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Example 13 There are three steel wires, one is 13 feet 5 inches long, one is 24 feet 3 inches long, and one is 55 feet 8 inches long. Now they need to be cut into equal small segments, with no remainder from any of the wires, and the segments should be as long as possible. What is the length of each small segment in in... | Since
$$135=3^{3} \times 5, \quad 243=3^{5}, \quad 558=2 \times 3^{2} \times 31,$$
thus $(135,243,558)=9$. Also,
$$\frac{135}{9}+\frac{243}{9}+\frac{558}{9}=15+27+62=104$$
Answer: The length of each small segment is 9 inches, and a total of 104 segments can be cut. | 9 \text{ inches, 104 segments} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,931 |
Example 14 Two gears, A and B, are meshed together. Gear A has 437 teeth, and gear B has 323 teeth. After a certain tooth on A and a certain tooth on B come into contact, how many minimum revolutions will each have to make before they contact each other again?
保留源文本的换行和格式，直接输出翻译结果。 | To find the minimum number of weeks each wheel must turn, we first need to determine the number of teeth each wheel (A and B) must pass. To do this, we need to find the least common multiple (LCM) of the number of teeth on wheel A (437) and wheel B (323). Since
\begin{tabular}{|c|c|c|}
\hline & \multicolumn{2}{|r|}{323... | 17 \text{ weeks for A, 23 weeks for B} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,932 |
Example 15 There are three workers transporting bricks from the brick pile to the scaffolding for wall construction. It takes Worker A 15.6 minutes, Worker B 16.8 minutes, and Worker C 18.2 minutes for a round trip. Now, all three start from the brick pile at the same time. What is the minimum number of minutes it will... | To find the minimum number of minutes required for the three people to return to the brick pile simultaneously, we need to find the least common multiple (LCM) of $15.6, 16.8$, and $18.2$. Since
$$\left.\begin{array}{c|c|c}
15.6 & 16.8 \\
15.6 & 15.6 \\
\hline 0 & 1.2
\end{array} \right\rvert\,$$
Thus, $(15.6, 16.8) =... | 218.4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,933 |
Lemma 11 Every integer $a$ greater than 1 can be decomposed into a product of prime factors, that is,
$$a=p_{i} \cdots p_{n} \quad n \geqslant 1$$
Here $p_{1}, \cdots, p_{n}$ are all prime numbers, some of which may be the same, for example $12=2 \times$
$$2 \times 3,18=2 \times 3 \times 3$$ | Prove that when $a$ is a prime number $p$, that is, $a=p$, then no further factorization is needed. If $a$ is a composite number, then by Lemma 5, its smallest factor greater than 1 is a prime. Let this prime be $p_{1}$. Since $a$ is a composite number and $p_{1}$ is a factor of $a$, we have $a=p_{1} a_{1}$, where $a_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,934 |
Lemma 12 If $p$ is a prime, then $p \nmid a$ implies $(p, a)=1$, and when $(p, a)=1$ it implies $p \nmid a$. | Since $p$ is a prime number, $p$ has only two positive divisors, which are 1 and $p$. If $p \nmid a$, then only $(p, a)=1$. Conversely, if $(p, a)=1$, then $p$ is not a common divisor of $p$ and $a$, so $p \nmid a$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,935 |
Lemma 13 If $a, b, c$ are all positive integers, then from $(a, b)=1$, $|a| b c$ we can get $a \mid c$. This means: when $\boldsymbol{a}$ and $b$ are coprime, but $a$ can divide $b c$, then it must be that $\boldsymbol{a}$ can divide $\boldsymbol{c}$. | Proof: Since $b \mid bc$ and $a \mid bc$, $bc$ is a common multiple of $a$ and $b$. Since $(a, b)=1$ and by Lemma 10, the least common multiple of $a$ and $b$ is $ab$. By Lemma 9, $ab \mid bc$, i.e., $\frac{bc}{ab}=\frac{c}{a}$ is an integer, so $a \mid c$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,936 |
Lemma 14 If $n \geqslant 2$ is an integer, and $a_{1}, a_{2}, \cdots, a_{n}$ and $a$ are all positive integers. When $a \mid a_{1} a_{2} \cdots a_{n}$ and
$$\left(a, a_{1}\right)=\left(a, a_{2}\right)=\cdots=\left(a, a_{n-1}\right)=1$$
then, it must be that $a \mid a_{n}$. | Given that $\left(a, a_{1}\right)=1, a \mid a_{1} a_{2} \cdots a_{n}$ and Lemma 13, we have $a \mid a_{2} \cdots a_{n}$. Therefore, when $n=2$, this lemma holds. If $n \geqslant 3$, then by $\left(a, a_{2}\right)=1, a \mid a_{2} a_{3} \cdots a_{n}$ and Lemma 13, we have $a \mid a_{3} \cdots a_{n}$, so when $n=3$, this ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,937 |
Lemma 15 If $a, b, c$ are all positive integers, and $(a, b)=1, c \mid a$, then we have
$$(b, c)=1$$ | Proof: If $(b, c)=d$, and $d>1$, then $d|b, d| c$. From $d \mid c$ and $c \mid a$ we have $d \mid a$. Since $d|a, d| b$, $d$ is a common divisor of $a$ and $b$, but $d>1$ and the greatest common divisor of $a$ and $b$ is 1, which contradicts Definition 5, so $(b, c)=1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,938 |
Lemma 4 If $a, b$ are two integers, $b \neq 0$, then there exist and are unique two integers $q, r$ such that
$$a=b q+r, \quad 0 \leqslant r<|b|$$
holds. | If $b>0$, then the multiples of $b$ when listed from negative to positive, in ascending order, are
$$\cdots,-4 b,-3 b,-2 b,-b, 0, b, 2 b, 3 b, 4 b, \cdots$$
If $b>0$, there exists an integer $q$ such that $q b \leqslant a<(q+1) b$. And when $b<0$, there exists an integer $q$ such that $q b \leqslant a<(q-1) b$. Theref... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,939 |
Lemma 16 If $a$ and $b$ are positive integers, and $(a, b)=1$, then $(a, bc)=(a, c)$ | Let $(a, c)=d_{1}$, and $(a, b c)=d_{2}$, then we have $d_{1}\left|a, d_{1}\right| c$, $d_{2}\left|a, d_{2}\right| b c$. From $d_{1} \mid c$ we get $d_{1} \mid b c$. From $d_{1}\left|a, a_{1}\right| b c$ we get that $d_{1}$ is also a common divisor of $a$ and $b c$, but $d_{2}$ is the greatest common divisor of $a$ and... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,940 |
Lemma 17 If $n \geqslant 2$ is an integer, and $b_{1}, b_{2}, \cdots, b_{n}$ and $a$ are all positive integers, when $\left(a, b_{1}\right)=\left(a, b_{2}\right)=\cdots=\left(a, b_{a}\right)=1$ then we have
$$\left(a, b_{1} b_{2} \cdots b_{n}\right)=1$$ | By $\left(a, b_{1}\right)=1$ and Lemma 16, we get
$$\left(a, b_{1} b_{2} \cdots b_{n}\right)=\left(a, b_{2} \cdots b_{n}\right)$$
Again, by $\left(a, b_{2}\right)=1$ and Lemma 16, we get $\left(a, b_{2} \cdots b_{n}\right)=\left(a, b_{3} \cdots b_{n}\right)$, so
$$\begin{aligned}
\left(a, b_{1} b_{2} \cdots b_{n}\righ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,941 |
Lemma 18 If $n \geqslant 2$ is an integer, $a_{1}, a_{2}, \cdots, a_{n}$ are all positive integers, and $p$ is a prime number, when $p \mid a_{1} a_{2} \cdots a_{n}$, then there exists at least one $a_{r}$ that is divisible by $p$, that is, $p \mid a_{r}$ | Assume that $p$ does not divide any $a_{i}(i=1,2, \cdots, n)$. From the fact that $p$ is a prime and Lemma 12, we get
$$\left(p, a_{1}\right)=\left(p, a_{2}\right)=\cdots=\left(p, a_{n}\right)=1$$
Therefore, by Lemma 17, we have $\left(p, a_{1} a_{2} \cdots a_{n}\right)=1$. From $\left(p, a_{1} a_{2} \cdots a_{n}\righ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,942 |
Lemma 19 If $n \geqslant 2$ is an integer, and $p_{1}, p_{2}, \cdots, p_{n}$ and $p$ are all prime numbers, when $p \mid p_{1} p_{2} \cdots p_{n}$, then there must be at least one $p_{r}$ where $r$ is one of the numbers $1,2, \cdots, n$, such that $p=p_{r}$. | Proof: From $p \mid p_{1} p_{2} \cdots p_{n}$ and Lemma 18, we know that there must be at least one prime $p_{r}$ such that $p \mid p_{r}$. Since $p_{r}$ is a prime, it has only two positive divisors, which are 1 and $p_{r}$. Given $p \neq 1$ and $p \mid p_{r}$, it follows that $p = p_{r}$.
