} \equiv b_{1}+b_{2}(\bmod m)$$
Thus, the lemma holds when $n=2$. Now suppose $n \geqslant 3$. By $a_{3} \equiv b_{3}(\bmod m)$, equation (10), and Lemma 4, we have
$$a_{1}+a_{2}+a_{3} \equiv b_{1}+b_{2}+b_{3}(\bmod m)$$
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,891 |
Example 3 Prove that 5874192 is divisible by 9.
untranslated text:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
translated text:
Translate the text above into English, please keep the line breaks and format of the source text, and output the translation result directly. | $$\begin{array}{l}
5874192=5 \times 10^{6}+8 \times 10^{5}+7 \times 10^{4}+4 \times 10^{3} \\
+10^{2}+9 \times 10+2
\end{array}$$
Since $10 \equiv 1(\bmod 9)$, using Lemma 7, we know that for any positive integer $\boldsymbol{n}$,
$$10^{n} \equiv 1(\bmod 9)$$
From (12), (13) and Lemma 8, we have
$$5874192 \equiv 5+8+... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,892 |
Example 4 Prove that 2221435693 cannot be divided by 9.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | $$\begin{aligned}
2221435693 & =2 \times 10^{9}+2 \times 10^{8}+2 \times 10^{7}+10^{5} \\
+ & 4 \times 10^{5}+3 \times 10^{4}+5 \times 10^{3}+6 \times 10^{2} \\
+9 & \times 10+3
\end{aligned}$$
By equations (13), (14) and Lemma 8, we have
$$\begin{array}{l}
2221435693 \equiv 2+2+2+1+4+3+5+6 \\
+9+3(\bmod 9)
\end{array... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,894 |
Lemma 9 According to the usual method, write a positive integer $a$ in the form of a decimal number, that is,
$$a=a_{n} 10^{n}+a_{n-1} 10^{n-1}+\cdots+a_{0}, \quad 0 \leqslant a_{i}<10$$
When 9 can divide $a_{n}+a_{n-1}+\cdots+a_{0}$, then we have 9 can divide $a$. And when 9 cannot divide $a_{n}+a_{n-1}+\cdots+a_{0}$,... | By (13) and Lemma 8, we have
$$a \equiv a_{n}+a_{n-1}+\cdots+a_{0}(\bmod 9)$$
When 9 can divide $a_{n}+a_{n-1}+\cdots+a_{0}$, then by (17) we get that 9 can divide $a$. And when 9 cannot divide $a_{n}+a_{n-1}+\cdots+a_{0}$, then by (17) we get that 9 cannot divide $a$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,895 |
Example 5 Prove that 9 can divide 221145236415. | Proof: By Lemma 9, we have
$$\begin{array}{c}
221145236415 \equiv 2+2+1+1+4+5+2+3 \\
+6+4+1+5(\bmod 9)
\end{array}$$
Also,
$$\begin{array}{l}
2+2+1+1+4+5+2+3+6+4+1 \\
\quad+5=36
\end{array}$$
By equations (18), (19), and since 9 divides 36, 9 can divide 221145236415. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,896 |
Example 6 Prove that $28997 \times 39459 \neq 1144192613$.
| $$\begin{array}{c}
\text { Proof: Since } 28997 \equiv 2+8+7 \equiv 8(\bmod 9), \\
39459 \equiv 3+4+5 \equiv 3(\bmod 9), \\
1144192613 \equiv 1+1+4+4+1+2+6+1+3 \\
\equiv 5(\bmod 9),
\end{array}$$
but $8 \times 3=24$, and $24 \neq 5(\bmod 9)$, hence
$$28997 \times 39459 \neq 1144192613$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,897 |
Example 7 Prove that $12345 \times 67891 \neq 838114385$.
