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Example: How many zeros are at the end of the decimal representation of 320! ?
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve for the positive integer $k$ such that $10^{k} \| 20$!. From equation (12), we know that we only need to find $\alpha(5,20)$, and from the previous example, we know that $k=4$, which is the power of 5. Therefore, there are four zeros at the end. | 4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
20. Find how many zeros are at the end of the decimal expression of 120!.
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 above is not part of the translation but is provided to clarify the instruction. The actual translation is above it. | 20. Find the highest power of 10 that divides 120!, which is also the highest power of 5 that divides 120!. Therefore, there are 28 zeros. | 28 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Proof: $101 x_{1}+37 x_{2}=3189$ has positive integer solutions. | Here, $c=3189<a_{1} a_{2}=101 \cdot 37$, so from the conclusion of Theorem 5, we cannot determine whether the equation has a positive solution (of course, it can be deduced that there is at most one). Therefore, we need to use formula (15) (or (14)). It can be found that $x_{1}=11 \cdot 3189, x_{2}=-30 \cdot 3189$ is a... | 1 | Number Theory | proof | Yes | Yes | number_theory | false |
16. There are five sailors and a monkey on a small island. During the day, they collected some coconuts as food. At night, one of the sailors woke up and decided to take his share of the coconuts. He divided the coconuts into five equal parts, with one left over, so he gave the extra one to the monkey and hid his share... | 16. This pile of coconuts has at least 3121. The number of coconuts each of the five people received in turn is $828, 703, 603$, 523, 459, and the monkey ate 5. | 3121 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the units digit of $3^{406}$ when it is written in decimal form. | Solve: According to the problem, we are required to find the smallest non-negative remainder $a$ when $3^{406}$ is divided by 10, i.e., $a$ satisfies
$$3^{406} \equiv a(\bmod 10), \quad 0 \leqslant a \leqslant 9$$
Obviously, we have $3^{2} \equiv 9 \equiv-1(\bmod 10), 3^{4} \equiv 1(\bmod 10)$, and thus
$$3^{404} \equ... | 9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find the last two digits of $3^{406}$ when written as a decimal number. | Solve: This is to find the smallest non-negative remainder $b$ when $3^{106}$ is divided by 100, i.e., $b$ satisfies
$$3^{406} \equiv b(\bmod 100), \quad 0 \leqslant b \leqslant 99$$
Notice that $100=4 \cdot 25, (4,25)=1$. Clearly, $3^{2} \equiv 1(\bmod 4), 3^{4} \equiv 1(\bmod 5)$. Note that 4 is the smallest power, ... | 29 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Let $m>n \geqslant 1$. Find the smallest $m+n$ such that
$$\text {1000| } 1978^{m}-1978^{n} \text {. }$$ | Solve: Using the congruence symbol, the problem is to find the smallest $m+n$ such that
$$1978^{m}-1978^{n} \equiv 0(\bmod 1000)$$
is satisfied. First, let's discuss what conditions $m, n$ must meet for the above equation to hold. Let $k=m-n$. Equation (9) becomes
$$2^{n} \cdot 989^{n}\left(1978^{k}-1\right) \equiv 0\... | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
18. Find the integer $n$ that satisfies $n \equiv 1(\bmod 4), n \equiv 2(\bmod 3)$. | 18. $n=4 k+1 \equiv 2(\bmod 3), k=1, n=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
7. Use the previous problem to solve (i) $3 x \equiv 1(\bmod 125)$; (ii) $5 x \equiv 1(\bmod 243)$. | 7. (i) $3 x \equiv 1\left(\bmod 5^{3}\right), 3 \cdot 2 \equiv 1(\bmod 5), 1-(1-3 \cdot 2)^{3}=126$, so $x=42$ is a solution to the original congruence equation. (ii) $5 x \equiv 1\left(\bmod 3^{5}\right) .5 \cdot(-1) \equiv 1(\bmod 3) .1-(1-5 \cdot(-1))^{5}=$ $-7775 . x=-1555$ is a solution. | -1555 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Solve the system of congruences
$$\left\{\begin{array}{l}
x \equiv 1(\bmod 3), \\
x \equiv-1(\bmod 5), \\
x \equiv 2(\bmod 7), \\
x \equiv-2(\bmod 11)
\end{array}\right.$$ | Let $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, satisfying the conditions of Theorem 1. In this case, $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$. We will find $M_{j}^{-1}$. Since $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$, we have
$$1 \equiv M_{1} M_{... | 394 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Calculate $\left(\frac{137}{227}\right)$. | To determine if 227 is a prime number, by Theorem 1 we get
$$\begin{aligned}
\left(\frac{137}{227}\right) & =\left(\frac{-90}{227}\right)=\left(\frac{-1}{227}\right)\left(\frac{2 \cdot 3^{2} \cdot 5}{227}\right) \\
& =(-1)\left(\frac{2}{227}\right)\left(\frac{3^{2}}{227}\right)\left(\frac{5}{227}\right) \\
& =(-1)\left... | -1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Let $n \geqslant 1$. Prove: $(n!+1,(n+1)!+1)=1$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | 5. $(n!+1,(n+1)!+1)=(n!+1,-n)=(1,-n)=1$.
The translation is as follows:
5. $(n!+1,(n+1)!+1)=(n!+1,-n)=(1,-n)=1$. | 1 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 1 Determine whether the congruence equation $2 x^{3}+5 x^{2}+6 x+1 \equiv 0(\bmod 7)$ has three solutions. | Solve: Here the coefficient of the first term is 2. By making an identity transformation, we can know that the original equation has the same solutions as
$$4\left(2 x^{3}+5 x^{2}+6 x+1\right) \equiv x^{3}-x^{2}+3 x-3 \equiv 0(\bmod 7)$$
Performing polynomial division, we get
$$x^{7}-x=\left(x^{3}-x^{2}+3 x-3\right)\l... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8. Let $p$ be a prime, $\delta_{p}(a)=3$. Prove: $\delta_{p}(1+a)=6$.
| 8. From $\delta_{p}(a)=3$, we get $a \neq \pm 1(\bmod p), a^{2}+a+1 \equiv 0(\bmod p)$. Therefore, $1+a \neq 1(\bmod p),(1+a)^{2} \equiv 1+2 a+a^{2} \equiv a \not \equiv 1(\bmod p),(1+a)^{3} \equiv-1(\bmod p)$. Hence, $\delta_{p}(1+a)=6$ | 6 | Number Theory | proof | Yes | Yes | number_theory | false |
15. Find all positive integer solutions \(a, b, c\) that satisfy \((a, b, c)=10,[a, b, c]=100\). | 15. $(a / 10, b / 10, c / 10)=1,[a / 10, b / 10, c / 10]=10 . a / 10, b / 10, c / 10$ can only take the values $1,2,5,10$ and satisfy the above two conditions. There are three possible scenarios: (i) $a / 10=b / 10=c / 10$, which is impossible; (ii) Two of the three are equal, and the other is uniquely determined by th... | 36 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the primitive root of $p=23$.