If a positive integer is wr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,943 |
Theorem 1 (Fundamental Theorem of Arithmetic) If the order of the prime factors is disregarded, there is only one way to express a positive integer \(a > 1\) as a product of prime factors. | If $a$ is a prime number $p$, that is, $a=p$, this theorem holds. If $a$ is a composite number, then by Lemma 11, $a$ can be decomposed into the product of prime factors. Let
$$a=p_{1} p_{2} \cdots p_{n}, \quad n \geqslant 2$$
where $p_{1}, p_{2}, \cdots, p_{n}$ are all prime numbers. Suppose $a$ can be decomposed int... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,944 |
Example 16 Find the standard factorization of 117. | $$3 \frac{117}{3 \frac{39}{13}}$$
So $117=3^{2} \times 13$.
| 117=3^{2} \times 13 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,945 |
Example 17 Find the standard factorization of 9828.
Find the standard factorization of 9828. | $$\begin{array}{r|r}
2 & 9828 \\
3 & 2457 \\
\hline 3^{2} & 819 \\
\hline 7 \mid 91 \\
\hline & 13
\end{array}$$
So $9828=2^{2} \times 3^{3} \times 7 \times 13$. | 9828=2^{2} \times 3^{3} \times 7 \times 13 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,946 |
Example 18 Find the standard factorization of 10725. | $$\begin{array}{r|r}
3 \lcm{10725} \\
5^{2} & 3575 \\
11 \frac{143}{13}
\end{array}$$
So $10725=3 \times 5^{2} \times 11 \times 13$. | 10725=3 \times 5^{2} \times 11 \times 13 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,947 |
1. Prove that when the unit digit of any integer $a$ is divisible by 2, then this integer is a multiple of 2. | 1. Proof: Any integer $a$ can be written as
$$a=10 n+b$$
where $n$ is an integer and $0 \leqslant b<10$. Since $2 \mid 10$, (1) and the assumption $2 \mid b$, it follows that $2 \mid a$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,948 |
Lemma 5 If $a$ is an integer greater than 1, then the smallest divisor of $a$ that is greater than 1 must be a prime number. | If $a$ is a prime number, then the only divisor of $a$ greater than 1 is $a$ itself, so the smallest divisor of $a$ greater than 1 is the prime number $a$.
If $a$ is a composite number, then $a$ must have other positive divisors besides 1 and $a$. Suppose $b$ is the smallest of these positive divisors, and we will pro... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,950 |
3. Prove that the square of any odd number $a$ minus 1 is a multiple of 8. | 3. Prove: Any odd number $a$ can be written in the form
$$a=4 n+b$$
where $n$ is an integer and $b$ is an odd number satisfying $1 \leqslant b \leqslant 3$. Also,
$$b^{2}-1=\left\{\begin{array}{ll}
0, & \text { when } b=1 . \\
8, & \text { when } b=3 .
\end{array}\right.$$
From (2), we can get $a^{2}-1=(4 n+b)^{2}-1$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,951 |
4. Prove that the product of any four consecutive integers plus 1 is definitely a square number. | 4. Proof: We use $a, a+1, a+2, a+3$ to represent four consecutive integers, where $a$ is an integer. Then
$$a(a+1)(a+2)(a+3)=\left(a^{2}+3 a\right)\left(a^{2}+3 a+2\right)=\left[\left(a^{2}+3 a+1\right)-1\right]\left[\left(a^{2}+3 a+1\right)+1\right]=\left(a^{2}+3 a+1\right)^{2}-1$$.
Therefore,
$$a(a+1)(a+2)(a+3)+1=\l... | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,952 |
5. Prove that when $a$ is an integer, $a(a-1)(2a-1)$ is a multiple of 6.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly. | 5. Proof: $a$ and $a-1$ are two consecutive integers, and among two consecutive integers, one must be even, so
$$2 \mid a(a-1)(2a-1)$$
Any integer $a$ can be written in the form
$$a=3n+b$$
where $n$ and $b$ are integers, and $1 \leqslant b \leqslant 3$.
When $b=1$, $a-1=3n$, so $3 \mid (a-1)$;
When $b=2$, $2a-1=6n+3$... | null | Algebra | proof | Yes | Yes | number_theory | false | 740,953 |
6. Prove that when $a$ is an odd number, $a\left(a^{2}-1\right)$ is a multiple of 24. | 6. Proof: Since $a\left(a^{2}-1\right)=(a-1) \cdot a \cdot(a+1)$ is the product of three consecutive integers, and among three consecutive integers, there is always one that is a multiple of 3, the product of three consecutive integers is a multiple of 3, hence
$$3 \mid a\left(a^{2}-1\right)$$
From problem 3, we know ... | 24 \mid a\left(a^{2}-1\right) | Number Theory | proof | Yes | Yes | number_theory | false | 740,954 |
7. Prove that if an integer $a$ is not divisible by 2 and 3, then $a^{2}+23$ must be divisible by 24. | 7. Proof: Any integer $a$ can be written in the form
$$a=12 n+b$$
where $n$ is an integer, and $1 \leqslant b \leqslant 12$. If $2 \nmid a, 3 \nmid a$, then since $2|12 n, 3| 12 n$, it follows that $2 \nmid b, 3 \nmid b$. Therefore, $b$ can only take the values 1, 5, 7, 11. When $b=1,5,7$, and 11, $a^{2}+23$ are respe... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,955 |
8. Find the greatest common divisor using the method of prime factorization:
(i) $48,84,120$.
(ii) $360,810,1260,3150$. | 8.
(i) Solution: Decomposing each number into prime factors, we get
$$\begin{array}{l}
48=2 \times 2 \times 2 \times 2 \times 3=2^{4} \times 3 \\
84=2 \times 2 \times 3 \times 7=2^{2} \times 3 \times 7 \\
120=2 \times 2 \times 2 \times 3 \times 5=2^{3} \times 3 \times 5
\end{array}$$
Therefore, $(48,84,120)=2^{2} \tim... | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,956 |
9. Use the Euclidean algorithm to find the greatest common divisor:
(i) 51425, 13310.
(ii) $353430, 530145, 165186$.