| Proof that since $12345 \equiv 1+2+3+4+5 \equiv 6(\bmod 9)$,
$$\begin{array}{l}
67891 \equiv 6+7+8+1 \equiv 4(\bmod 9) \\
838114385 \equiv 8+3+8+1+1+4+3 \\
+8+5 \equiv 5(\bmod 9)
\end{array}$$
But $4 \times 6=24$, and $24 \not \equiv 5(\bmod 9)$, hence we get
$$12345 \times 67891 \neq 838114385$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,898 |
Lemma 10 If $c$ is an integer that satisfies equation (28), i.e., $a c +$ $b \equiv 0 \quad(\bmod m)$, then all integers $x$ that satisfy $x \equiv c(\bmod m)$ can also satisfy equation (28). | Given $x \equiv c(\bmod m)$, we get $m \mid(x-c)$, i.e., $x-c=m n$, where $n$ is an integer. From $x=m n+c$ and $a c+b \equiv 0(\bmod m)$, we obtain
$$a x+b \equiv a(m n+c)+b \equiv a c+b \equiv 0 \quad(\bmod m)$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,899 |
Lemma 11 When the greatest common divisor of $a, m$ (i.e., $(a, m)$) does not divide $b$ (i.e., $(a, m) \nmid b)$, the linear congruence
$$a x+b \equiv 0(\bmod m), \quad a \not \equiv 0(\bmod m)$$
has no integer solutions. | Proof: Suppose there exists an integer $c$ such that $a c+b \equiv 0(\bmod m)$, i.e., $m \mid(a c+b)$, hence there exists an integer $n$ such that $a c+b=m n$, and we get
$$a c-m n=b \text {. }$$
Let $(a, m)=l$, then we have $a=l d, m=l e$, where $d$ and $e$ are integers. Substituting them into (29), we get
$$b=a c-m ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,900 |
Lemma 12 When $(a, m)=1$, the linear congruence $a x+b \equiv 0 \quad(\bmod m), \quad a \neq 0 \quad(\bmod m)$
has integer solutions. | By Lemma 2 of Chapter 3, we know that there exist two integers $x, y$ such that
$$a x+m y=-b$$
holds. That is, $m \mid(a x+b)$, hence we get $a x+b \equiv 0(\bmod m)$.
We know that the integer $x$ satisfying equation (28) is also the value of $x$ in the solution of the indeterminate equation
$$a x+m y=-b$$
Therefore,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,901 |
Example 8 Find the integer solution of $2 x \equiv 179(\bmod 562)$. | Since $(2,562)=2,2 \nmid(-179)$, by Lemma 11 we know that $2 x \equiv 179(\bmod 562)$ has no integer solutions. | no integer solutions | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,902 |
Example 9 Find the integer solution of $256 x \equiv 179(\bmod 337)$. | Since $(256,337)=1$, by Lemma 12 we know that $256 x \equiv$ $179(\bmod 337)$ has integer solutions. Since $337=256+81,256=81 \times$ $3+13,81=13 \times 6+3,13=4 \times 3+1$, we get $1=13$ $4 \times 3=13-4 \times(81-13 \times 6)=25 \times 13-4 \times 81=$ $25 \times(256-81 \times 3)-4 \times 81=25 \times 256-79 \times ... | x \equiv 81(\bmod 337) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,903 |
Example 7 Find the least common multiple of 108, 28, and 42. | Since
$$108=2^{2} \times 3^{3}, 28=2^{2} \times 7, 42=2 \times 3 \times 7$$
we get
$$\{108,28,42\}=2^{2} \times 3^{3} \times 7=756$$ | 756 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,904 |
Example 10 Find the integer solution of $1215 x \equiv 560(\bmod 2755)$. | Solving: Since $(1215,2755)=5,5 \mid 560$, from the original equation we get
$$243 x \equiv 112(\bmod 551)$$
