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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 particular solution (please verify this yourself). By formula (15), the number of solutions is
$$\begin{array}{l}
-[-11 \cdot 3189 / 37]-[30 \cdot 3189 / 101]-1 \\
\quad=949-947-1=1
\end{array}$$
That is, the equation has exactly one positive solution. Please find this solution yourself.
|
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, then went back to sleep. After a while, the second sailor woke up and did the same thing. When he divided the remaining coconuts into five equal parts, there was also one left over, which he gave to the monkey, and then he hid his share and went back to sleep. The other three sailors also did the same thing in turn. The next morning, they woke up and pretended nothing had happened, dividing the remaining coconuts into five equal parts, with none left over. How many coconuts were there originally, and how many coconuts did each of them get in total?
|
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
$$3^{404} \equiv 1(\bmod 10)$$
Therefore, $3^{406} \equiv 3^{404} \cdot 3^{2} \equiv 9(\bmod 10)$. So the unit digit is 9.
|
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, by Example 5 in Chapter 1, §4, we know that for $3^{d} \equiv 1(\bmod 25)$ to hold, $4 \mid d$ must be true. Therefore, calculate:
$$\begin{array}{c}
3^{4} \equiv 81 \equiv 6(\bmod 25), \quad 3^{8} \equiv 36 \equiv 11(\bmod 25) \\
3^{12} \equiv 66 \equiv-9(\bmod 25), \quad 3^{16} \equiv-54 \equiv-4(\bmod 25) \\
3^{20} \equiv-24 \equiv 1(\bmod 25)
\end{array}$$
From this and $3^{20} \equiv 1(\bmod 4)$, by Property IX, we derive $3^{20} \equiv 1(\bmod 100), 3^{400} \equiv 1(\bmod 100)$. Therefore, $3^{406} \equiv 3^{400} \cdot 3^{6} \equiv 3^{6} \equiv 29(\bmod 100)$. So, the unit digit is 9, and the tens digit is 2.
If we do not use the property $4 \mid d$, we have to calculate the remainder $b_{j}$ of $3^{j}$ modulo 25 (to facilitate the calculation of $b_{j}$, the absolute smallest remainder should be taken), as shown in the table below:
\begin{tabular}{|c|cccccccccc|}
\hline$j$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$b_{j}$ & 3 & 9 & 2 & 6 & -7 & 4 & 12 & 11 & 8 & -1 \\
\hline
\end{tabular}
From the $3^{10} \equiv-1(\bmod 25)$ obtained here, we conclude $3^{20} \equiv 1(\bmod 25)$.
|
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
$$2^{n} \cdot 989^{n}\left(1978^{k}-1\right) \equiv 0\left(\bmod 2^{3} \cdot 5^{3}\right)$$
By Property VII and Property IX, it is equivalent to
$$\left\{\begin{array}{l}
2^{n} \equiv 0\left(\bmod 2^{3}\right) \\
1978^{k}-1 \equiv 0\left(\bmod 5^{3}\right)
\end{array}\right.$$
From (10), we know $n \geqslant 3$. Next, we find the $k$ that satisfies (11). First, find the smallest $l$ such that
$$1978^{t}-1 \equiv 0(\bmod 5)$$
holds, denoted as $d_{1}$. Since
$$1978 \equiv 3(\bmod 5)$$
we have $d_{1}=4$. Next, find the smallest $h$ such that
$$1978^{h}-1 \equiv 0\left(\bmod 5^{2}\right)$$
holds, denoted as $d_{2}$. From Chapter 1, §4, Example 5, we know $4 \mid d_{2}$. Noting that
$$1978 \equiv 3\left(\bmod 5^{2}\right)$$
From the calculation in Example 2, we have $d_{2}=20$. Finally, find the smallest $k$ such that
$$1978^{k}-1 \equiv 0\left(\bmod 5^{3}\right)$$
holds, denoted as $d_{3}$. From Chapter 1, §4, Example 5, we know $20 \mid d_{3}$. Noting that
$$\begin{array}{l}
1978 \equiv-22\left(\bmod 5^{3}\right) \\
\begin{aligned}
(-22)^{20} & \equiv(25-3)^{20} \equiv 3^{20} \equiv(243)^{4} \\
& \equiv 7^{4} \equiv(50-1)^{2} \equiv 26\left(\bmod 5^{3}\right)
\end{aligned}
\end{array}$$
Through calculation, we get
$$\begin{array}{c}
1978^{20} \equiv 26\left(\bmod 5^{3}\right), \quad 1978^{40} \equiv(25+1)^{2} \equiv 51\left(\bmod 5^{3}\right) \\
1978^{60} \equiv(25+1)(50+1) \equiv 76\left(\bmod 5^{3}\right) \\
1978^{80} \equiv(50+1)^{2} \equiv 101\left(\bmod 5^{3}\right) \\
1978^{100} \equiv(100+1)(25+1) \equiv 1\left(\bmod 5^{3}\right)
\end{array}$$
Therefore, $d_{3}=100$. So, from Chapter 1, §4, Example 5, we know $100 \mid k$, and the smallest $k=100$. Thus, the necessary and sufficient conditions for (10) and (11), i.e., (9) to hold are
$$n \geqslant 3, \quad 100 \mid m-n$$
Therefore, the smallest $m+n=(m-n)+2 n=106$. If the congruence concept and its properties are not used to solve this problem, it would be very complicated.