保留源文本的换行和格式，直接输出翻译结果。
(Note: The last sentence is a note for the translator and should not be included in the final translation. The correct output should be as follows:)
Example 1 Find the primitive root of $p=23$.
| Since the index of $a$ modulo $p$ must be a divisor of $p-1$, to find the index, we just need to compute the residue of $a^{d}$ modulo $p$, where $d \mid p-1$.
Here $p-1=22=2 \cdot 11$, its divisors $d=1,2,11,22$. First, find the index of $a=2$ modulo 23:
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 23) \\
2^{11} \equiv\le... | 5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find a primitive root modulo 41. | Solve $41-1=40=2^{3} \cdot 5$, divisors $d=1,2,4,8,5,10,20,40$. Now we will find the orders of $a=2,3, \cdots$.
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 41), 2^{4} \equiv 16(\bmod 41), 2^{5} \equiv-9(\bmod 41) \\
2^{10} \equiv-1(\bmod 41), \quad 2^{20} \equiv 1(\bmod 41)
\end{array}$$
So $\delta_{41}(2) \mid 20$. Since... | 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
7. Let $1975 \leqslant n \leqslant 1985$, ask which of these $n$ have primitive roots. | 7. Only $n=1979$ is a prime number. | 1979 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
11. Under the notation of the previous question, find $j$ that satisfies
(i) $0 \bmod 3 \cap 0 \bmod 5=j \bmod 15$;
(ii) $1 \bmod 3 \cap 1 \bmod 5=j \bmod 15$;
(iii) $-1 \bmod 3 \cap -2 \bmod 5=j \bmod 15$. | 11. (i) $j=0$;
(ii) $j=1$;
(iii) $j=8$. | 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find the greatest common divisor of 6731 and 2809 | Solve:
\begin{aligned}
6731 & =2809 \times 2+1113 \\
2809 & =1113 \times 2+583 \\
1113 & =583 \times 1+530 \\
583 & =530+53 \\
530 & =53 \times 10+0
\end{aligned}
So $(6731,2809)=53$. For convenience in writing, the series of calculations above can be abbreviated as follows:
2 \begin{tabular}{|r|r|r}
6731 & 2809 & 2 \... | 53 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Take a total of ten coins of one cent, two cents, and five cents, to pay eighteen cents. How many different ways are there to do this?
保留源文本的换行和格式，直接输出翻译结果如下：
5. Take a total of ten coins of one cent, two cents, and five cents, to pay eighteen cents. How many different ways are there to do this? | 5. Solution: Let $x, y, z$ represent the number of 1-cent, 2-cent, and 5-cent coins, respectively. Therefore, we have the following equations:
$$\begin{array}{l}
x+2 y+5 z=18 \\
x+y+z=10
\end{array}$$
Subtracting the second equation from the first, we get $y+4 z=8$.
We need to find the non-negative integer solutions t... | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find the least common multiple of 108, 28, and 42. | Since
$$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$$ | 756 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 16 Han Xin Counts the Soldiers: There is a troop of soldiers. If they form a column of five, there is one person left over at the end. If they form a column of six, there are five people left over at the end. If they form a column of seven, there are four people left over at the end. If they form a column of el... | Let $x$ be the number of soldiers we are looking for. According to the problem,
In Theorem 1, take $m_{1}=5, m_{2}=6, m_{3}=7, m_{4}=11, b_{1}=1$, $b_{2}=5, b_{3}=4, b_{4}=10$. Then we have
$$\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 |
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
$$\begin{array}{c}
b_{1}=3, \quad b_{2}=5, \quad m_{1}=7, \quad m_{2}=11, \\
m=m_{1} \cdot m_{2}=7 \times 11=77, \\
M_{1}=\frac{77}{7}=11, \quad M_{2}=\frac{77}{11}=7 . \\
11 M_{1}^{\prime}=1 \quad(\bmod 7), \text { so } M_{1}^{\prime}=2 . \\
7 M_{2}^{\prim... | 299 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6. Solve the following problems: (Yang Hui: "Xu Gu Zhai Qi Suan Fa" (1275))
(i) When divided by 7, the remainder is 1; when divided by 8, the remainder is 2; when divided by 9, the remainder is 4. What is the original number?
(ii) When divided by 2, the remainder is 1; when divided by 5, the remainder is 2; when divide... | 6.
(i) Solution: Let the number be $x$, then according to the problem, we have
Here
By
By
By
$$\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 |
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 |
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
$$\left\{\begin{array}{l... | 3300 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 9 Find the least common multiple of 24871 and 3468.
The text above is translated into English, preserving the original text's line breaks and format. | Since
thus $(24871,3468)=17$. By Lemma 10 we have
$$\{24871,3468\}=\frac{24871 \times 3468}{17}=5073684 .$$ | 5073684 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 10 Find the least common multiple of 513, 135, and 3114. | Since
$1\left|\begin{array}{r|r}513 & 135 \\ 405 & 108 \\ \hline 108 & 27 \\ 108 & \\ \hline 0 & 27\end{array}\right|^{3}$
Therefore, $(513,135)=27$, by Lemma 10 we have
$$\{513,135\}=\frac{513 \times 135}{27}=2565 .$$
Since
\begin{tabular}{|c|c|c|}
\hline 1 & 2565 & \multirow{2}{*}{\begin{tabular}{l}
3114 \\
2565
\e... | 887490 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 11 Find the least common multiple of $8127, 11352, 21672$ and 27090. | Since
\begin{tabular}{|c|c|c|}
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
8127 \\
6450
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{r}
11352 \\
8127
\end{tabular}
\end{tabular} \\
\hline 1 & 1677 & 3225 \\
\hline & 1548 & 1677 \\
\hline 12 & 129 & 1548 \\
\hline & & 1548 \\
\hline & 129 & 0 \\... | 3575880 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 12 A steel plate, 1丈 3尺 5寸 long and 1丈 5寸 wide. Now it is to be cut into equally sized squares, the squares must be as large as possible, and no steel plate should be left over. Find the side length of the square.