(iii) $81719, 52003, 33649, 30107$. | 9
(i) Solution: Since
1 \begin{tabular}{r|r|r}
51425 & 13310 \\
39930 & 11495 \\
\hline 11495 & 1815 & 6 \\
10890 & 1815 \\
\hline 605 & 0
\end{tabular}
Therefore, $(51425,13310)=605$.
(ii) Solution: Since
We get
$$(353430,530145)=176715$$
Also,
$14 \left\lvert\,$\begin{tabular}{r|r|}
176715 & 165186 \\
165186 & 161... | 23 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,957 |
11. If $a, b, n$ are positive integers, prove:
(i) $\left(a^{n}, b^{n}\right)=(a, b)^{n}$.
(ii) $(n a, n b)=n(a, b)$. | 11.
(i) Proof: Suppose $(a, b)=d$, then it must be that $\left(\frac{a}{d}, \frac{b}{d}\right)=1$, hence $\left(\left(\frac{a}{d}\right)^{n},\left(\frac{b}{d}\right)^{n}\right)=1$. That is, $\left(\frac{a^{n}}{d^{n}}, \frac{b^{n}}{d^{n}}\right)=1$, so $\left(a^{n}, b^{n}\right)=d^{n}$.
Since we assume $(a, b)=d$, we o... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,959 |
12. Use the properties of the greatest common divisor from the previous question to find the greatest common divisor of the following:
(i) $216,64,1000$.
(ii) $24000,36000,144000$. | 12.
(i) Solution: The property of the greatest common divisor (GCD) from the previous problem can obviously be extended to more than two numbers, so
$$\begin{array}{c}
(216,64,1000)=\left(6^{3}, 4^{3}, 10^{3}\right) \\
=(6,4,10)^{3}=2^{3}=8
\end{array}$$
(ii) Solution:
$$\begin{array}{l}
(24000,36000,144000) \\
=1000 \... | 12000 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,960 |
Lemma 6 If $a$ is an integer greater than 1, and all primes $\leqslant \sqrt{a}$ do not divide $a$, then $a$ is a prime. | First, prove that if $a$ is not divisible by any integer $>1$ and $\leqslant \sqrt{a}$, then $a$ is a prime number. Assume $a$ is a composite number and $a=b c$, where $b$ and $c$ are both integers greater than 1. Since $a$ is not divisible by any integer $>1$ and $\leqslant \sqrt{a}$, it follows that $b>\sqrt{a}$ and ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,961 |
14. There is a rectangular room that is 5.25 meters long and 3.25 meters wide. Now, square tiles are used to cover the floor, and it is required to exactly cover the entire room. What is the maximum side length of the square tiles used? | 14. Solution: To pave the entire room with square tiles, the maximum side length of the tiles should be the greatest common divisor (GCD) of the room's length and width. To eliminate decimals, we use centimeters as the unit of length, with the room's length being 525 cm and the width 325 cm. White text
$$\begin{array}{... | 25 | Geometry | math-word-problem | Yes | Yes | number_theory | false | 740,963 |
15. A rectangular piece of wood is 3 feet 5 inches 7 points long, 1 foot and 5 points wide, and 8 inches 4 points thick. It needs to be cut into equally sized cubic blocks, with the volume of each block being as large as possible. What is the edge length of the block? | 15. Solution: The side length of the wooden block is clearly the greatest common divisor of the length, width, and thickness of the wood. We use fractions of an inch as the unit of length, since
$$\begin{aligned}
357 & =3 \times 7 \times 17 \\
105 & =3 \times 5 \times 7 \\
84 & =2^{2} \times 3 \times 7
\end{aligned}$$
... | 2 \text{ inches and } 1 \text{ fraction of an inch} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,964 |
16. Class A, B, and C have 54 students, 48 students, and 72 students respectively. Now, it is planned to organize physical exercise groups within each class, but the number of students in each group must be the same. What is the maximum number of students in each exercise group? How many groups will there be in Class A... | 16. Solution: Since students from each class must be organized into exercise groups, and the number of students in each group is the same, the number of students in each group must be a common factor of the number of students in the three classes. Therefore, finding the maximum number of students in a group is equivale... | 6, 9, 8, 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,965 |
17. A box of grenades, assuming the weight of each grenade is an integer greater than one pound, the net weight after removing the weight of the box is 201 pounds. After taking out several grenades, the net weight is 183 pounds. Prove that the weight of each grenade is 3 pounds. | 17. Proof: Since 201 jin and 183 jin are both the weights of an integer number of grenades, the weight of each grenade must be a common divisor of them. Their greatest common divisor is
$$(201,183)=3 \times(67,61)=3,$$
Since 3 is a prime number, and the divisors of 3 are only 1 and 3, the weight of each grenade must b... | 3 | Number Theory | proof | Yes | Yes | number_theory | false | 740,966 |
18. Venus and Earth are at a certain position relative to the Sun at a certain moment. It is known that Venus orbits the Sun in 225 days, and Earth orbits the Sun in 365 days. How many days at least will it take for both planets to return to their original positions simultaneously? | 18. Solution: The time required for both planets to return to their original positions must be a common multiple of the time each planet takes to orbit the sun once. To find the least amount of time for them to return to their original positions, we need to find the least common multiple (LCM) of their orbital periods.... | 16425 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,967 |
19. Design a packaging box with a square base to transport four different sizes of chess sets, where the base of each chess box is also square, with side lengths of 21 cm, 12 cm, 14 cm, and 10.5 cm, respectively. To ensure that the packaging box can completely cover the base regardless of which size of chess set it is ... | 19. Solution: To ensure that each type of chess box can completely cover the bottom of the packaging box, the side length of the bottom of the packaging box should be a common multiple of the side lengths of the bottom of each chess box. Therefore, the smallest side length of the box bottom is the least common multiple... | 84 | Geometry | math-word-problem | Yes | Yes | number_theory | false | 740,968 |
20. In the performance of a group gymnastics, it is required that when the formation changes to 10 rows, 15 rows, 18 rows, and 24 rows, the formation can always form a rectangle. How many people are needed at minimum for the group gymnastics performance? | 20. Solution: Since the formation needs to be a rectangle, the number of people must be a multiple of the number of rows. Finding the minimum number of people is essentially finding the least common multiple (LCM) of the row numbers. Using the method of prime factorization, we get
$$\begin{array}{ll}
10=2 \times 5, & 1... | 360 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,969 |
21. There are four gears, A, B, C, and D, with the number of teeth being 84, 36, 60, and 48, respectively. How many revolutions will each gear make before the teeth that are currently engaged mesh together again? | 21. Solution: When the teeth that are engaged simultaneously come back to engage again, each gear has turned an integer number of revolutions. Therefore, the number of teeth turned is the least common multiple of the number of teeth on each gear. Since
$$\begin{array}{l}
84=2^{2} \times 3 \times 7 \\
36=2^{2} \times 3^... | 60, 140, 84, 105 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,970 |
22. Find the standard factorization of the following numbers:
(i) 16500.
(ii) 1452990. | 22.
(i) Solution:
\begin{tabular}{|c|c|c|}
\hline $2 \mid 1$ & \multicolumn{2}{|l|}{16500} \\
\hline 2 & & 250 \\
\hline 3 & 41 & \\
\hline 5 & & \\
\hline 5 & 52 & 275 \\
\hline & 5 & 55 \\
\hline & & 11 \\
\hline
\end{tabular}
So $16500=2^{2} \times 3 \times 5^{3} \times 11$.