|
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
$$1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$$
Therefore, we can take $M_{1}^{-1}=1$. From $M_{2} \equiv(-2) \cdot(2) \cdot 1 \equiv 1(\bmod 5)$, we have
$$1 \equiv M_{2} M_{2}^{-1} \equiv M_{2}^{-1}(\bmod 5)$$
Therefore, we can take $M_{2}^{-1}=1$. From $M_{3} \equiv 3 \cdot 5 \cdot 4 \equiv 4(\bmod 7)$, we have
$$1 \equiv M_{3} M_{3}^{-1} \equiv 4 M_{3}^{-1}(\bmod 7)$$
Therefore, we can take $M_{3}^{-1}=2$. From $M_{4} \equiv 3 \cdot 5 \cdot 7 \equiv 4 \cdot 7 \equiv 6(\bmod 11)$, we have
$$1 \equiv M_{4} M_{4}^{-1} \equiv 6 M_{4}^{-1}(\bmod 11)$$
Therefore, we can take $M_{4}^{-1}=2$. Thus, by Theorem 1, the solution to the system of congruences is
$$\begin{aligned}
x \equiv & (5 \cdot 7 \cdot 11) \cdot 1 \cdot 1+(3 \cdot 7 \cdot 11) \cdot 1 \cdot(-1)+(3 \cdot 5 \cdot 11) \cdot 2 \cdot 2 \\
& +(3 \cdot 5 \cdot 7) \cdot 2 \cdot(-2)(\bmod 3 \cdot 5 \cdot 7 \cdot 11),
\end{aligned}$$
i.e., $\square$
$$x \equiv 385-231+660-420 \equiv 394(\bmod 1155)$$
|
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
$$\begin{aligned}
\left(\frac{137}{227}\right) & =\left(\frac{-90}{227}\right)=\left(\frac{-1}{227}\right)\left(\frac{2 \cdot 3^{2} \cdot 5}{227}\right) \\
& =(-1)\left(\frac{2}{227}\right)\left(\frac{3^{2}}{227}\right)\left(\frac{5}{227}\right) \\
& =(-1)\left(\frac{2}{227}\right)\left(\frac{5}{227}\right) .
\end{aligned}$$
By Theorem 3, we get
$$\left(\frac{2}{227}\right)=-1$$
By Theorem 5, Theorem 1, and Theorem 3, we get
$$\left(\frac{5}{227}\right)=\left(\frac{227}{5}\right)=\left(\frac{2}{5}\right)=-1$$
From the above three equations, we get
$$\left(\frac{137}{227}\right)=-1$$
This indicates that the congruence equation $x^{2} \equiv 137(\bmod 227)$ has no solution.
|
-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
$$x^{7}-x=\left(x^{3}-x^{2}+3 x-3\right)\left(x^{3}+x^{2}-2 x-2\right) x+7 x\left(x^{2}-1\right) .$$
Therefore, the number of solutions to the original congruence equation is 3.
|
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 the conditions, they are $\{10,10,1\}$, $\{5,5,2\},\{2,2,5\},\{1,1,10\}$, and permutations, giving a total of $3 \cdot 4=12$ solutions; (iii) All three are distinct, any three distinct values can be chosen, they are $\{10,5,2\},\{10,5,1\},\{10,2,1\},\{5,2,1\}$ and permutations, giving a total of $6 \cdot 4=24$ solutions. Multiplying each number by 10 gives the solutions to the original problem, with a total of 36 solutions.
|
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:
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 23) \\
2^{11} \equiv\left(2^{4}\right)^{2} \cdot 2^{3} \equiv(-7)^{2} \cdot 8 \equiv 3 \cdot 8 \equiv 1(\bmod 23)
\end{array}$$
So $\delta_{23}(2)=11,2$ is not a primitive root modulo 23. Next, find $\delta_{23}(3)$:
$$\begin{array}{c}
3^{2} \equiv 9(\bmod 23), \quad 3^{3} \equiv 4(\bmod 23) \\
3^{11} \equiv\left(3^{3}\right)^{3} \cdot 3^{2} \equiv 4^{3} \cdot 9 \equiv(-5) \cdot 9 \equiv 1(\bmod 23)
\end{array}$$
So, $\delta_{23}(3)=11,3$ is not a primitive root modulo 23. Next, find $\delta_{23}(4)$.
$$4^{2} \equiv-7(\bmod 23), \quad 4^{11} \equiv\left(4^{4}\right)^{2} \cdot 4^{3} \equiv 3^{2} \cdot(-5) \equiv 1(\bmod 23)$$
So $\delta_{23}(4)=11,4$ is not a primitive root modulo 23. Next, find $\delta_{23}(5)$.
$$\begin{aligned}
5^{2} & \equiv 2(\bmod 23) \\
5^{11} & \equiv\left(5^{4}\right)^{2} \cdot 5^{3} \equiv 4^{2} \cdot 10 \\
& \equiv 4 \cdot(-6) \equiv-1(\bmod 23) \\
5^{22} & \equiv 1(\bmod 23)
\end{aligned}$$
So, $\delta_{23}(5)=22,5$ is a primitive root modulo 23, and it is the smallest positive primitive root.