Note: In traditional Chinese units, 1丈 = 10尺, and 1尺 = 10寸. | Solution: Since we want the largest square, we need to find the largest side length of the square. To find the largest side length of the square, we need to find the greatest common divisor (GCD) of 135 inches and 105 inches. Since
$$135=3^{3} \times 5, \quad 105=3 \times 5 \times 7$$
we have $(135,105)=15$.
Answer: T... | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8. Find the greatest common divisor using the method of prime factorization:
(i) $48,84,120$.
(ii) $360,810,1260,3150$. | 8.
(i) Solution: Decomposing each number into prime factors, we get
$$\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}$$
Therefore, $(48,84,120)=2^{2} \tim... | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
9. Use the Euclidean algorithm to find the greatest common divisor:
(i) 51425, 13310.
(ii) $353430, 530145, 165186$.
(iii) $81719, 52003, 33649, 30107$. | 9
(i) Solution: Since
1 \begin{tabular}{r|r|r}
51425 & 13310 \\
39930 & 11495 \\
\hline 11495 & 1815 & 6 \\
10890 & 1815 \\
\hline 605 & 0
\end{tabular}
Therefore, $(51425,13310)=605$.
(ii) Solution: Since
We get
$$(353430,530145)=176715$$
Also,
$14 \left\lvert\,$\begin{tabular}{r|r|}
176715 & 165186 \\
165186 & 161... | 23 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12. Use the properties of the greatest common divisor from the previous question to find the greatest common divisor of the following:
(i) $216,64,1000$.
(ii) $24000,36000,144000$. | 12.
(i) Solution: The property of the greatest common divisor (GCD) from the previous problem can obviously be extended to more than two numbers, so
$$\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 |
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
$$\begin{array}{... | 25 | Geometry | math-word-problem | Yes | Yes | number_theory | false |
17. A box of grenades, assuming the weight of each grenade is an integer greater than one pound, the net weight after removing the weight of the box is 201 pounds. After taking out several grenades, the net weight is 183 pounds. Prove that the weight of each grenade is 3 pounds. | 17. Proof: Since 201 jin and 183 jin are both the weights of an integer number of grenades, the weight of each grenade must be a common divisor of them. Their greatest common divisor is
$$(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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
Example 3 Find the greatest common divisor of $1008, 1260, 882$ and 1134. | Solve: Decompose these four numbers into prime factors
$$\begin{array}{l}
1008=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7=2^{4} \times 3^{2} \times 7 \\
1260=2 \times 2 \times 3 \times 3 \times 5 \times 7=2^{2} \times 3^{2} \times 5 \times 7 \\
882=2 \times 3^{2} \times 7^{2} \\
1134=2 \times 3^{4} \times ... | 126 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Prove that the remainder of $\left(12371^{16}+34\right)^{23+7+c}$ divided by 111 is equal to 70, where $c$ is any non-negative integer. | Example 2's proof: From $(70,111)=1$ and Example 8's $\left(12371^{56}+\right.$ $34)^{28} \equiv 70(\bmod 111)$, we get $\left(12371^{56}+34,111\right)=1$. Since $111=3 \times 37$ and Lemma 14, we have $\varphi(111)=2 \times 36=72$. Since $c$ is a positive integer and by Theorem 1, we have $\left(12371^{56}+\right.$ $3... | 70 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 26 Find $\sigma(450)=$ ? | Since $45!=2 \times 3 \times 5^{2}$, by Lemma 10 we have
$$\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}$$ | 1209 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 27 Find $\sigma_{2}(28)=$ ? | Since the factors of 28 are $1,2,4,7,14,28$, we have
$$\sigma_{2}(28)=1+2^{2}+4^{2}+7^{2}+14^{2}+28^{2}=1050$$ | 1050 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
side 28 find $\sigma_{3}(62)=$ ? | Since the factors of 62 are $1,2,31,62$, we have
$$\sigma_{3}(62)=1+2^{3}+31^{3}+62^{3}=268128$$ | 268128 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Find the remainder when $\left(12371^{55}+34\right)^{28}$ is divided by 111. | From $12371=111^{2}+50$, we get $12371 \equiv 50(\bmod 111)$. By Lemma 3, we have
$$12371^{56} \equiv 50^{56}(\bmod 111)$$
We also have $(50)^{28}=(125000)^{9}(50), 125000 \equiv 14(\bmod 111)$, so by Lemma 3, we get
$$(50)^{28} \equiv(14)^{9}(50)(\bmod 111)$$
Furthermore, $14^{3} \equiv 80(\bmod 111),(80)^{3} \equiv... | 70 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8. (i) Let $N=9450$, find $\varphi(N)$.