Since $(243,551)=1$, by Lemma 12, we know that $243 x \equiv 112(\bmod 551)$ has integer solutions. Since
$$\begin{array}{l}
551=2 \times 243+65, \quad 243=65 \times 3+48 \\
65=48+17, \quad 48=17 \times 2+14, \... | x \equiv 200,751,1302,1853,2404 \quad(\bmod 2755) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,906 |
Example 11 Find the integer solution of $1296 x \equiv 1125(\bmod 1935)$. | Since $(1296,1935)=9,9 \mid 1125$, from the original equation we get
$$144 x \equiv 125(\bmod 215)$$
Since $(144,215)=1$, by Lemma 12 we know that $144 x \equiv 125(\bmod 215)$ has integer solutions. Since $215=144+71,144=71 \times 2+2$, and $71=2 \times 35+1$, we get
$$\begin{aligned}
1= & 71-2 \times 35=71-(144-71 \... | x \equiv 80,295,510,725,940,1155,1370,1585,1800(\bmod 1935) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,907 |
Theorem 1 If $k \geqslant 2$, and
$$m_{1}, m_{2}, \cdots, m_{k}$$
are $k$ pairwise coprime positive integers, that is, any two of these $k$ positive integers are coprime. Let
$$M=m_{1} m_{2} \cdots m_{k}=m_{1} M_{1}=m_{2} M_{2}=\cdots=m_{k} M_{k}$$
then the positive integer solution that simultaneously satisfies the ... | Because $m_{1}, m_{2}, \cdots, m_{k}$ are pairwise coprime, when $i \neq j$, we have $\left(m_{i}, m_{j}\right)=1$. Since $M_{i}=\frac{M}{m_{i}}$, we get $\left(M_{i}, m_{i}\right)=1$, so
$$\left(M_{1}, m_{1}\right)=\left(M_{2}, m_{2}\right)=\cdots=\left(M_{k}, m_{k}\right)=1$$
From $\left(M_{1}, m_{1}\right)=1$ and L... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,908 |
Example 12: When divided by 7, the remainder is 1; when divided by 8, the remainder is 1; when divided by 9, the remainder is 3. Find the number. (Translation: Find a positive integer $x$ such that when $x$ is divided by 7, the remainder is 1; when $x$ is divided by 8, the remainder is 1; and when $x$ is divided by 9, ... | According to the problem, we have
\( x \equiv 1 \pmod{7}, x \equiv 1 \pmod{8}, x \equiv 3 \pmod{9} \).
In Theorem 1, let \( m_1 = 7, m_2 = 8, m_3 = 9, b_1 = b_2 = 1, b_3 = 3 \); at this point, we have
\[
\begin{array}{l}
M = 7 \times 8 \times 9 = 504 \\
M_1 = \frac{504}{7} = 72 \\
M_2 = \frac{504}{8} = 63 \\
M_3 = \fra... | x = 57 + 504k, \quad k = 0, 1, 2, \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,909 |
Example 13: When divided by 11, the remainder is 3; when divided by 12, the remainder is 2; when divided by 13, the remainder is 1. What is the original number? (Translation: Find a positive integer $x$ such that when divided by 11, the remainder is 3; when divided by 12, the remainder is 2; when divided by 13, the rem... | According to the problem, we have
$$x \equiv 3(\bmod 11), \quad x \equiv 2(\bmod 12), \quad x \equiv 1(\bmod 13)$$
In Theorem 1, let $m_{1}=11, m_{2}=12, m_{3}=13, b_{1}=3, \dot{b}_{2}=2$, $b_{3}=1$, then we have
$$\begin{array}{l}
M=11 \times 12 \times 13=1716 \\
M_{1}=\frac{1716}{11}=156 \\
M_{2}=\frac{1716}{12}=143... | x=14+1716k, \quad k=0,1,2, \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,910 |
Find a positive integer $x$, when divided by 2 the remainder is 1, when divided by 5 the remainder is 2, when divided by 7 the remainder is 3, when divided by 9 the remainder is 4.