To further find the primitive roots modulo $p^{\alpha}, 2 p^{\alpha}$, we need to verify whether equation (8) holds when $g=5, p=23$. This is essentially finding the residue of $g^{p-1}$ modulo $p^{2}$. Here $23^{2}=529$.
$$\begin{aligned}
5^{2} & \equiv 25\left(\bmod 23^{2}\right) \\
5^{8} & \equiv(23+2)^{4} \equiv 4 \cdot 23 \cdot 2^{3}+2^{4} \equiv 10 \cdot 23-7\left(\bmod 23^{2}\right) \\
5^{10} & \equiv(10 \cdot 23-7)(23+2) \equiv 13 \cdot 23-14 \\
& \equiv 12 \cdot 23+9\left(\bmod 23^{2}\right) \\
5^{20} & \equiv(12 \cdot 23+9)^{2} \equiv 216 \cdot 23+81 \\
& \equiv 13 \cdot 23-11\left(\bmod 23^{2}\right) \\
5^{22} & \equiv(13 \cdot 23-11)(23+2) \equiv 15 \cdot 23-22 \\
& \equiv 1+14 \cdot 23\left(\bmod 23^{2}\right)
\end{aligned}$$
Since $23 \nmid 14$ and 5 is odd, it proves that 5 is a primitive root for all moduli $23^{a}, 2 \cdot 23^{a}$.
|
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 $d \mid 20, d<20$ when $2^{d} \not \equiv 1(\bmod 41)$, we have $\delta_{41}(2)=20$. Therefore, 2 is not a primitive root.
$$3^{2} \equiv 9(\bmod 41), \quad 3^{4} \equiv-1(\bmod 81), \quad 3^{8} \equiv 1(\bmod 41)$$
Similarly, $\delta_{41}(3)=8$. So 3 is also not a primitive root.
Notice that $\left[\delta_{41}(2), \delta_{41}(3)\right]=[20,8]=40$. Therefore, we do not need to calculate the orders of $4,5, \cdots$ one by one, but can use property XI in $\S 1$ to find the primitive root. Notice that
$$\delta_{41}(2)=4 \cdot 5, \quad \delta_{41}(3)=1 \cdot 8$$
By property XI, $c=2^{4} \cdot 3=48$ is a primitive root. Therefore, 7 is also a primitive root modulo 41. This method of finding primitive roots is essentially the proof 1 of Theorem 2, so proof 1 can be a method to find primitive roots in certain cases.
To find the primitive roots of $41^{\alpha}, 2 \cdot 41^{\alpha}$, we need to calculate $7^{40}$ modulo $41^{2}=1681$.
$$\begin{array}{l}
7^{2} \equiv 41+8\left(\bmod 41^{2}\right) \\
7^{4} \equiv(41+8)^{2} \equiv 18(41-1)\left(\bmod 41^{2}\right) \\
7^{5} \equiv 3 \cdot(41+1)(41-1) \equiv-3\left(\bmod 41^{2}\right) \\
7^{10} \equiv 9\left(\bmod 41^{2}\right) \\
7^{20} \equiv 81(\bmod 41)^{2} \\
7^{40} \equiv(2 \cdot 41-1)^{2} \equiv 1+(-4) \cdot 41\left(\bmod 41^{2}\right)
\end{array}$$
Since $41 \nmid-4$ and 7 is odd, it follows that 7 is a primitive root for all moduli $41^{\alpha}, 2 \cdot 41^{\alpha}$.
|
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 \\
5618 & 2226 \\
\hline 1113 & 583 & 1 \\
583 & 530 \\
\hline 530 & 53 & \\
530 & $(0)$ & \\
\hline 0 & 53 &.
\end{tabular}
Let $n \geqslant 3$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}$ be positive integers. Here we introduce a method to find the greatest common divisor of these $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$: We first find the greatest common divisor of $a_{1}$ and $a_{2}$, and if the greatest common divisor of $a_{1}$ and $a_{2}$ is $b_{1}$. Then we find the greatest common divisor of $b_{1}$ and $a_{3}$, and if the greatest common divisor of $b_{1}$ and $a_{3}$ is $b_{2}$, then $b_{2}$ is the greatest common divisor of $a_{1}, a_{2}$, and $a_{3}$. When $n \geqslant 4$, we find the greatest common divisor of $b_{2}$ and $a_{4}$, and if the greatest common divisor of $b_{2}$ and $a_{4}$ is $b_{3}$, then $b_{3}$ is the greatest common divisor of $a_{1}, a_{2}, a_{3}$, and $a_{4}$. When $n \geqslant 5$, we find the greatest common divisor of $b_{3}$ and $a_{5}$, $\cdots$.
|
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 to the above equations. First, solve
$$u+4 v=1$$
Since $4=3+1$, we get $1=-3+4$. Therefore, $u=-3, v=1$ is a set of integer solutions to $u+4 v=1$. Thus,
$$y=8 \times(-3)=-24, z=8 \times 1=8$$
is a set of integer solutions to $y+4 z=8$. The complete set of integer solutions is
$$y=-24-4 t, \quad z=8+t, \quad t=0, \pm 1, \pm 2, \cdots$$
Therefore,
$$x=10-y-z=26+3 t$$
According to the problem, we need $x \geqslant 0, y \geqslant 0, z \geqslant 0$. From $x=26+3 t \geqslant 0$, we get $t \geqslant-\frac{26}{3}$; from $y=-24-4 t \geqslant 0$, we get $t \leqslant-6$; and from $z=8+t \geqslant 0$, we get $t \geqslant-8$. Therefore, $-8 \leqslant t \leqslant-6$ (taking $t=-8, -7, -6$) corresponds to the following three sets of solutions:
$$\left\{\begin{array} { l }
{ x = 2 } \\
{ y = 8 } \\
{ z = 0 , }
\end{array} \quad \left\{\begin{array}{l}
x=5 \\
y=4 \\
z=1
\end{array},\left\{\begin{array}{l}
x=8 \\
y=0 \\
z=2
\end{array}\right.\right.\right.$$
Therefore, there are three different ways to take the coins, which are the three sets of solutions above.