(ii) Find the sum of all positive integers not greater than 9450 and coprime with 9450. | 8. (i) Solution: Given $9450=2 \cdot 3^{3} \cdot 5^{2} \cdot 7$, and by Lemma 14, we have
$$\begin{aligned}
\varphi(N) & =9450\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \\
& =\frac{2 \cdot 3^{3} \cdot 5^{2} \cdot 7 \cdot 2 \cdot 4 \cdot 6}{2 \cdot 3 \cdot 5 ... | 10206000 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12. Find positive integers $\boldsymbol{n}$ and $m, n>m \geqslant 1$, such that the last three digits of $1978^{n}$ and $1978^{m}$ are equal, and make $n+m$ as small as possible. (20th International Mathematical Olympiad Problem) | 12. Solution: $1978^{n}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right)$
$$=2^{m} \cdot 989^{m}\left(1978^{n-m}-1\right)$$
Since the last three digits of $1978^{n}$ and $1978^{m}$ are the same, the last three digits of $1978^{n}-1978^{m}$ are all 0. Therefore, $1978^{n}-1978^{m}$ is divisible by 1000. And $1000=2^{3} \cdot... | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
11. If $3^{k} \mid 1000!$ and $3^{k+1} \nmid 1000!$, find $k$. | 11. Solution: From the result of the previous problem, we have.
$$\begin{aligned}
k= & {\left[\frac{1000}{3}\right]+\left[\frac{1000}{9}\right]+\left[\frac{1000}{27}\right]+\left[\frac{1000}{81}\right] } \\
& +\left[\frac{1000}{243}\right]+\left[\frac{1000}{729}\right] \\
= & 333+111+37+12+4+1=498
\end{aligned}$$ | 498 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Given $p=29$, try to find a primitive root of $p^{2}=841$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let's verify that 14 is a primitive root of 29. Since
$$\varphi(p)=28=(4)(7)=(2)(14)$$
To verify that 14 is indeed a primitive root of \( p = 29 \), it is not necessary to check for all \( m (1 \leqslant m < 28) \) that
$$14^{m} \equiv 1 \pmod{29}$$
but only the following conditions need to be verified (see Theorem 1... | 43 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Theorem 9 We have $g(8) \leqslant 42273$. | First, by applying the method of comparing coefficients or direct expansion, it is easy to verify that the following identity holds:
$$\begin{array}{l}
5040\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{4} \\
\quad=6 \sum(2 a)^{8}+60 \sum(a \pm b)^{8} \\
\quad+\sum(2 a \pm b \pm c)^{8}+6 \sum(a \pm b \pm c+d)^{8}
\end{array}$$
... | 42273 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 1 Find the number of integers from 1 to 1000 that are not divisible by 5, nor by 6 and 8. | We use the notation $\operatorname{LCM}\left\{a_{1}, \cdots, a_{n}\right\}$ to represent the least common multiple of $n$ integers $a_{1}, \cdots, a_{n}$. Let $S$ be the set consisting of the natural numbers from 1 to 1000. Property $P_{1}$ is "an integer is divisible by 5", property $P_{2}$ is "an integer is divisible... | 600 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1. Find the number of positive integers among the first $10^{5}$ that are not divisible by $7, 11, 13$. | 1. Solution: According to Theorem 1 of this chapter, the number of integers sought is
$$\begin{array}{c}
10^{5}-\left[\frac{10^{5}}{7}\right]-\left[\frac{10^{5}}{11}\right]-\left[\frac{10^{5}}{13}\right]+\left[\frac{10^{5}}{7 \times 11}\right] \\
+\left[\frac{10^{5}}{7 \times 13}\right]+\left[\frac{10^{5}}{11 \times 13... | 71929 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. A school organized three extracurricular activity groups in mathematics, Chinese, and foreign language. Each group meets twice a week, with no overlapping schedules. Each student can freely join one group, or two groups, or all three groups simultaneously. A total of 1200 students participate in the extracurricular ... | 2. Solution: Since all 1200 students have joined at least one extracurricular group, the number of students who did not join any group is 0. We use $A_{1}, A_{2}, A_{3}$ to represent the sets of students who joined the math group, the Chinese group, and the English group, respectively. Thus, by the problem statement, w... | 80 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Discuss the congruence equation
$$x^{2} \equiv -286 \pmod{4272943}$$
whether it has a solution, where 4272943 is a prime number. | Let $p=4272943$, by Lemma 7 we have
$$\left(\frac{-286}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\left(\frac{143}{p}\right)$$
Since $4272943 \equiv 7(\bmod 8)$, we have
$$\left(\frac{-1}{p}\right)=-1,\left(\frac{2}{p}\right)=1$$
Thus,
$$\left(\frac{-286}{p}\right)=-\left(\frac{143}{p}\right)$$
Sinc... | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
15. Let $p \geqslant 3$, try to calculate the value of the following expression:
$$\left(\frac{1 \cdot 2}{p}\right)+\left(\frac{2 \cdot 3}{p}\right)+\cdots+\left(\frac{(p-2)(p-1)}{p}\right)$$ | 15. Solution: To solve this problem, we need to study the properties of the Legendre symbol with the general term $\left(\frac{n(n+1)}{p}\right)$, where $(n, p)=1$.
By $(n, p)$ being coprime, we know there must exist an integer $r_{n}$ such that $p \nmid r_{n}$ and $n r_{n} \equiv 1(\bmod p)$. This $r_{n}$ is called t... | -1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Find the smallest positive integer $a$, such that there exists a positive odd integer $n$, satisfying
$$2001 \mid\left(55^{n}+a \cdot 32^{n}\right)$$ | 2. From $2001=3 \times 23 \times 29$ and the conditions, we have
$$\left\{\begin{array}{ll}
a \equiv 1 & (\bmod 3) \\
a \equiv 1 & (\bmod 29), \\
a \equiv-1 & (\bmod 23)
\end{array}\right.$$
From the first two equations, we can set $a=3 \times 29 \times k+1$, substituting into the last equation gives
$$k \equiv 5(\bmo... | 436 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Find the smallest prime $p$ such that there do not exist $a, b \in \mathbf{N}$, satisfying
$$\left|3^{a}-2^{b}\right|=p$$ | 3. Notice that, $2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, 7=2^{3}-3^{0}, 11=3^{3}-$ $2^{4}, 13=2^{4}-3^{1}, 17=3^{4}-2^{6}, 19=3^{3}-2^{3}, 23=3^{3}-2^{2}, 29=2^{5}-3^{1}, 31=$ $2^{5}-3^{0}, 37=2^{6}-3^{3}$. Therefore, the required $p \geqslant 41$.
On the other hand, if $\left|3^{a}-2^{b}\right|=41$, there are tw... | 41 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
29. Let $p$ be a prime, $a, b \in \mathbf{N}^{*}$, satisfying: $p>a>b>1$. Find the largest integer $c$, such that for all $(p, a, b)$ satisfying the conditions, we have
$$p^{c} \mid\left(\mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}\right)$$ | 29. When taking $p=5, a=3, b=2$, we should have $5^{c} \mid 3000$, so $c \leqslant 3$. Below, we prove that for any $p, a, b$ satisfying the conditions, we have $p^{3} \mid \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}$.