Ask for the original number. (Trans: Find a positive integer $x$, when divided by 2 the remainder is 1, when divided by 5 the remainder is... | Solve According to the problem, we have
$$\begin{array}{lll}
x \equiv 1 & (\bmod 2), & x \equiv 2 \quad(\bmod 5) \\
x \equiv 3 & (\bmod 7), & x \equiv 4 \quad(\bmod 9)
\end{array}$$
In Theorem 1, take $m_{1}=2, m_{2}=5, m_{3}=7, m_{4}=9, b_{1}=1, b_{2}$ ? $=2, b_{3}=3, b_{4}=4$, at this point we have
$$\begin{array}{l... | x=157+630k, \quad k=0,1,2, \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,911 |
Example 16 Han Xin Counts the Soldiers: There is a troop of soldiers. If they form a column of five, there is one person left over at the end. If they form a column of six, there are five people left over at the end. If they form a column of seven, there are four people left over at the end. If they form a column of el... | Let $x$ be the number of soldiers we are looking for. According to the problem,
In Theorem 1, take $m_{1}=5, m_{2}=6, m_{3}=7, m_{4}=11, b_{1}=1$, $b_{2}=5, b_{3}=4, b_{4}=10$. Then we have
$$\begin{array}{l}
M=5 \times 6 \times 7 \times 11=2310 \\
M_{1}=\frac{2310}{5}=462 \\
M_{2}=\frac{2310}{6}=385 \\
M_{3}=\frac{23... | 2111 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,912 |
1. In 1978, "August 1st" was a Tuesday, August 1978 had 31 days, and September 1978 had 30 days. What day of the week was National Day in 1978? | 1. Solution: From "August 1st" to National Day there are 61 days, since $61 \equiv 5(\bmod 7)$, and "August 1st" is a Tuesday, therefore National Day is a Sunday. | Sunday | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 740,913 |
2. Use the casting out nines method to check whether the following calculations are correct:
(i) $4568 \times 7391=30746529$.
(ii) $2368 \times 846=2003328$.
(iii) $16 \times 937 \times 1559=23373528$,
(iv) $17^{4}=83521$.
(v) $23372428 \div 6236=3748$. | 2.
(i) Solution: Since
$$\begin{array}{l}
4568 \equiv 4+5+6+8 \equiv 5(\bmod 9) \\
7391 \equiv 7+3+1 \equiv 2(\bmod 9) \\
30746529 \equiv 3+7+4+6+5+2 \equiv 0(\bmod 9)
\end{array}$$
and
$$2 \times 5 \neq 0 \quad(\bmod 9)$$
the original calculation is incorrect.
(ii) Solution: Since
$$\begin{array}{l}
2368 \equiv 2+3+... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,914 |
Example 8 Find the least common multiple of 198,240 and 360.
保留源文本的换行和格式，直接输出翻译结果。
Example 8 Find the least common multiple of 198,240 and 360. | Since
$$198=2 \times 3^{2} \times 11, 240=2^{4} \times 3 \times 5, 360=2^{3} \times 3^{2} \times 5,$$
we get
$$\{198,240,360\}=2^{4} \times 3^{2} \times 5 \times 11=7920$$ | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,915 |
4. Solve the following congruences:
(i) $258 x \equiv 131 \pmod{348}$.
(ii) $3 x \equiv 10 \pmod{29}$.
(iii) $47 x \equiv 89 \pmod{111}$.
(iv) $660 x \equiv 595 \pmod{1385}$. | 4.
(i) Solution: Since
$$(258,348)=6,$$
and $6 \nmid 131$, so by Lemma 11, the congruence has no solution.
(ii) Solution: Since
$$29=9 \times 3+2, \quad 3=2+1$$
Therefore,
$$1=3-2=3-(29-9 \times 3)=10 \times 3-29$$
That is,
$$3 \times 10 \equiv 1(\bmod 29)$$
From the original equation,
$$3 \times 10 x \equiv 10 \ti... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,917 |
5. Solve the following systems of congruences:
(i) $\left\{\begin{array}{l}x \equiv 3 \quad(\bmod 7) \\ x \equiv 5 \quad(\bmod 11)\end{array}\right.$
(ii) $\left\{\begin{array}{ll}x \equiv 2 & (\bmod 11), \\ x \equiv 5 & (\bmod 7), \\ x \equiv 4 & (\bmod 5) .\end{array}\right.$
(iii) $\left\{\begin{array}{ll}x \equiv 1... | 5.