(ii) Solution:
So $1452990=2 \times 3 \... | 1452990=2 \times 3 \times 5 \times 7 \times 11 \times 17 \times 37 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,971 |
Lemma 7 There are infinitely many prime numbers | Assume the number of prime numbers is finite, with a total of $n$ primes, which are $p_{1}, p_{2}$, $p_{3}, \cdots, p_{n}$. Among them, $p_{1}=2, p_{2}=3, p_{3}=5, \cdots$. Let $a=p_{1} \cdots p_{n}+1$. If $a$ is a prime number, then because $a$ is not equal to any of $p_{1}, p_{2}, \cdots, p_{n}$, the number of prime ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,972 |
23. Suppose $n$ and $A$ are positive integers, and $\sqrt[n]{A}$ is not an integer. Prove that $\sqrt[n]{A}$ must not be a rational fraction. (A rational fraction is a fraction that is not an integer, i.e., the numerator cannot be divided by the denominator.) | 23. Proof: We use proof by contradiction, assuming that $\sqrt[n]{A}$ is a rational fraction, that is,
$$\sqrt[n]{A}=\frac{p}{q}, \quad q>1, \text { and }(p, q)=1 .$$
This leads to a contradiction. Raising both sides to the $n$-th power, we get
$$A=\frac{p^{n}}{q^{n}}$$
Since $(p, q)=1$, it follows that $\left(p^{n},... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,973 |
24. Suppose the coefficients $a_{1}, a_{2}, \cdots, a_{n}$ of the $n$-th degree algebraic equation
$$x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}=0$$
are all integers. If it has a rational root, prove that this root must be an integer. | 24. Proof: Suppose $\frac{p}{q}$ is a root of the equation, $|p|$ and $q$ are positive integers, and $(|p|, q)=1$. According to the problem, we only need to prove that $q=1$.
Since $\frac{p}{q}$ is a root of the equation, it satisfies the equation. Substituting it into the equation, we get
$$\frac{p^{n}}{q^{n}}+a_{1} ... | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,974 |
25. Prove: When $n$ goes through all natural numbers, numbers of the form $4 n-1$ contain infinitely many primes. | 25. Let: We will prove this problem step by step.
First step proof: Numbers of the form \(4n-1\) must have a prime factor of the form \(4n-1\). This is because odd primes can be written in the form \(4n-1\) or \(4n+1\), where \(n\) is an integer. And since
\[
\begin{array}{l}
\left(4n_{1}+1\right)\left(4n_{2}+1\right)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,975 |
27. Find the greatest common divisor of the following:
(i) $435785667,131901878$.
(ii) $15959989,7738$. | 27.
(i) Solution: Since
$3\left|\begin{array}{r|r|r}435785667 \\ 395705634 & 131901878 \\ \hline 40080033 & 120240099 \\ 34985337 & 101861779 \\ \hdashline & 5094696 & 1472387 \\ 4417161 & 1355070\end{array}\right| 3$
\begin{tabular}{|c|c|c|c|}
\hline 2 & \begin{tabular}{l}
\begin{tabular}{l}
677535 \\
586585
\end{tabu... | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,977 |
28. Find the standard factorization of the following numbers:
(i) 174530187.
(ii) 710352035484.
(iii) 40528613317500. | 28. (i) Solution: Since
$3 \boxed{174530187}$
$3 \boxed{58176729}$
$3 \boxed{13392243}$
$13 \boxed{6464081}$
$23 \boxed{38249}$
13663
Therefore, $174530187=3^{3} \times 13^{2} \times 23 \times 1663$.
(ii) Solution: Since
Therefore,
$$\begin{array}{c}
710352035484=2^{2} \times 3 \times 7^{3} \times 13 \times 17 \times... | 710352035484=2^{2} \times 3 \times 7^{3} \times 13 \times 17 \times 19 \times 23 \times 1787 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,978 |
29. Please prove that $F_{5}=2^{32}+1$ is not a prime number. | 29. Proof: We have $F_{5}=2^{32}+1=2^{4} \times\left(2^{7}\right)^{4}+1=(1+$
$$\begin{array}{l}
\left.2^{7} \times 5-1\right)\left(2^{7}\right)^{4}+1=\left(1+2^{7} \times 5\right)\left(2^{7}\right)^{4}+1 \cdots\left(2^{7}\right)^{4}= \\
\left(1+2^{7} \times 5\right)\left\{\left(2^{7}\right)^{4}+\left(1-5 \times 2^{7}\r... | 641 \times 6700417 | Number Theory | proof | Yes | Yes | number_theory | false | 740,979 |
Create 1 using binary numbers to represent the decimal numbers $0,1,2,3,4,5,6,7$, 8,9.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | \[
\begin{array}{l}
0=(0)_{2}, \quad 1=(1)_{2}, \quad 2=(10)_{2} \\
3=2+1=(10)_{2}+(1)_{2}=(11)_{2} \\
4=(100)_{2} \\
5=4+1=(100)_{2}+(1)_{2}=(101)_{2} \\
6=4+2=(100)_{2}+(10)_{2}=(110)_{2} \\
7=6+1=(110)_{2}+(1)_{2}=(111)_{2} \\
8=(1000)_{2} \\
9=8+1=(1000)_{2}+(1)_{2}=(1001)_{2}
\end{array}
\] | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,980 |
Example 224 converted to a binary number equals what?
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | From the values column in Table 1, we know that the largest number not exceeding 24 is 16, and $2^{4}=16$. From $24-16=8$ and from the values column in Table 1, we know that the largest number not exceeding 8 is 8, and $2^{3}=8$. Since
$$\begin{aligned}
24=16+8 & =1 \times 2^{4}+1 \times 2^{3}+0 \times 2^{2}+0 \times 2... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,981 |
Example 3 converted to a binary number equals what?
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | From the values column in Table 1, we know that the largest number not greater than 92 is 64, and $2^{6}=64$. By $92-64=28$, and from the values column in Table 1, we know that the largest number not greater than 28 is 16, and $2^{4}=16$. By $28-16=12$, and from the values column in Table 1, we know that the largest nu... | 92=(1011100)_{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,982 |
Example 1 Find the greatest common divisor of 36 and 24. | Solve: Decompose these two numbers into prime factors
$$36=2 \times 2 \times 3 \times 3, \quad 24=2 \times 2 \times 2 \times 3$$
By comparing the prime factors of these two numbers, we can see that the prime factors $2,2,3$ are common to both numbers. Their product is the greatest common divisor of these two numbers:
... | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,983 |
Example $5(10011)_{2}$ converted to a decimal number equals what?
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | \[
\begin{array}{c}
(10011)_{2}=1 \times 2^{4}+0 \times 2^{3}+0 \times 2^{2}+1 \\
\times 2+1=19 .
\end{array}
\] | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,985 |
Example $6(110101)_{2}$ converted to a decimal number equals what?
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。 | \begin{array}{c}(110101)_{2}=1 \times 2^{5}+1 \times 2^{4}+0 \times 2^{3}+1 \\ \times 2^{2}+0 \times 2+1=32+16+4 \\ +1=53 .\end{array} | null | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 740,986 |
Example $7(736)_{8}$ converted to a decimal number equals what?
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve $\begin{aligned}(736)_{8} & =7 \times 8^{2}+3 \times 8+6=448+24 \\ & +6=478 .\end{aligned}$ | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,987 |
Example $8(1712)_{8}$ converted to a decimal number equals what?
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | \begin{aligned}(1712)_{8} & =1 \times 8^{3}+7 \times 8^{2}+1 \times 8+2 \\ & =970 .\end{aligned} | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,988 |
Example 9 Convert 117 to an octal number equals what?
Convert the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: From $\frac{117}{8}=14+\frac{5}{8}$ we get $b_{0}=5$, from $\frac{14}{8}=1+\frac{6}{8}$ we get $b_{1}=6, b_{2}=1$,
that is
$117=(165)_{8}$ | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,989 |
Example 10 Convert 92 to an octal number equals what?