In fact, notice that
$$\begin{aligned}
& \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}=\frac{(a p)(a p-1) ... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find the smallest positive integer $n$, such that the indeterminate equation
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$$
has integer solutions $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. | Note that for any $x \in \mathbf{Z}$, if $x$ is even, then $x^{4} \equiv 0(\bmod 16)$; if $x$ is odd, then $x^{2} \equiv 1(\bmod 8)$, and in this case, $x^{4} \equiv 1(\bmod 16)$.
The above discussion shows that $x_{i}^{4} \equiv 0$ or $1(\bmod 16)$, thus the remainder of $x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}$ modulo ... | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
7. Let $a, b, c, d$ all be prime numbers, satisfying: $a>3 b>6 c>12 d$, and
$$a^{2}-b^{2}+c^{2}-d^{2}=1749$$
Find the value of $a^{2}+b^{2}+c^{2}+d^{2}$. | 7. From the conditions, we know that $a, b, c$ are all odd numbers. If $d$ is odd, then $a^{2}-b^{2}+c^{2}-d^{2}$ is even, which is a contradiction. Therefore, $d$ is even, and thus $d=2$. Consequently, $a^{2}-b^{2}+c^{2}=1753$. From the conditions, we also know that $a \geqslant 3 b+2, b \geqslant 2 c+1, c \geqslant 5... | 1999 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Let $m, n \in \mathbf{N}^{*}, m$ be an odd number. Prove: $\left(2^{m}-1,2^{n}+1\right)=1$.
| 2. Let $\left(2^{m}-1,2^{n}+1\right)=d$, then
$$1 \equiv\left(2^{m}\right)^{n}=\left(2^{n}\right)^{m} \equiv(-1)^{m}=-1(\bmod d)$$
This leads to $d \mid 2$. Combining this with $d$ being odd, we conclude $d=1$. | 1 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 3 Let the function $f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$ be defined as follows: Let
$$\frac{(2 n)!}{n!(n+1000)!}=\frac{A(n)}{B(n)}$$
where $n \in \mathbf{N}^{*}, A(n), B(n)$ are coprime positive integers. If $B(n)=1$, then $f(n)=1$; if $B(n)>1$, then $f(n)$ is the largest prime factor of $B(n)$.
Prov... | Proof:
The idea of the proof is to find a constant such that for any $n \in \mathbf{N}^{*}$, the product of $\frac{A(n)}{B(n)}$ and this constant is a positive integer, thereby deducing that $f$ is a bounded function.
First, we prove a lemma: For any non-negative real numbers $x, y$,
$$[2 x]+[2 y] \geqslant[x]+[x+y]$$... | 1999 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 5 Let the set of all integer points (points with integer coordinates) in the plane be denoted as $S$. It is known that for any $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ in $S$, there exists another point $P$ in $S$ such that the segments $A_{i} P(i=1,2, \cdots, n)$ do not contain any points from $S$ internally. ... | The maximum value of the required $n$ is 3.
On one hand, when $n \geqslant 4$, take 4 points $A(0,0), B(1,0), C(0,1)$, $D(1,1)$ in $S$ (the remaining $n-4$ points are chosen arbitrarily), then for any other point $P$ in $S$, analyzing the parity of the coordinates of $P$, there are only 4 cases: (odd, odd), (odd, even)... | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
2. In decimal notation, how many $m \in\{1,2, \cdots, 2009\}$ are there such that there exists $n \in \mathbf{N}^{*}$, satisfying: $S\left(n^{2}\right)=m$? Here $S(x)$ denotes the sum of the digits of the positive integer $x$. | 2. Notice that, perfect squares $\equiv 0,1,4,7(\bmod 9)$, and for $k \in \mathbf{N}^{*}$, let $n=$ $\underbrace{11 \cdots 1} \underbrace{22 \cdots 25}$, then $S(n)=3 k+4$, and
$k-1$ 个 $\underbrace{2+0}_{k \text { 个 }}$
$$\begin{aligned}
n & =\frac{10^{k-1}-1}{9} \times 10^{k+1}+\frac{2 \times\left(10^{k}-1\right)}{9} ... | 893 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6. Let \( x=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{100000}} \). Find the value of \([x]\). | 6. Notice that, when $k \geqslant 2$, we have
$$\begin{array}{l}
\frac{1}{\sqrt{k}}=\frac{2}{2 \sqrt{k}}\frac{2}{\sqrt{k}+\sqrt{k+1}}=2(\sqrt{k+1}-\sqrt{k}) .
\end{array}$$
Therefore,
$$\begin{aligned}
x & =1+\sum_{k=2}^{10^{6}} \frac{1}{\sqrt{k}}2 \sum_{k=1}^{10^{6}}(\sqrt{k+1}-\sqrt{k}) \\
& =2\left(\sqrt{10^{6}+1}-... | 1998 | Calculus | math-word-problem | Yes | Yes | number_theory | false |
25. Find the largest positive integer $m$, such that for $k \in \mathbf{N}^{*}$, if $1<k<m$ and $(k, m)=1$, then $k$ is a power of some prime.
---
The translation maintains the original format and line breaks as requested. | 25. The required maximum positive integer $m=60$.
On the one hand, it can be directly verified that $m=60$ meets the requirement (it is only necessary to note that the product of the smallest two primes coprime with 60, 7 and 11, is greater than 60).
On the other hand, let $p_{1}, p_{2}, \cdots$ represent the sequenc... | 60 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the number of positive integers $a$ that satisfy the following condition: there exist non-negative integers $x_{0}$, $x_{1}, \cdots, x_{2009}$, such that
$$a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2009}} .$$ | Solution: Clearly, $a$ that satisfies the condition is greater than 1. Taking both sides of (1) modulo $(a-1)$, we get a necessary condition as
$$1^{x_{0}} \equiv 1^{x_{1}}+1^{x_{2}}+\cdots+1^{x_{2009}}(\bmod a-1)$$
This indicates: $(a-1) \mid 2008$, the number of such $a$ is $=d(2008)=8$.