(i) Solution: By the Chinese Remainder Theorem,
Given
Given
$$\begin{array}{c}
b_{1}=3, \quad b_{2}=5, \quad m_{1}=7, \quad m_{2}=11, \\
m=m_{1} \cdot m_{2}=7 \times 11=77, \\
M_{1}=\frac{77}{7}=11, \quad M_{2}=\frac{77}{11}=7 . \\
11 M_{1}^{\prime}=1 \quad(\bmod 7), \text { so } M_{1}^{\prime}=2 . \\
7 M_{2}^{\prim... | 299 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,918 |
6. Solve the following problems: (Yang Hui: "Xu Gu Zhai Qi Suan Fa" (1275))
(i) When divided by 7, the remainder is 1; when divided by 8, the remainder is 2; when divided by 9, the remainder is 4. What is the original number?
(ii) When divided by 2, the remainder is 1; when divided by 5, the remainder is 2; when divide... | 6.
(i) Solution: Let the number be $x$, then according to the problem, we have
Here
By
By
By
$$\left.\begin{array}{c}
\left\{\begin{array}{l}
x \equiv 1 \quad(\bmod 7), \\
x \equiv 2 \quad(\bmod 8), \\
x \equiv 4 \quad(\bmod 9) .
\end{array}\right. \\
b_{1}=1, \quad b_{2}=2, \quad b_{3}=4, \\
m_{1}=7, \quad m_{2}=8, \... | 274 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,919 |
7. Prove: The system of congruences
$$\left\{\begin{array}{ll}
x \equiv a & \left(\bmod m_{1}\right) \\
x \equiv a & \left(\bmod m_{2}\right)
\end{array}\right.$$
has all solutions
$$x \equiv a\left(\bmod \left\{m_{1}, m_{2}\right\}\right)$$ | 7. Proof: From $x \equiv a\left(\bmod \left\{m_{1}, m_{2}\right\}\right)$, we know
$$\left\{m_{1}, m_{2}\right\} \mid(x-a)$$
Since $m_{1}\left|\left\{m_{1}, m_{2}\right\}, m_{2}\right|\left\{m_{1}, m_{2}\right\}$, by Lemma 2 of Chapter 1, we have
$$\begin{array}{c}
m_{1}\left|(x-a), \quad m_{2}\right|(x-a), \\
x \equi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,920 |
8. Prove:
(i) If $\left(m_{1}, m_{2}\right)=d$, then when $d \mid\left(b_{1}-b_{2}\right)$, the system of congruences
$$\left\{\begin{array}{ll}
x \equiv b_{1} & \left(\bmod m_{1}\right) \\
x \equiv b_{2} & \left(\bmod m_{2}\right)
\end{array}\right.$$
must have a solution, and all its solutions are
$$x \equiv x_{0} \... | 8 .
(i) Proof: Let $\left\{\begin{array}{l}x \equiv b_{1}\left(\bmod m_{1}\right) \\ x \equiv b_{2}\left(\bmod m_{2}\right)\end{array}\right.$
From equation (4), we know
$$\begin{array}{l}
x=b_{1}+m_{1} t_{1} \\
x=b_{2}+m_{2} t_{2}
\end{array}$$
Here, $t_{1}, t_{2}$ are integers. Equation (4) has a solution if there ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,921 |
9. Try to solve:
(i) $\left\{\begin{array}{l}x \equiv 2 \quad(\bmod 7), \\ x \equiv 5 \quad(\bmod 9), \\ x \equiv 11 \quad(\bmod 15) .\end{array}\right.$
(ii) There is a number whose total is unknown; when reduced by multiples of five, there is no remainder; when reduced by multiples of seven hundred and fifteen, there... | 9.