|
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 eleven, there are ten people left over at the end. Find the number of soldiers.
|
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
$$\begin{array}{l}
M=5 \times 6 \times 7 \times 11=2310 \\
M_{1}=\frac{2310}{5}=462 \\
M_{2}=\frac{2310}{6}=385 \\
M_{3}=\frac{2310}{7}=330 \\
M_{4}=\frac{2310}{11}=210
\end{array}$$
Let $M_{1}^{\prime}$ be a positive integer that satisfies $M_{1}^{\prime} M_{1} \equiv 1(\bmod 5)$, then $1 \equiv$ $M_{1}^{\prime} M_{1} \equiv 462 M_{1}^{\prime} \equiv 2 M_{1}^{\prime}(\bmod 5)$, so we get $M_{1}^{\prime}=3$. Let $M_{2}^{\prime}$ be a positive integer that satisfies $M_{2}^{\prime} M_{2} \equiv 1(\bmod 6)$, then $1 \equiv M_{2}^{\prime} M_{2} \equiv$ $385 M_{2}^{\prime} \equiv M_{2}^{\prime}(\bmod 6)$, so we get $M_{2}^{\prime}=1$. Let $M_{3}^{\prime}$ be a positive integer that satisfies $M_{3}^{\prime} M_{3} \equiv 1(\bmod 7)$, then $1 \equiv M_{3}^{\prime} M_{3} \equiv 330 M_{3}^{\prime} \equiv$ $M_{3}^{\prime}(\bmod 7)$, so we get $M_{3}^{\prime}=1$. Let $M_{4}^{\prime}$ be a positive integer that satisfies $M_{4}^{\prime} M_{4} \equiv 1 \quad(\bmod 11)$, then $1 \equiv M_{4}^{\prime} M_{4} \equiv 210 M_{4}^{\prime} \equiv M_{4}^{\prime} (\bmod$ 11). Therefore, by (46) we get
$$\begin{aligned}
x & \equiv 3 \times 462+5 \times 385+4 \times 330+10 \times 210 \\
& \equiv 6731 \equiv 2111(\bmod 2310)
\end{aligned}$$
Thus, we have
$$x=2111+2310 k, \quad k=0,1,2, \cdots$$
$$\begin{array}{l}
x \equiv 1(\bmod 5), x \equiv 5(\bmod 6), \\
x \equiv 4 \quad(\bmod 7), \quad x \equiv 10 \quad(\bmod 11).
\end{array}$$
|
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) $\left\{\begin{array}{ll}x \equiv 1 & (\bmod 7) \\ 3 x \equiv 4 & (\bmod 5) \\ 8 x \equiv 4 & (\bmod 9)\end{array}\right.$
|
5.
(i) Solution: By the Chinese Remainder Theorem,
Given
Given
$$\begin{array}{c}
b_{1}=3, \quad b_{2}=5, \quad m_{1}=7, \quad m_{2}=11, \\
m=m_{1} \cdot m_{2}=7 \times 11=77, \\
M_{1}=\frac{77}{7}=11, \quad M_{2}=\frac{77}{11}=7 . \\
11 M_{1}^{\prime}=1 \quad(\bmod 7), \text { so } M_{1}^{\prime}=2 . \\
7 M_{2}^{\prime}=1 \quad(\bmod 11), \text { so } M_{2}^{\prime}=8 .
\end{array}$$
Therefore, the solution is
$$\begin{aligned}
x & \equiv 3 \times 11 \times 2+5 \times 7 \times 8 \\
& \equiv 346 \equiv 38(\bmod 77)
\end{aligned}$$
(ii) Solution: The moduli are pairwise coprime, so the Chinese Remainder Theorem can be used.
$$\begin{array}{c}
b_{1}=2, \quad b_{2}=5, \quad b_{3}=4 \\
m_{1}=11, \quad m_{2}=7, \quad m_{3}=5 \\
m=m_{1} \cdot m_{2} \cdot m_{3}=11 \times 7 \times 5=385 \\
M_{1}=\frac{385}{11}=35, \quad M_{2}=\frac{385}{7}=55 \\
M_{3}=\frac{385}{5}=77
\end{array}$$
By
That is
$$35 M_{1}^{\prime} \equiv 1 \quad(\bmod 11)$$
So
$$\begin{array}{c}
(11 \times 3+2) M_{1}^{\prime} \equiv 1 \quad(\bmod 11) \\
2 M_{1}^{\prime} \equiv 1(\bmod 11)
\end{array}$$
We get
$$M_{1}^{\prime}=-5$$
Similarly, by
$$55 M_{2}^{\prime} \equiv 1(\bmod 7) ， \quad \text { i.e., } 6 M_{2}^{\prime} \equiv 1(\bmod 7) ，$$
We get
$$M_{2}^{\prime}=-1$$
By
$$77 M_{3}^{\prime} \equiv 1(\bmod 5) ， \text { i.e., } 2 M_{3}^{\prime} \equiv 1(\bmod 5) \text { ， }$$
We get
$$M_{3}^{\prime}=3$$
By the Chinese Remainder Theorem, the solution is
$$\begin{aligned}
x \equiv & 2 \times 35 \times(-5)+5 \times 55 \times(-1) \\
& +4 \times 77 \times 3 \equiv 299 \quad(\bmod 385)
\end{aligned}$$
(iii) Solution: From the second and third equations, we get
$$x \equiv 3(\bmod 5), \quad x \equiv 5(\bmod 9)$$
Combining the first equation with the above two equations, we can use the Chinese Remainder Theorem to solve.