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。 | Solve: From $\frac{92}{8}=11+\frac{4}{8}$ we get $b_{0}=4$, from $\frac{11}{8}=1+\frac{3}{8}$ we get $b_{1}=3, b_{2}=1$, i.e., $\square$ $92=(134)_{8}$. | null | Geometry | math-word-problem | Yes | Yes | number_theory | false | 740,990 |
Example 12 Convert $(111001101010)_2$ to an octal number.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly. | The first step is to divide this binary number into groups of three digits: $(111001101010)_{2}=\left(\begin{array}{llll}111 & 001 & 101 & 010\end{array}\right)_{2}$. Then, use the formula in (2.2):
$$\begin{array}{ll}
(111)_{2}=(7)_{8}, & (001)_{2}=(1)_{8} \\
(101)_{2}=(5)_{8}, & (010)_{2}=(2)_{8}
\end{array}$$
Thus... | (7152)_8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,992 |
Example 13 Convert $(573)_{8}$ to a binary number. | Using the formula in (2.2), we have
$$\begin{array}{l}
(5)_{8}=(101)_{2} \\
(7)_{8}=(111)_{2} \\
(3)_{8}=(011)_{2}
\end{array}$$
Thus, we get
$$(573)_{8}=(101111011)_{2}$$ | (101111011)_{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,993 |
Example 2 Find the greatest common divisor of $48$, $60$, and $72$. | Solve: Decompose the three numbers into prime factors respectively:
$$\begin{array}{l}
48=2 \times 2 \times 2 \times 2 \times 3=2^{4} \times 3 \\
60=2 \times 2 \times 3 \times 5=2^{2} \times 3 \times 5 \\
72=2 \times 2 \times 2 \times 3 \times 3=2^{3} \times 3^{2}
\end{array}$$
By comparing the prime factors of the ab... | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,994 |
Convert $(2175)_{8}$ to a binary number. | Using the formula in (2.2), we have
$$\begin{array}{ll}
(2)_{8}=(010)_{2}, & (1)_{8}=(001)_{2}, \\
(7)_{8}=(111)_{2}, & (5)_{8}=(101)_{2},
\end{array}$$
thus we get
$$(2175)_{8}=(010001111101)_{2}=(10001111101)_{2}$$ | (10001111101)_{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,995 |
Example 17 Find $(11101)_{2}-(1011)_{2}=$ ? | Solve: First, find the complement of $(1011)_{2}$, which is $(0101)_{2}=(101)_{2}$. Then add this complement to the minuend (11101) to get
Then subtract $(10000)_{2}$ to obtain the answer of this subtraction:
$$\begin{array}{l}
\left(\begin{array}{lllll}
1 & 1 & 1 & 0 & 1
\end{array}\right)_{2} \\
\frac{+\quad(101)_{2... | (10010)_{2} | Other | math-word-problem | Yes | Yes | number_theory | false | 740,998 |
Example 18 Find $(1111)_{2} \div(101)_{2}=$ ? | First, find the complement of the divisor $(101)_{2}$, which is $(011)_{2}=(11)_{2}$. Then add the complement $(11)_{2}$ to the dividend $(1111)_{2}$ to get
$$\begin{array}{r}
(1111)_{2} \\
+\quad(11)_{2} \\
\hline(10010)_{2}
\end{array}$$
Then subtract $(1000)_{2}$ to obtain the remainder after the first subtraction ... | (11)_{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,999 |
Example 19 Find $(101101)_{2} \div(1111)_{2}=$ ? | First, find the complement of the divisor $(1111)_2$, which is $(0001)_2 = (1)_2$. Then add the complement $(1)_2$ to the dividend $(101101)_2$ and subtract $(10000)_2$ to obtain the remainder after the first subtraction of the divisor, i.e.,
Then add the complement $(1)_2$ to the remainder $(11110)_2$ and subtract $(... | (11)_2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,000 |
1. Convert the following decimal numbers to binary numbers:
(i) 420,
(ii) 2640. | 1.
(i) Solution: From the value column in Table 1, we know that the largest number not exceeding 420 is 256, and $2^{8}=256$. From $420-256=164$, and from the value column in Table 1, we know that the largest number not exceeding 164 is 128, and $2^{7}=128$. From $164-128=36$, and from the value column in Table 1, we k... | 420=(110100100)_{2}, 2640=(101001010000)_{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,001 |
3. Convert the following decimal numbers to octal numbers:
(i) 420,
(ii) 2640. | 3 .
(i) Solution: From $\frac{420}{8}=52+\frac{4}{8}$, we get $b_{0}=4$. From $\frac{52}{8}=6+\frac{4}{8}$, we get $b_{1}=4, b_{2}=6$, which means $420=(644)_{8}$.
(ii) Solution: From $\frac{2640}{8}=330$, we get $b_{0}=0$. From $\frac{330}{8}=41+\frac{2}{8}$, we get $b_{1}=2$. From $\frac{41}{8}=5+\frac{1}{8}$, we get... | 644 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,003 |
Example 3 Find the greatest common divisor of $1008, 1260, 882$ and 1134. | Solve: Decompose these four numbers into prime factors
$$\begin{array}{l}
1008=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7=2^{4} \times 3^{2} \times 7 \\
1260=2 \times 2 \times 3 \times 3 \times 5 \times 7=2^{2} \times 3^{2} \times 5 \times 7 \\
882=2 \times 3^{2} \times 7^{2} \\
1134=2 \times 3^{4} \times ... | 126 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,005 |
5. Convert the following binary numbers to octal numbers:
(i) $(101101101)_{2}$,
(ii) $(101011001101001)_{2}$. | 5 .
(i) Solution: The first step is to divide this binary number into groups of three digits, $(101101101)_{2}=\left(\begin{array}{ll}101 & 101101\end{array}\right)_{2}$, then use the formula in (2.2) $(101)_{2}=(5)_{8}$, which gives $(101101101)_{2}=(555)_{8}$.
(ii) Solution: The first step is to divide this binary nu... | (555)_{8}, (53151)_{8} | Other | math-word-problem | Yes | Yes | number_theory | false | 741,006 |
7. Find the result of the addition operation for the following binary numbers:
(i) $(1001)_{2}+(101)_{2}+(1111)_{2}+(111)_{2}=$ ?
(ii) $(101011)_{2}+(10011)_{2}+(1111)_{2}=$ ? | 7.
(i) Solution: Since
Therefore, we have
$$(1001)_{2}+(101)_{2}+(1111)_{2}+(111)_{2}=(100100)_{2}$$
(ii) Solution: Since
Therefore, we have
$$(101011)_{2}+(10011)_{2}+(1111)_{2}=(1001101)_{2}$$ | (100100)_{2} | Other | math-word-problem | Yes | Yes | number_theory | false | 741,008 |
8. Find the result of the following binary multiplication:
(i) $(111)_{2} \times(101)_{2}=$ ?
(ii) $(1001)_{2} \times(111)_{2} \times(101)_{2}=$ ? | 8.
(i) Solution: Since
therefore, $(111)_{2} \times(101)_{2}=(100011)_{2}$.
(ii) Solution: Since
therefore, $(1001)_{2} \times(111)_{2} \times(101)_{2}=(100111011)_{2}$.
$$\begin{array}{l}
\left(\begin{array}{lll}
1 & 1 & 1
\end{array}\right)_{2} \\
\frac{\times(101)_{2}}{(111)_{2}}
\end{array}$$
$$\begin{array}{l}
\l... | (100011)_{2} | Other | math-word-problem | Yes | Yes | number_theory | false | 741,009 |
9. Find the result of the following binary subtraction operations:
(i) $(1010111)_{2}-(11001)_{2}-(11110)_{2}=$ ?