On the other hand, let $a \i... | 8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 By Fermat's Little Theorem, for any odd prime $p$, we have $2^{p-1} \equiv 1(\bmod p)$. Question: Does there exist a composite number $n$ such that $2^{n-1} \equiv 1(\bmod n)$ holds? | $$\begin{array}{l}
\text { Hence } \\
\text { Therefore }
\end{array}$$
$$\begin{array}{l}
2^{10}-1=1023=341 \times 3 \\
2^{10} \equiv 1(\bmod 341) \\
2^{340} \equiv 1^{34} \equiv 1(\bmod 341)
\end{array}$$
Hence 341 meets the requirement.
Furthermore, let $a$ be an odd composite number that meets the requirement, the... | 341 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 It is known that among 4 coins, there may be counterfeit coins, where genuine coins each weigh 10 grams, and counterfeit coins each weigh 9 grams. Now there is a balance scale, which can weigh the total weight of the objects on the tray. Question: What is the minimum number of weighings needed to ensure that ... | At least 3 weighings can achieve this.
In fact, let the 4 coins be $a, b, c, d$. Weigh $a+b+c, a+b+d$, and $a+c+d$ three times. The sum of these three weights is $3a + 2(b+c+d)$, so if the sum of these three weights is odd, then $a$ is the counterfeit coin; otherwise, $a$ is genuine. Once $a$ is determined, solving the... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false |
Example 6 A cube with a side length of 3 is divided into 27 unit cubes. The numbers $1, 2, \cdots$, 27 are randomly placed into the unit cubes, one number in each. Calculate the sum of the 3 numbers in each row (horizontal, vertical, and column), resulting in 27 sum numbers. Question: What is the maximum number of odd ... | Solve: To calculate the sum $S$ of these 27 sums, since each number appears in exactly 3 rows, we have
$$S=3 \times(1+2+\cdots+27)=3 \times 27 \times 14$$
Thus, $S$ is an even number, so the number of odd numbers among these 27 sums must be even.
If 26 of these 27 sums are odd, let's assume that the even number is the... | 24 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Consider the following sequence:
$$101,10101,1010101, \cdots$$
Question: How many prime numbers are there in this sequence? | It is easy to know that 101 is a prime number. Next, we prove that this is the only prime number in the sequence.
Let $a_{n}=\underbrace{10101 \cdots 01}_{n \uparrow 01}$, then when $n \geqslant 2$, we have
$$\begin{aligned}
a_{n} & =10^{2 n}+10^{2(n-1)}+\cdots+1 \\
& =\frac{10^{2(n+1)}-1}{10^{2}-1} \\
& =\frac{\left(1... | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Find the smallest positive integer $n$, such that there exist integers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$$ | From property 1, for any integer $a$, we know that
$$a^{2} \equiv 0(\bmod 4) \text { or } a^{2} \equiv 1(\bmod 8),$$
From this, we can deduce that
$$a^{4} \equiv 0 \text { or } 1(\bmod 16).$$
Using this conclusion, if $n<15$, let
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4} \equiv m(\bmod 16),$$
then
and
$$\begin{array}{c... | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1 Let $p$ and $q$ both be prime numbers, and $7p + q$, $pq + 11$ are also prime numbers. Find the value of $\left(p^{2} + q^{p}\right)\left(q^{2} + p^{q}\right)$. | 1. If $p$ and $q$ are both odd, then $7p + q$ is even, and it is not a prime number. Therefore, one of $p$ and $q$ must be even.
Case one: Suppose $p$ is even, then $p = 2$. In this case, since $7p + q$ is a prime number, $q$ must be an odd prime. If $q \neq 3$, then $q \equiv 1$ or $2 \pmod{3}$.
If $q \equiv 1 \pmod... | 221 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 Let $p_{1}<p_{2}<p_{3}<p_{4}<p_{5}$ be 5 prime numbers, and $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}$ form an arithmetic sequence. Find the minimum value of $p_{5}$. | 2. Let $d$ be the common difference, then $p_{1}, p_{1}+d, p_{1}+2 d, p_{1}+3 d, p_{1}+4 d$ are all primes. If $2 \nmid d$, i.e., $d$ is odd, then one of $p_{1}+d$ and $p_{1}+2 d$ is even, and it is not a prime. If $3 \nmid d$, then one of $p_{1}+d, p_{1}+2 d, p_{1}+3 d$ is a multiple of 3 (they form a complete residue... | 29 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4 Find the largest positive integer $k$, such that there exists a positive integer $n$, satisfying $2^{k} \mid 3^{n}+1$.
| 4. Notice that, when $n$ is even, let $n=2 m$, we have
$$3^{n}=9^{m} \equiv 1(\bmod 8)$$
When $n=2 m+1$,
$$3^{n}=9^{m} \times 3 \equiv 3(\bmod 8)$$
Therefore, for any positive integer $n$, we have
$$3^{n}+1 \equiv 2 \text { or } 4(\bmod 8),$$
so $k \leqslant 2$. Also, $2^{2} \mid 3^{1}+1$, hence, the maximum value o... | 2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8 Fibonacci sequence $\left\{F_{n}\right\}$ is defined as follows: $F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n}, n=1$, $2, \cdots$.
(1) Prove: The sum of any 10 consecutive terms of this sequence is a multiple of 11;
(2) Find the smallest positive integer $k$, such that the sum of any $k$ consecutive terms of this sequence is... | 8. Consider the sequence $\left\{F_{n}\right\}$ where each term is taken modulo 11 (or 12).
(1) $\left\{F_{n}(\bmod 11)\right\}: 1,1,2,3,5,-3,2,-1,1,0,1,1, \cdots$, so $\left\{F_{n}(\bmod 11)\right\}$ is a purely periodic sequence with a period of 10. Therefore,
the sum of any 10 consecutive terms in $\left\{F_{n}\righ... | 36 | Number Theory | proof | Yes | Yes | number_theory | false |
14 Labeled as $1,2, \cdots, 100$, there are some matches in the matchboxes. If each question allows asking about the parity of the sum of matches in any 15 boxes, then to determine the parity of the number of matches in box 1, at least how many questions are needed? | 14. At least 3 questions are needed.