(i) Solution: Since $(7,9)=1,(7,15)=1,(9,15)=3,11-5=6$, and $3 \mid(11-5)$, by problem 8(i, ii), we know the system of congruences has a solution and is equivalent to
$$\left\{\begin{array}{l}
x \equiv 2 \quad(\bmod 7) \\
x \equiv 5 \quad(\bmod 9) \\
x \equiv 11 \equiv 1 \quad(\bmod 5)
\end{array}\right.$$
Solving ... | 10020 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,922 |
10. The distance between Port A and Port B does not exceed 5000 kilometers. Today, three ships depart from Port A to Port B at midnight simultaneously. Assuming the three ships sail at a constant speed for 24 hours a day, the first ship arrives at midnight several days later, the second ship arrives at 18:00 several da... | 10. Solution: Let the distance between ports A and B be $x$ kilometers. The distance the second ship travels in 18 hours is $240 \times \frac{18}{24}=180$ kilometers, and the distance the third ship travels in 8 hours is $180 \times \frac{8}{24}=60$ kilometers. According to the problem, we have
$$\left\{\begin{array}{l... | 3300 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,923 |
Lemma 9 Assume that $a$ and $b$ are both positive integers, and the least common multiple of $a$ and $b$ is $m$, i.e., $\{a, b\}=m$. If $m^{\prime}$ is a common multiple of $a$ and $b$, then
$$m \mid m^{\prime} \text {. }$$ | Proof: Since $m^{\prime}$ is a common multiple of $a$ and $b$, and $m$ is the least common multiple of $a$ and $b$, we have $1 \leqslant m \leqslant m^{\prime}$. By Lemma 4, we have
$$m^{\prime}=m q+r$$
where $a$ and $r$ are non-negative integers, and $0 \leqslant r<m$. Since $m$ is the least common multiple of $a$ an... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,924 |
Example 9 Find the least common multiple of 24871 and 3468.
The text above is translated into English, preserving the original text's line breaks and format. | Since
thus $(24871,3468)=17$. By Lemma 10 we have
$$\{24871,3468\}=\frac{24871 \times 3468}{17}=5073684 .$$ | 5073684 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,926 |
Example 10 Find the least common multiple of 513, 135, and 3114. | Since
$1\left|\begin{array}{r|r}513 & 135 \\ 405 & 108 \\ \hline 108 & 27 \\ 108 & \\ \hline 0 & 27\end{array}\right|^{3}$
Therefore, $(513,135)=27$, by Lemma 10 we have
$$\{513,135\}=\frac{513 \times 135}{27}=2565 .$$
Since
\begin{tabular}{|c|c|c|}
\hline 1 & 2565 & \multirow{2}{*}{\begin{tabular}{l}
3114 \\
2565
\e... | 887490 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,927 |
Example 11 Find the least common multiple of $8127, 11352, 21672$ and 27090. | Since
\begin{tabular}{|c|c|c|}
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
8127 \\
6450
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{r}
11352 \\
8127
\end{tabular}
\end{tabular} \\
\hline 1 & 1677 & 3225 \\
\hline & 1548 & 1677 \\
\hline 12 & 129 & 1548 \\
\hline & & 1548 \\
\hline & 129 & 0 \\... | 3575880 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,929 |
Example 12 A steel plate, 1丈 3尺 5寸 long and 1丈 5寸 wide. Now it is to be cut into equally sized squares, the squares must be as large as possible, and no steel plate should be left over. Find the side length of the square.
Note: In traditional Chinese units, 1丈 = 10尺, and 1尺 = 10寸. | Solution: Since we want the largest square, we need to find the largest side length of the square. To find the largest side length of the square, we need to find the greatest common divisor (GCD) of 135 inches and 105 inches. Since
$$135=3^{3} \times 5, \quad 105=3 \times 5 \times 7$$
we have $(135,105)=15$.
Answer: T... | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,930 |
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