$$\begin{array}{c}
b_{1}=1, \quad b_{2}=3, \quad b_{3}=5 \\
m_{1}=7, \quad m_{2}=5, \quad m_{3}=9 \\
m=m_{1} \cdot m_{2} \cdot m_{3}=7 \times 5 \times 9=315
\end{array}$$
$$M_{1}=\frac{315}{7}=45, \quad M_{2}=\frac{315}{5}=63, \quad M_{3}=\frac{315}{9}=35$$
By
$45 M_{1}^{\prime} \equiv 1 \quad(\bmod 7)$ , we get $M_{1}^{\prime}=-2$ .
By $\quad 63 M_{2}^{\prime} \equiv 1(\bmod 5)$ , we get $M_{2}^{\prime}=2$ .
By $\quad 35 M_{3}^{\prime} \equiv 1(\bmod 9)$ , we get $M_{3}^{\prime}=-1$ .
Therefore,
$$\begin{aligned}
x \equiv & 1 \times 45 \times(-2)+3 \times 63 \times 2 \\
& +5 \times 35 \times(-1) \equiv 113(\bmod 315)
\end{aligned}$$
|
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 divided by 7, the remainder is 3; when divided by 9, the remainder is 5. What is the original number?
(iii) When divided by 11, the remainder is 3; when divided by 72, the remainder is 2; when divided by 13, the remainder is 1. What is the original number?
|
6.
(i) Solution: Let the number be $x$, then according to the problem, we have
Here
By
By
By
$$\left.\begin{array}{c}
\left\{\begin{array}{l}
x \equiv 1 \quad(\bmod 7), \\
x \equiv 2 \quad(\bmod 8), \\
x \equiv 4 \quad(\bmod 9) .
\end{array}\right. \\
b_{1}=1, \quad b_{2}=2, \quad b_{3}=4, \\
m_{1}=7, \quad m_{2}=8, \quad m_{3}=9, \\
m=7 \times 8 \times 9=504,
\end{array}\right\}$$
Therefore,
$$\begin{array}{c}
x \equiv 1 \times 72 \times 4+2 \times 63 \times(-1)+4 \times 56 \\
\times(-4) \equiv-734 \equiv 274(\bmod 504)
\end{array}$$
(ii) Solution: Let the number be $x$, according to the problem, we have
$$\left\{\begin{array}{l}
x \equiv 1 \quad(\bmod 2) \\
x \equiv 2 \quad(\bmod 5) \\
x \equiv 3 \quad(\bmod 7) \\
x \equiv 5 \quad(\bmod 9)
\end{array}\right.$$
Here
$$\begin{array}{c}
b_{1}=1, \quad b_{2}=2, \quad b_{3}=3, \quad b_{4}=5 \\
m_{1}=2, \quad m_{2}=5, \quad m_{3}=7, \quad m_{4}=9 \\
m=m_{1} \cdot m_{2} \cdot m_{3} \cdot m_{4}=2 \times 5 \times 7 \times 9=630 \\
M_{1}=\frac{630}{2}=315, \quad M_{2}=\frac{630}{5}=126 \\
M_{3}=\frac{630}{7}=90, \quad M_{4}=\frac{630}{9}=70
\end{array}$$
By
By
By
By
Therefore,
$$\begin{array}{l}
315 M_{1}^{\prime} \equiv 1 \quad(\bmod 2), \text { get } M_{1}^{\prime}=1 \\
126 M_{2}^{\prime} \equiv 1 \quad(\bmod 5), \text { get } M_{2}^{\prime}=1 \\
90 M_{3}^{\prime} \equiv 1 \quad(\bmod 7), \\
70 M_{4}^{\prime} \equiv 1 \quad(\bmod 9), M_{3}^{\prime}=-1 \\
x=315+2 \times 126+3 \times 90 \times(-1)+5 \\
\times 70 \times 4 \equiv 1697 \equiv 437(\bmod 630)
\end{array}$$
(iii) Solution: Let the number be $x$, according to the problem, we have
$$\left\{\begin{array}{ll}
x \equiv 3 & (\bmod 11) \\
x \equiv 2 & (\bmod 72) \\
x \equiv 1 & (\bmod 13)