First, prove that "3 questions are sufficient." For example:
The first question is: $a_{1}, a_{2}, \cdots, a_{15}$;
The second question is: $a_{1}, a_{2}, \cdots, a_{8}, a_{16}, a_{17}, \cdots, a_{22}$;
The third question is: $a_{1}, a_{9}, a_{10}, \cdots, a_{22}$.
Here, $a_{i}$ re... | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
17 Find the number of all positive integers $a$ that satisfy the following condition: there exist non-negative integers $x_{0}, x_{1}, x_{2}, \cdots$, $x_{2001}$, such that $a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}}$. | 17. If $a$ is a number that satisfies the condition, then $a^{x_{0}}>1$, so $a>1$. At this point, taking
both sides modulo $a-1$, we know
so $\square$
$$\begin{array}{c}
a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}} \\
1 \equiv \underbrace{1+\cdots+1}_{2001 \uparrow}(\bmod a-1), \\
a-1 \mid 2000 .
\end{array}$$
O... | 20 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
20 Find the smallest positive integer $a$, such that for any integer $x$, we have $65 \mid\left(5 x^{13}+13 x^{5}+9 a x\right)$. | 20. From the condition, we know $65 \mid (18 + 9a)$ (take $x=1$), and since $(9,65)=1$, it follows that $65 \mid a+2$, hence $a \geq 63$.
When $a=63$, using Fermat's Little Theorem, we know that for any integer $x$, we have
$$\begin{aligned}
& 5 x^{13} + 13 x^{5} + 9 a x \\
\equiv & 13 x + 9 a x \\
\equiv & (3 + (-1) ... | 63 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
23 Find the number of integer pairs $(a, b)$ that satisfy the following conditions: $0 \leqslant a, b \leqslant 36$, and $a^{2}+b^{2}=$ $0(\bmod 37)$. | 23. Notice that, $a^{2}+b^{2} \equiv a^{2}-36 b^{2}(\bmod 37)$, so from the condition we have
$$37 \mid a^{2}-36 b^{2},$$
which means
$$37 \mid(a-6 b)(a+6 b),$$
thus
$37 \mid a-6 b$ or $37 \mid a+6 b$.
Therefore, for each $1 \leqslant b \leqslant 36$, there are exactly two $a(a \equiv \pm 6 b(\bmod 37))$ that satisfy... | 73 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
26 Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$.
Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$. | 26. Let $n=1000 x+y$, where $x$ is a positive integer, $y$ is an integer, and $0 \leqslant y \leqslant 999$. According to the problem,
$$x^{3}=1000 x+y$$
From $0 \leqslant y \leqslant 999$, we know
$$1000 x \leqslant x^{3}<1000 x+1000=1000(x+1)$$
Thus,
$$x^{2} \geqslant 1000, x^{3}+1 \leqslant 1000(x+1)$$
Therefore,... | 32768 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
28 Find the smallest positive integer $n$, such that in decimal notation $n^{3}$ ends with the digits 888. | 28. From the condition, we know $n^{3} \equiv 888(\bmod 1000)$, hence
$$n^{3} \equiv 888(\bmod 8), n^{3} \equiv 888(\bmod 125)$$
From the former, we know $n$ is even, let $n=2 m$, then
$$m^{3} \equiv 111(\bmod 125)$$
Therefore
$$m^{3} \equiv 111 \equiv 1(\bmod 5)$$
Noting that when $m=0,1,2,3,4(\bmod 5)$, correspond... | 192 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
33 Find the largest positive integer that cannot be expressed as the sum of a positive multiple of 42 and a composite number.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 33. For any positive integer $n$ that cannot be expressed as a positive multiple of 42 and a composite number, consider the remainder $r$ when $n$ is divided by 42. If $r=0$ or $r$ is a composite number, then $n \leqslant 42$.
Now consider the case where $r=1$ or $r$ is a prime number.
If $r \equiv 1(\bmod 5)$, then
$... | 215 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
34 Find a positive integer $n$, such that each number $n, n+1, \cdots, n+20$ is not coprime with 30030. | 34. Since $30030=2 \times 3 \times 5 \times 7 \times 11 \times 13$, if we take $N=210 k$, then $N$ and $N \pm r$ are not coprime with 30030, where $r$ is a number in $2,3, \cdots, 10$. Now consider the numbers $N \pm 1$. We choose $k$ such that
$$210 k \equiv 1(\bmod 11) \text { and } 210 k \equiv-1(\bmod 13),$$
The f... | 9440 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
37 points are arranged on a circle, and one of the points is labeled with the number 1. Moving clockwise, label the next point with the number 2 after counting two points, then label the next point with the number 3 after counting three points, and so on, labeling the points with the numbers $1,2, \cdots, 2000$. In thi... | 37. Equivalent to finding the smallest positive integer $n$, such that
$$1+2+\cdots+n \equiv 1+2+\cdots+2000(\bmod 2000)$$
i.e. $\square$
$$\frac{n(n+1)}{2} \equiv 1000(\bmod 2000),$$
which is equivalent to
$$n(n+1) \equiv 2000(\bmod 4000),$$
This requires
$$2000 \mid n(n+1)$$
Notice that
$$(n, n+1)=1$$
and
$$2000... | 624 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
38 There are 800 points on a circle, labeled $1,2, \cdots, 800$ in a clockwise direction, dividing the circle into 800 gaps. Now, choose one of these points and color it red, then perform the following operation: if the $k$-th point is colored red, then move $k$ gaps in a clockwise direction and color the point reached... | 38. This is equivalent to finding the maximum number of distinct numbers in the sequence $a, 2a, 2^2a, 2^3a, \cdots$ under modulo 800, where $a$ takes values in $1, 2, \cdots, 800$.
Notice that when $2^n \not\equiv 2^m \pmod{800}$, it is not necessarily true that $2^n a \not\equiv 2^m a \pmod{800}$. Conversely, when $... | 24 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the number of positive integer solutions to the indeterminate equation
$$7 x+19 y=2012$$ | Solve: First, find a particular solution of (1).
$$x=\frac{1}{7}(2012-19 y)=287-3 y+\frac{1}{7}(3+2 y) .$$
Therefore, $\frac{1}{7}(3+2 y)$ must be an integer. Taking $y_{0}=2$, then $x_{0}=282$.