(ii) $(10110001)_{2}-(1101100)_{2}-(11110)_{2}=$ ? | 9.
(i) Solution: We first use the conventional subtraction operation, to get
Therefore, we have $(1010111)_{2}-(11001)_{2}-(11110)_{2}=(100000)_{2}$.
Now we will use the method of finding complements to perform the operation. First, we find the complement of $(11001)_{2}$ to get $(00111)_{2}=(111)_{2}$, then add the c... | (100111)_{2} | Other | math-word-problem | Yes | Yes | number_theory | false | 741,010 |
10. Find the result of the following binary division operations:
(i) $(1100011)_{2} \div(100001)_{2}=$ ?
(ii) $(110001)_{2} \div(111)_{2}=$ ? | 10.
(i) Solution: We first use the conventional division operation to get
Therefore, we have $(1100011)_{2} \div(100001)_{2}=(11)_{2}$.
Now we will use the method of finding complements to perform the operation. We first find the complement of the divisor $(100001)_{2}$ to get $(011111)_{2}=(11111)_{2}$, then add the ... | (11)_{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,011 |
Since $(36,83)=1$, by Lemma $1-$ there must exist two integers $x$, $y$ such that
$$36 x+83 y=1$$
holds, find $x, y$. | Since
$$\begin{array}{l}
93=2 \times 36+11, \quad 36=3 \times 11+3 \\
11=3 \times 3+2, \quad 3=2+1
\end{array}$$
we get
$$\begin{aligned}
1 & =3-2=3-(11-3 \times 3)=4 \times 3-11 \\
& =4 \times(36-3 \times 11)-11=4 \times 36-13 \times 11 \\
& =4 \times 36-13 \times(93-2 \times 36) \\
& =30 \times 36-13 \times 93 \\
& ... | x=30, y=-13 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,013 |
Example 2 Since $(72,157)=1$, by Lemma 1 there must be two integers $x$, $y$ that satisfy
$$72 x+157 y=1$$
established, find $x, y$. | Since
$$\begin{array}{l}
157=2 \times 72+13, \quad 72=5 \times 13+7 \\
13=7+6, \quad 7=6+1
\end{array}$$
we get
$$\begin{aligned}
1 & =7-6=7-(13-7)-2 \times 7-13 \\
& =2 \times(72-5 \times 13)-13=2 \times 72-11 \times 13 \\
& =2 \times 72-11 \times(157-2 \times 72) \\
& =24 \times 72-11 \times 157=72 \times 24-157 \ti... | x=24, y=-11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,014 |
Lemma 2 If $a$ and $b$ are two coprime positive integers, and $c$ is an integer, then there must exist two integers $x, y$ such that
$$a x+b y=c$$
holds. | By Lemma 1, we know there exist two integers $s, t$ such that
$$a s+b t=1$$
holds. Let $x=s c, y=t c$, then we get $a x+b y=c(a s+b t)=c$. Therefore, equation (11) has integer solutions $x, y$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,015 |
Example 2 Prove that the remainder of $\left(12371^{16}+34\right)^{23+7+c}$ divided by 111 is equal to 70, where $c$ is any non-negative integer. | Example 2's proof: From $(70,111)=1$ and Example 8's $\left(12371^{56}+\right.$ $34)^{28} \equiv 70(\bmod 111)$, we get $\left(12371^{56}+34,111\right)=1$. Since $111=3 \times 37$ and Lemma 14, we have $\varphi(111)=2 \times 36=72$. Since $c$ is a positive integer and by Theorem 1, we have $\left(12371^{56}+\right.$ $3... | 70 | Number Theory | proof | Yes | Yes | number_theory | false | 741,017 |
Lemma 7 Let $m$ be an integer greater than 1, and $b$ be an integer satisfying $(b, m)=1$. If $a_{1}, a_{2}, \cdots, a_{m}$ is a complete residue system modulo $m$, then $b a_{1}, b a_{2}, \cdots, b a_{m}$ is also a complete residue system modulo $m$. | Proof: Suppose in $b a_{1}, b a_{2}, \cdots, b a_{m}$ there exist two integers $b a_{k}, b a_{\lambda}$ (where $1 \leqslant k<\lambda \leqslant m$), such that
$$b a_{k} \equiv b a_{\lambda}(\bmod m)$$
holds, then by $(b, m)=1$ and Lemma 2 we have
$$a_{k} \equiv a_{2}(\bmod m)$$
By Lemma 4 and $a_{1}, a_{2}, \cdots, a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,018 |
Example 17 Prove that $\sqrt{5}=2.236067977499 \ldots$.
| Proof In (40), taking $b=9$, by Lemma 5 and Example 15 we have
$$\frac{p_{11}}{q_{11}}\frac{32 \times 9^{6}-16 \times 9^{5}-48 \times 9^{4}+20 \times 9^{3}+18 \times 9^{2}-46}{4\left(32 \times 9^{5}-16 \times 9^{4}-32 \times 9^{3}+12 \times 9^{2}+54-1\right)} \\
=\frac{15762392}{4(1762289)}>2.2360679774997
\end{array}$... | 2.2360679774997 | Number Theory | proof | Yes | Yes | number_theory | false | 741,019 |
Example 19 Prove that $\sqrt{11}=3.316624790355 \cdots$.