\end{array}\right.$$
Here
$$\begin{array}{c}
b_{1}=3, \quad b_{2}=2, \quad b_{3}=1, \\
m_{1}=11, \quad m_{2}=72, \quad m_{3}=13, \\
m=m_{1} \cdot m_{2} \cdot m_{3}=11 \times 72 \times 13=10296, \\
M_{1}=\frac{10296}{11}=936, \quad M_{2}=\frac{10296}{72}=143, \\
M_{3}=\frac{10296}{13}=792 . \\
936 M_{1}^{\prime} \equiv 1 \quad(\bmod 11), \text { get } M_{1}^{\prime}=1 . \\
143 M_{2}^{\prime} \equiv 1 \quad(\bmod 72), \text { get } M_{2}^{\prime}=-1 . \\
792 M_{3}^{\prime} \equiv 1 \quad(\bmod 13), \text { get } M_{3}^{\prime}=-1 .
\end{array}$$
Therefore,
$$\begin{array}{l}
x \equiv 3 \times 936+2 \times 143 \times(-1)+1 \times 792 \\
\times(-1) \equiv 1730(\bmod 10296)
\end{array}$$
|
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 is a remainder of ten; when reduced by multiples of two hundred and forty-seven, there is a remainder of one hundred and forty; when reduced by multiples of three hundred and ninety-one, there is a remainder of two hundred and forty-five; when reduced by multiples of one hundred and eighty-seven, there is a remainder of one hundred and nine. What is the total number? (Huang Zongxian: "General Solution of the Method of Seeking One," answer: 10020.)
|
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 the above using the Chinese Remainder Theorem:
$$\begin{array}{c}
b_{1}=2, \quad b_{2}=5, \quad b_{3}=1 \\
m_{1}=7, \quad m_{2}=9, \quad m_{3}=5 \\
m=m_{1} \cdot m_{2} \cdot m_{3}=7 \times 9 \times 5=315
\end{array}$$
$$M_{1}=\frac{315}{7}=45, \quad M_{2}=\frac{315}{9}=35, \quad M_{3}=\frac{315}{5}=63$$
From $45 M_{1}^{\prime} \equiv 1(\bmod 7)$, we get $M_{1}^{\prime}=-2$.
From $35 M_{2}^{\prime} \equiv 1(\bmod 9)$, we get $M_{2}^{\prime}=-1$.
From $63 M_{3}^{\prime} \equiv 1(\bmod 5)$, we get $M_{3}^{\prime}=2$.
Thus, the solution is
$$\begin{array}{c}
x \equiv 2 \times 45 \times(-2)+5 \times 35 \times(-1)+1 \times 63 \\
\times 2 \equiv-229 \equiv 86(\bmod 315) .
\end{array}$$
(ii) Solution: Let the total number be $x$. According to the problem, we have
$$\left\{\begin{array}{l}
x \equiv 0 \quad(\bmod 5) \\
x \equiv 10 \quad(\bmod 715) \\
x \equiv 140 \quad(\bmod 247) \\
x \equiv 245 \quad(\bmod 391) \\
x \equiv 109 \quad(\bmod 187)
\end{array}\right.$$
First, check if there is a solution. Since $715=5 \times 11 \times 13, 247=13 \times 19, 391=17 \times 23, 187=11 \times 17$, we have $(5,715)=5$, and $5 \mid(10-0)$; $(715,247)=13$, and $13 \mid(140-10)$; $(715,187)=11$, and $11 \mid(109-10)$; $(391,187)=17$, and $17 \mid(245-109)$; thus, by problem 8(i), the system of congruences has a solution. By problem 8(ii), appropriately choose the moduli so that the original system of congruences is equivalent to
$$\left\{\begin{array}{l}
x \equiv 0 \quad(\bmod 5) \\
x \equiv 10(\bmod 11 \times 13) \\
x \equiv 140 \equiv 7(\bmod 19) \\
x \equiv 245 \equiv 15(\bmod 23) \\
x \equiv 109 \equiv 7(\bmod 17)
\end{array}\right.$$
The moduli in the above system of congruences are pairwise coprime, so we can use the Chinese Remainder Theorem to solve it.
From $\square$
$$\begin{array}{c}
1062347 M_{1}^{\prime} \equiv 2 M_{1}^{\prime} \equiv 1 \quad(\bmod 5), \text { we get } M_{1}^{\prime}=3 . \\
37145 M_{2}^{\prime} \equiv 108 M_{2}^{\prime} \equiv 1 \quad(\bmod 11 \times 13) \\
\text { and } 11 \times 13 \equiv 143
\end{array}$$
Since
we have
$$\begin{array}{l}
143=108+35, \quad 108=3 \times 35+3, \\
35=11 \times 3+2, \quad 3=2+1,
\end{array}$$
$$\begin{aligned}
1 & =3-2=3-(35-11 \times 3)=12 \times 3-35 \\
& =12 \times(108-3 \times 35)-35 \\
& =12 \times 108-37 \times 35 \\
& =12 \times 108-37 \times(143-108) \\
& =49 \times 108-37 \times 143 .
\end{aligned}$$
Thus, $\square$
$108 \times 49 \equiv 1(\bmod 143)$, we get $M_{2}^{\prime}=49$.
From $279565 M_{3}^{\prime} \equiv 18 M_{3}^{\prime} \equiv 1(\bmod 19)$, we get $M_{3}^{\prime}=-1$.
From $230945 M_{4}^{\prime} \equiv 2 M_{4}^{\prime} \equiv 1(\bmod 23)$, we get $M_{4}^{\prime}=-11$.