Using the conclusion of Theorem 2, the general solution of equation (1) is
$$\left\{\begin{array}{l}
x=282-19 t, \\
y=2+7 t ... | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find the number of positive integer solutions to the indeterminate equation
$$x+2 y+3 z=2012$$ | Let $(x, y, z)$ be a positive integer solution to (1), then $3 z \leqslant 2009$, i.e., $1 \leqslant z \leqslant 669$, which respectively yield
$$x+2 y=2009,2006, \cdots, 5$$
Correspondingly, the range of values for $y$ are
$$\begin{array}{l}
1 \leqslant y \leqslant 1004,1 \leqslant y \leqslant 1002,1 \leqslant y \leq... | 336340 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 12 Let $m, n$ be positive integers, and $n>1$. Find the minimum value of $\left|2^{m}-5^{n}\right|$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note is for you, the assistant, and should not be included... | Since $\left|2^{m}-5^{n}\right|$ is an odd number, and when $m=7, n=3$, $\left|2^{m}-5^{n}\right|=3$, if we can prove that when $n>1$, $\left|2^{m}-5^{n}\right| \neq 1$, then the minimum value sought is 3.
If there exist positive integers $m, n$, such that $n>1$, and $\left|2^{m}-5^{n}\right|=1$, then
$$2^{m}-5^{n}=1 ... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1 Given that the ages of A, B, and C are all positive integers, A's age does not exceed twice B's age, B is 7 years younger than C, the sum of the three people's ages is a prime number less than 70, and the sum of the digits of this prime number is 13. Question: What is the maximum age of A? | 1. Let Jia's age be $x$ years, and Yi's age be $y$ years, then Bing is $y+7$ years old, and $x \leqslant 2 y$.
Since the positive integers less than 70 and with a digit sum of 13 are only $49$, $58$, and $67$, the sum of the three people's ages (which is a prime number) can only be 67 years, i.e.,
$$x+y+(y+7)=67$$
Th... | 30 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4. Positive integers $a, b, c, d$ satisfy: $1<a<b<c<d<1000$, and $a+d=b+c$, $bc-ad=2004$. Find the number of all such positive integer tuples $(a, b, c, d)$. | 4. Let $b=a+x, c=a+y$, then $x<y$, and $d=a+x+y$ (this is obtained from $a+d=b+c$), thus
$$b c-a d=(a+x)(a+y)-a(a+x+y)=x y$$
That is
$$x y=2004$$
Combining $a+x+y<1000$ and $2004=2^{2} \times 3 \times 167$, we know that $(x, y)=(3,668),(4,501),(6,334),(12,167)$.
Accordingly, $1<a<329,1<a<495,1<a<660,1<a<821$. Thus, ... | 2297 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5 Find the smallest positive integer $c$, such that the indeterminate equation $x y^{2}-y^{2}-x+y=c$ has exactly three sets of positive integer solutions.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. Factorizing the left side of the equation, we get
$$(y-1)(x y+x-y)=c$$
Notice that, for any positive integer $c$, there is a solution
$$(x, y)=(1, c+1)$$
When $c$ is a prime number, there is at most one additional set of positive integer solutions. Therefore, to make the equation have exactly 3 sets of positive in... | 10 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
7 rectangles can be divided into $n$ identical squares, and it can also be divided into $n+76$ identical squares, find the value of the positive integer $n$.
---
Note: The translation keeps the original format and line breaks as requested. However, the phrase "7-个长方形" is translated as "7 rectangles" for clarity in En... | 7. Let the side length of the squares when divided into $n$ squares be $x$, and the side length when divided into $n+76$ squares be $y$, then
$$n x^{2}=(n+76) y^{2}$$
Since both divisions are performed on the same rectangle, $\frac{x}{y}$ is a rational number (this can be seen by considering one side of the rectangle)... | 324 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8 Positive integers $x, y, z$ satisfy $\left\{\begin{array}{l}7 x^{2}-3 y^{2}+4 z^{2}=8, \\ 16 x^{2}-7 y^{2}+9 z^{2}=-3 .\end{array}\right.$ Find the value of $x^{2}+y^{2}+z^{2}$. | 8. Let $x, y, z$ satisfy
$$\left\{\begin{array}{l}
7 x^{2}-3 y^{2}+4 z^{2}=8 \\
16 x^{2}-7 y^{2}+9 z^{2}=-3
\end{array}\right.$$
Multiplying (1) by 7 and (2) by 3, we get
Substituting back into (1) yields
$$\begin{array}{c}
x^{2}+z^{2}=65 \\
3 x^{2}-3 y^{2}=8-260
\end{array}$$
Thus,
$$\begin{aligned}
y^{2}-x^{2} & =... | 165 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
15 Positive integers $a, b, c$ satisfy: $[a, b]=1000,[b, c]=2000,[c, a]=2000$. Find the number of such ordered positive integer triples $(a, b, c)$. | 15. From the conditions, we can set
$$a=2^{\alpha_{1}} \cdot 5^{\beta_{1}}, b=2^{\alpha_{2}} \cdot 5^{\beta_{2}}, c=2^{\alpha_{3}} \cdot 5^{\beta_{3}}$$
Then
$$\begin{array}{l}
\max \left\{\alpha_{1}, \alpha_{2}\right\}=3, \max \left\{\alpha_{2}, \alpha_{3}\right\}=4, \max \left\{\alpha_{3}, \alpha_{1}\right\}=4 \\
\m... | 70 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
16. Let $x, y$ be positive integers, and $y>3, x^{2}+y^{4}=2\left((x-6)^{2}+(y+1)^{2}\right)$. Prove: $x^{2}+y^{4}=1994$. | 16. Transpose and expand, we get
that is
$$\begin{array}{c}
y^{4}-2 y^{2}-4 y-2=x^{2}-24 x+72 \\
(x-12)^{2}=y^{4}-2 y^{2}-4 y+70
\end{array}$$
Notice that
$$\begin{aligned}
\left(y^{2}-2\right)^{2} & =y^{4}-4 y^{2}+4 \\
& 3 \text{ holds, and } y^{4}-2 y^{2}-4 y+70 \text{ is a perfect square, so it can only be or }
$$... | 1994 | Algebra | proof | Yes | Yes | number_theory | false |
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