| Proof: In (40), taking $b=10$, by Lemma 5 and Example 15, we have
$$\frac{p_{11}}{q_{11}}<\sqrt{99}<\frac{p_{12}}{q_{12}}$$
Thus, we have
$$\begin{array}{l}
\sqrt{11}=\frac{\sqrt{99}}{3} \leqslant \frac{32 \times 10^{6}-48 \times 10^{4}+18 \times 10^{2}-1}{3\left(32 \times 10^{5}-32 \times 10^{3}+60\right)} \\
=\frac... | 3.316624790355 | Number Theory | proof | Yes | Yes | number_theory | false | 741,021 |
Example 21 We have
$$\begin{array}{l}
{[x]+[y] \leqslant[x+y], \quad\{x\}+\{y\} \geqslant\{x+y\} .} \\
{[-x]=\left\{\begin{array}{ll}
-[x]+1, & \text { when } x \text { is not an integer; } \\
-[x], & \text { when } x \text { is an integer. }
\end{array}\right.}
\end{array}$$ | Let $x=n+a, y=m+b$, where $n$ and $m$ are integers and $0 \leqslant a<1,0 \leqslant b<1$. By properties (II) and (III) derived from Definition 3, we have
$$\begin{aligned}
{[x]+[y] } & =n+m+[a]+[b] \\
& =m+n \leqslant m+n+[a+b] \\
& =[x+y]
\end{aligned}$$
We also have
$$\begin{aligned}
\{x\}+\{y\} & =x-[x]+y-[y] \geqs... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,023 |
Example 22 Suppose $n$ is any positive integer and $\alpha$ is a real number, then
$$[\alpha]+\left[\alpha+\frac{1}{n}\right]+\cdots+\left[\alpha+\frac{n-1}{n}\right] \Rightarrow[n \alpha]$$
holds. | Let $\alpha=m+a$, where $m$ is an integer and $0 \leqslant a<1$. Then by property (II) we have
$$\begin{array}{c}
{[\alpha]+\left[\alpha+\frac{1}{n}\right]+\cdots+\left[\alpha+\frac{n-1}{n}\right]=[m+a]} \\
+\left[m+a+\frac{1}{n}\right]+\cdots+\left[m+a+\frac{n-1}{n}\right] \\
=m n+[a]+\left[a+\frac{1}{n}\right]+\cdots... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,024 |
Example 23 Let $a, b$ be two integers, $b>0$, then we have
$$a=b\left[\frac{a}{b}\right]+b\left\{\frac{a}{b}\right\}, 0 \leqslant b\left\{\frac{a}{b}\right\} \leqslant b-1$$ | Proof: By property (I), we have
$$\frac{a}{b}=\left[\frac{a}{b}\right]+\left\{\frac{a}{b}\right\}$$
Thus, we get
$$a=b\left[\frac{a}{b}\right]+b\left\{\frac{a}{b}\right\}$$
By Example 7 and $b>0$, we have $b\left\{\frac{a}{b}\right\} \geqslant 0$. Let $a=b m+r$, where $m$ is an integer and $0 \leqslant r \leqslant b-... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,025 |
Example 24 we have
$$[2 x]+[2 y] \geqslant[x]+[y]+[x+y]$$ | Let $x=m+a$, where $m$ is an integer and $0 \leqslant a<1$; $y=n+b$, where $n$ is an integer and $0 \leqslant b<1$. When $a \geqslant b$, we have by properties (II) and (III)
$$\begin{aligned}
{[2 x] } & +[2 y]=[2 m+2 a]+[2 n+2 b]=2 m+2 n \\
& +[2 a]+[2 b] \geqslant m+n+m+n+[a+b] \\
= & {[m+a]+[n+b]+[m+n+a+b] } \\
& =[... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,026 |
Lemma 6 Let $n=p_{1}^{a_{1}} \cdots p_{m}^{a_{m}}$,
where $p_{1}, \cdots, p_{m}$ are all distinct primes, and $\alpha_{1}, \cdots, \alpha_{m}$ are all positive integers, then we have
$$d(n)=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{m}+1\right)$$ | To prove that any divisor of $n$ has the form
$$p_{1}^{\beta_{1}} \cdots p_{m}^{\beta_{m}} \cdot$$
Here,
$$\begin{array}{l}
0 \leqslant \beta_{1} \leqslant \alpha_{1} \\
\cdots \cdots \cdots \\
0 \leqslant \beta_{m} \leqslant \alpha_{m}
\end{array}$$
Since $\beta_{1}$ can take $\alpha_{1}+1$ different integers, $\cdo... | d(n)=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{m}+1\right) | Number Theory | proof | Yes | Yes | number_theory | false | 741,027 |
Lemma 7 Let $a, b$ be two positive integers and $(a, b)=1$, then we have $d(a b)=d(a) d(b)$. | Let $a=p_{1}^{a_{1}} \cdots p_{n}^{a_{n}}, \quad b=q_{1}^{\beta_{1}} \cdots q_{m}^{\beta_{m}}$, where $p_{1}, \cdots, p_{n}, q_{1}, \cdots, q_{m}$ are all prime numbers and $\alpha_{1}, \cdots, \alpha_{n}, \beta_{1}, \cdots, \beta_{m}$ are all positive integers. Since $(a, b)=1$, we know that any $p_{i}(i=1, \cdots, n)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,028 |
Lemma 8 Let $m$ be an integer greater than 1, and $b, c$ be two arbitrary integers satisfying the condition $(b, m)=1$. If $a_{1}, a_{2}, \cdots, a_{m}$ is a complete residue system modulo $m$, then $b a_{1}+c, b a_{2}+c, \cdots, b a_{m}+c$ is also a complete residue system modulo $m$. | Given that $a_{1}, a_{2}, \cdots, a_{m}$ is a complete residue system modulo $m$, from Lemma 7 and $(b, m)=1$ we know that $b a_{1}, b a_{2}, \cdots, b a_{m}$ is also a complete residue system modulo $m$. Since $b a_{1}, b a_{2}, \cdots, b a_{m}$ is a complete residue system modulo $m$, from Lemma 6 and $c$ being an in... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,029 |
Lemma 8 When $l, m$ are positive integers and $m \geqslant 2$, we have
$$1+m+\cdots+m^{l}=\frac{m^{l+1}-1}{m-1}$$ | Proof When $l=1$, we have
$$1+m=\frac{(m+1)(m-1)}{m-1}=\frac{m^{2}-1}{m-1}$$
Thus, the lemma holds when $l=1$. Now assume $k \geqslant 2$, and the lemma holds for $l$ equal to $1,2, \cdots, k-1$, then we have
$$\begin{aligned}
1 & +m+\cdots+m^{k-1}+m^{k}=\frac{m^{k}-1}{m-1}+m^{k} \\
& =\frac{m^{k+1}-1}{m-1}
\end{align... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,031 |
$p_{1}, \cdots, p_{m}$ are $m$ distinct prime numbers, $\alpha_{1}, \cdots, \alpha_{m}$ are $m$ positive integers, then we have
$$\sigma(n)=\frac{p_{1}^{\alpha_{1}+1}-1}{p_{1}-1} \cdots \frac{p_{m}^{\alpha_{m}+1}-1}{p_{m}-1} .$$ | When $m=1$, by Lemma 9, we know that this lemma holds. When $m=2$, then we have $n=p_{1}^{\alpha_{1}} p_{2}^{a_{2}}$. Let $p_{1}^{0}=p_{2}^{0}=1$. If we multiply each number $p_{1}^{i}$ (where $i=0,1, \cdots, \alpha_{1}$) in $1+p_{1}+\cdots+p_{1}^{\alpha_{1}}$ with each number $p_{2}^{j}$ (where $j=0,1, \cdots, \alpha_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,033 |
Example 26 Find $\sigma(450)=$ ? | Since $45!=2 \times 3 \times 5^{2}$, by Lemma 10 we have
$$\begin{aligned}
\sigma(450) & =\frac{2^{2}-1}{2-1} \cdot \frac{3^{3}-1}{3-1} \cdot \frac{5^{3}-1}{5-1} \\
& =3 \times 13 \times 31=1209
\end{aligned}$$ | 1209 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,034 |
Lemma 12 If $n$ is an integer $\geqslant 2$ and $2^{n}-1$ is a prime, then
$$2^{n-1}\left(2^{n}-1\right)$$
is a perfect number. | Since $\left(2^{n-1}, 2^{n}-1\right)=1$ holds, by Lemma 11 we have
$$\sigma\left(2^{n-1}\left(2^{n}-1\right)\right)=\sigma\left(2^{n-1}\right) \cdot \sigma\left(2^{n}-1\right)$$
Since $2^{n}-1$ is a prime number, the divisors of $2^{n}-1$ are $1, 2^{n}-1$, thus we get
$$\sigma\left(2^{n}-1\right)=2^{n}-1+1=2^{n}$$
Si... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,036 |
Example 27 Find $\sigma_{2}(28)=$ ? | Since the factors of 28 are $1,2,4,7,14,28$, we have
$$\sigma_{2}(28)=1+2^{2}+4^{2}+7^{2}+14^{2}+28^{2}=1050$$ | 1050 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,037 |
side 28 find $\sigma_{3}(62)=$ ? | Since the factors of 62 are $1,2,31,62$, we have
$$\sigma_{3}(62)=1+2^{3}+31^{3}+62^{3}=268128$$ | 268128 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,038 |
Lemma 13 If $m, n$ are two positive integers and $(m, n)=1$, then we have
$$\mu(m n)=\mu(m) \cdot \mu(n)$$ | If $m$ or $n$ is divisible by the square of a prime, then $mn$ is also divisible by the square of this prime. Hence, we get
$$\mu(m n)=0=\mu(m) \cdot \mu(n)$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,039 |
Lemma 14 We have
$$\sum_{d \mid n} \mu(d)=\left\{\begin{array}{l}
1, \text { when } n=1 \text { ; } \\
0, \text { when } n>1 \text { . }
\end{array}\right.$$ | To prove when $n=1$, then since $\sum_{d \mid n} \mu(d)=\mu(1)=1$, the lemma holds.