From $312455 M_{5}^{\prime} \equiv 12 M_{5}^{\prime} \equiv 1(\bmod 17)$, we get $M_{5}^{\prime}=-7$.
Finally, the solution is
$$\begin{aligned}
x \equiv 10 & \times 37145 \times 49+7 \times 279565 \times(-1) \\
& +15 \times 230945 \times(-11)+7 \times 312455 \\
& \times(-7) \equiv 18201050-1956955-38105925 \\
& -15310295 \equiv-37172125 \\
\equiv & 10020(\bmod 5311735) .
\end{aligned}$$
Answer: The smallest total number is 10020.
$$\begin{array}{l}
b_{1}=0, \quad b_{2}=10, \quad b_{3}=7, \quad b_{4}=15, \quad b_{5}=7, \\
m_{1}=5, \quad m_{2}=11 \times 13, \quad m_{3}=19, \quad m_{4}=23, \\
m_{5}=17, \quad m=5 \times 11 \times 13 \times 19 \times 23 \times 17 \\
=5311735, \\
M_{1}=\frac{5311735}{5}=1062347, \\
M_{2}=\frac{5311735}{11 \times 13}=37145, \\
M_{3}=\frac{5311735}{19}=279565, \\
M_{4}=\frac{5311735}{23}=230945, \\
M_{5}=\frac{5311735}{17}=312455.
\end{array}$$
|
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 days after that, and the third ship arrives at 8:00 a few days after the second ship. If the first ship travels 300 kilometers per day, the second ship travels 240 kilometers per day, and the third ship travels 180 kilometers per day, what is the actual distance between Port A and Port B in kilometers, and how long did each of the three ships travel?
|
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
$$\left\{\begin{array}{l}
x \equiv 0(\bmod 300) \\
x \equiv 180(\bmod 240) \\
x \equiv 60(\bmod 180)
\end{array}\right.$$
Since
$$\begin{array}{l}
(300,240)=60, \text { and } 60 \mid(180-0) ; \\
(300,180)=60, \text { and } 60 \mid(60-0) ; \\
(240,180)=60, \text { and } 60 \mid(180-60) ;
\end{array}$$
the system of congruences has a solution. Because
$$\begin{aligned}
300 & =2^{2} \times 3 \times 5^{2} \\
240 & =2^{4} \times 3 \times 5 \\
180 & =2^{2} \times 3^{2} \times 5
\end{aligned}$$
the original system of congruences is equivalent to
$$\left\{\begin{array}{l}
x \equiv 0 \quad\left(\bmod 5^{2}\right) \\
x \equiv 180 \equiv 4 \quad\left(\bmod 2^{4}\right) \\
x \equiv 60 \equiv 6 \quad\left(\bmod 3^{2}\right)
\end{array}\right.$$
We solve this using the Chinese Remainder Theorem. Here,
$$\begin{array}{c}
b_{1}=0, \quad b_{2}=4, \quad b_{3}=6 \\
m_{1}=5^{2}, \quad m_{2}=2^{4}, \quad m_{3}=3^{2} \\
m=5^{2} \times 2^{4} \times 3^{2}=3600 \\
M_{1}=\frac{3600}{5^{2}}=144, \quad M_{2}=\frac{3600}{2^{4}}=225 \\
M_{3}=\frac{3600}{9}=400
\end{array}$$
From $144 M_{1}^{\prime} \equiv 1\left(\bmod 5^{2}\right)$, i.e., $19 M_{1}^{\prime} \equiv 1\left(\bmod 5^{2}\right)$, we get $M_{1}^{\prime}=4$. From $225 M_{2}^{\prime} \equiv 1\left(\bmod 2^{4}\right)$, i.e., $M_{2}^{\prime} \equiv 1\left(\bmod 2^{4}\right)$, we get $M_{2}^{\prime}=1$. From $400 M_{3}^{\prime} \equiv 1\left(\bmod 3^{2}\right)$, i.e., $4 M_{3}^{\prime} \equiv 1(\bmod 9)$, we get $M_{3}^{\prime}=-2$. Therefore,
$$\begin{aligned}
x & \equiv 4 \times 225 \times 1+6 \times 400 \times(-2) \\
& \equiv-3900 \equiv 3300(\bmod 3600)
\end{aligned}$$
Since the distance between ports A and B does not exceed 5000 kilometers, the actual distance is 3300 kilometers.
Also,
$$\begin{array}{l}
\frac{3300}{300}=11 \\
\frac{3300}{240}=13 \frac{18}{24} \\
\frac{3300}{180}=18 \frac{6}{18}
\end{array}$$
Answer: The distance between ports A and B is 3300 kilometers. The first ship takes 11 days, the second ship takes 13 days and 18 hours, and the third ship takes 18 days and 8 hours.