Now suppose $n \geqslant 2$ is an integer. When $m$ is a positive integer and $m \mid n$, we use the notation $\sum_{m \mid d \mid n}$ to denote a sum over all divisors $d$ of $n$ that are divisible by $m$. In particular, when $m=1$, $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,041 |
Lemma 15 Let $n=p_{1}^{a_{1}} \cdots p_{m}^{\alpha_{m}}$, where $p_{1}, \cdots, p_{m}$ are $m$ distinct primes, and $\alpha_{1}, \cdots, \alpha_{m}$ are all positive integers, then we have
$$\sum_{d \mid n}|\mu(d)|=2^{m}$$ | Proof: Since $\mu(d) = 0$ when $d$ is divisible by the square of a prime, we have
$$\sum_{d \mid n}|\mu(d)|=\sum_{d \mid p_{1} \cdots p_{m}}|\mu(d)|$$
We will prove that when $m \geqslant 1$, we have
$$\sum_{d \mid p_{1} \cdots p_{m}}|\mu(d)|=2^{m}$$
When $m=1$, since
$$\sum_{d \mid p}|\mu(d)|=1+|\mu(p)|=2,$$
(60) h... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,042 |
2. Convert the following rational fractions into continued fractions:
(i) $\frac{50}{13}$,
(ii) $-\frac{53}{25}$. | 2. (i) Solution:
$$\begin{aligned}
\frac{50}{13} & =3+\frac{11}{13}=3+\frac{1}{\frac{13}{11}}=3+\frac{1}{1+\frac{2}{11}} \\
& =3+\frac{1}{1+\frac{1}{\frac{11}{2}}}=3+\frac{1}{1+\frac{1}{5+\frac{1}{2}}} \\
& =[3,1,5,2] .
\end{aligned}$$
(ii) Solution:
$$\begin{aligned}
-\frac{53}{25} & =-3+\frac{22}{25}=-3+\frac{1}{\fra... | [3,1,5,2] | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,044 |
3. Calculate the approximate value of $\sqrt{41}$ using continued fractions.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the text provided is already in English, here is the equivalent text in the requeste... | 3. Solution: Since $6<\sqrt{41}<7$, we have
$$\begin{aligned}
\sqrt{41} & =6+(\sqrt{41}-6)=6+\frac{1}{\frac{1}{\sqrt{41}-6}} \\
& =6+\frac{1}{\frac{\sqrt{41}+6}{5}}
\end{aligned}$$
Since $2<\frac{\sqrt{41}+6}{5}<3$, we have
$$\begin{aligned}
\frac{\sqrt{41}+6}{5} & =2+\frac{\sqrt{41}-4}{5}=2+\frac{1}{\frac{5}{\sqrt{41... | 6.403100<\sqrt{41}<6.403125 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,045 |
4. It is known that the continued fraction of $\pi$ is
$$x=[3,7,15,1,292,1,1, \ldots]$$
Try to find its first seven convergents and their approximate values. | 4. Solution: By Lemma 1, we get
$$\begin{aligned}
\frac{p_{1}}{q_{1}} & =\frac{3}{1}, \frac{p_{2}}{q_{2}}=\frac{3 \times 7+1}{7} \\
& =\frac{22}{7}=3.14285714 \cdots \\
\frac{p_{3}}{q_{3}} & =\frac{22 \times 15+3}{7 \times 15+1}=\frac{333}{106}=3.141509433 \cdots \\
\frac{p_{4}}{q_{4}} & =\frac{333 \times 1+22}{106 \ti... | 3.1415926534<\pi<3.1415926540 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,046 |
5. Suppose the binary linear Diophantine equation $a x+b y=c, a>0$, and $(a,|b|)=1, \frac{a}{|b|}$ has $k$ convergents.
Prove:
$$\left\{\begin{array}{l}
x_{0}=(-1)^{k} c q_{k-1}, \\
y_{0}=(-1)^{k+1} c p_{k-1} \cdot \frac{|b|}{b}
\end{array}\right.$$
is a solution to the equation. Here $\frac{p_{k-1}}{q_{k-1}}$ is the... | 5. Proof: Since $\frac{a}{|b|}=\frac{p_{k}}{q_{k}}$, by Lemma 2 we get
$$a q_{k-1}-|b| p_{k-1}=(-1)^{k}$$
Multiplying both sides by $(-1)^{k} c$, we obtain
$$a\left[(-1)^{k} c q_{k-1}\right]+|b|\left[(-1)^{k+1} c p_{k-1}\right]=c$$
That is $\square$
$$a\left[(-1)^{k} c q_{k-1}\right]+b \cdot \frac{|b|}{b}\left[(-1)^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,047 |
6. Use the result from the previous question to find the integer solutions of the following equations:
(i) $43 x+15 y=8$.
(ii) $10 x-37 y=3$ | 6. (i) Solution: Converting $\frac{43}{15}$ into a continued fraction, we get
$$\frac{43}{15}=2+\frac{1}{1+\frac{\Gamma}{6+\frac{1}{2}}}$$
Therefore,
$$k=4 . \quad \frac{p_{1}}{q_{1}}=2, \quad \frac{p_{2}}{q_{2}}=3, \quad \frac{p_{3}}{q_{3}}=\frac{6 \times 3+2}{6 \times 1+1}=\frac{20}{7}$$
So,
$$\left\{\begin{array}{... | \begin{array}{l}
x=56-15 t, \\
y=-160+43 t
\end{array} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,048 |
7. Prove:
(i) $\sum_{k=1}^{n}\left[\frac{k}{2}\right]=\left[\frac{n^{2}}{4}\right]$.
(ii) $\sum_{k=1}^{n}\left[\frac{k}{3}\right]=\left[\frac{n(n-1)}{\sigma}\right]$.
(iii) When $0<a<8$, there must exist an integer $b$ such that
$$\sum_{k=1}^{n}\left[\frac{k}{a}\right]=\left[\frac{(2 n+b)^{2}}{8 a}\right] .$$ | 7. (i) Proof: Let $n=2m+c, 0 \leqslant c \leqslant 1$, then
$$\begin{array}{l}
\sum_{k=1}^{n}\left[\frac{k}{2}\right]=\sum_{k=1}^{2 m+c} \frac{k}{2}-\sum_{k=1}^{2 m+c}\left\{\frac{k}{2}\right\} \\
\quad=\frac{1}{4}(2 m+c)(2 m+c+1)-\frac{1}{2}(m+c) \\
\quad=\frac{1}{4}\left\{(2 m+c)^{2}+(2 m+c)-2(m+c)\right\} \\
\quad=\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,049 |
Lemma 9 If $m$ is an integer greater than 1 and $a, b$ are any two integers such that
$$a \equiv b(\bmod m)$$
holds, then $(a, m)=(b, m)$. | Prove that from $a \equiv b(\bmod m)$ we get $a=b+m t$, where $t$ is an integer, hence $(b, m) \mid a$. Also, from $(b, m) \mid m$ we get $(b, m) \mid(a, m)$. From $b=a-m t$ we have $(a, m) \mid b$. Also, from $(a, m) \mid m$ we get $(a, m) \mid(b, m)$. Therefore, from $(b, m) \mid(a, m)$ and $(a, m) \mid(b, m)$ we get... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,051 |
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