|
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
\end{tabular}} \\
\hline & 2196 & \\
\hline 1 & 369 & 549 \\
\hline & 360 & 369 \\
\hline 20 & 9 & 180 \\
\hline & & 180 \\
\hline & & 0 \\
\hline
\end{tabular}
Therefore, $(2565,3114)=9$, by Lemma 10 we have
$$\{2565,3114\}=\frac{2565 \times 3114}{9}=887490$$
Therefore,
$$\{513,135,3114\}=887490 .$$
|
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 \\
\hline
\end{tabular}
Therefore, \((8127,11352)=129\), and by Lemma 10 we have
\[
\{8127,11352\}=\frac{8127 \times 11352}{129}=715176
\]
Since
\[
\begin{array}{|r|l|}
715176 & 21672 \\
715176 & \\
\hline 0 & 21672
\end{array}
\]
Therefore, \((715176,21672)=21672\), and by Lemma 10 we have
\[
\{715176,21672\}=\frac{715176 \times 21672}{21672}=715176
\]
Since
\[
2\left|\begin{array}{r|r|}715176 & 27090 \\ 704340 & 21672 \\ \hline 10836 & 5418 \\ 10836 & \\ \hline 0 & 5418\end{array}\right|
\]
Therefore, \((715176,27090)=5418\), and by Lemma 10 we have
\[
\{715176,27090\}=\frac{715176 \times 27090}{5418}=3575880
\]
Thus, we have
\[
\{8127,11352,21672,27090\}=3575880.
\]
|
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: The side length of the square is 1 foot 5 inches.
|
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, $(48,84,120)=2^{2} \times 3=12$.
(ii) Solution: Decomposing each number into prime factors, we get
$$\begin{array}{l}
360=2^{3} \times 3^{2} \times 5 \\
810=2 \times 3^{4} \times 5 \\
1260=2^{2} \times 3^{2} \times 5 \times 7 \\
3150=2 \times 3^{2} \times 5^{2} \times 7
\end{array}$$
Therefore, $(360,810,1260,3150)=2 \times 3^{2} \times 5=90$.
|
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 & 161406 \\
\hline 11529 & 3780 \\
11340 & 3780 \\
\hline 189 & 0
\end{tabular}\right.,
We get
$(176715,165186)=189$.
Therefore, $(353430,530145,165186)=189$.
(iii) Solution: Since
$1\left|\begin{array}{r|r|r}81719 & 52003 \\ 52003 & 29716 \\ \hline 29716 & 22287 \\ 22287 & 1 \\ \hline 7429 & 0\end{array}\right|$
We get $(81719,52003)=7429$
Since
\begin{tabular}{|c|c|c|}
\hline & 33649 & 7429 \\
\hline & 29716 & 3933 \\
\hline 1 & 3933 & 3496 \\
\hline & 3496 & 3496 \\
\hline 8 & 437 & 0 \\
\hline
\end{tabular}
We get
$(33649,7429)=437$.
Since
\begin{tabular}{|c|c|c|c|}
\hline & 30107 & 437 & 68 \\
\hline & 29716 & 391 & \\
\hline 1 & 391 & 46 & 8 \\
\hline & 368 & 46 & \\
\hline 2 & 23 & 0 & \\
\hline
\end{tabular}
We get
$(30107,437)=23$.
Therefore, $(81719,52003,33649,30107)=23$.
|
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:
$$\begin{array}{l}
(24000,36000,144000) \\
=1000 \times(24,36,144) \\
=1000 \times 12 \times(2,3,12) \\
=1000 \times 12=12000
\end{array}$$
|
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
$$\begin{array}{l}
(525,325)=5 \times(105,65) \\
\quad=5 \times 5 \times(21,13)=5 \times 5=25
\end{array}$$
So $(525,325)=25$.
Answer: The maximum side length of the tiles is 25 cm.
|
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 be 3 jin.
|
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. Since
$$\begin{array}{l}
225=3^{2} \times 5^{2} \\
365=5 \times 73
\end{array}$$
the LCM of $\{225,365\}=3^{2} \times 5^{2} \times 73=16425$.
Answer: The two planets will simultaneously return to their original positions after at least 16425 days.
|
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 transporting, what is the minimum side length of the packaging box's base in centimeters?
|
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 of the side lengths of the bottoms of the various chess boxes. We use millimeters as the unit of length to ensure that all side lengths are integers. Using the method of prime factorization, we get
$$\begin{array}{l}
210=2 \times 3 \times 5 \times 7 \\
120=2^{3} \times 3 \times 5 \\
140=2^{2} \times 5 \times 7 \\
105=3 \times 5 \times 7
\end{array}$$
So $\{210,120,140,105\}=2^{3} \times 3 \times 5 \times 7=840$.
Answer: The minimum side length of the box bottom is 840 millimeters, which is 84 centimeters.
|
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
$$\begin{array}{ll}
10=2 \times 5, & 15=3 \times 5 \\
18=2 \times 3^{2}, & 24=2^{3} \times 3
\end{array}$$
Therefore, $\{10,15,18,24\}=2^{3} \times 3^{2} \times 5=360$.
Answer: The minimum number of people required is 360.
|
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{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{r}
117317 \\
90950
\end{tabular}
\end{tabular} & 5 \\
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
90950 \\
79101
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
26367 \\
23698
\end{tabular}
\end{tabular} & 3 \\
\hline 2 & \begin{tabular}{l}
\begin{tabular}{l}
11849 \\
10676
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
2669 \\
2346
\end{tabular}
\end{tabular} & 4 \\
\hline 2 & \begin{tabular}{l}
\begin{tabular}{r}
1173 \\
969
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
323 \\
204
\end{tabular}
\end{tabular} & 3 \\
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
204 \\
119
\end{tabular}
\end{tabular} & \begin{tabular}{l}
119 \\
85
\end{tabular} & 1 \\
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
85 \\
68
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
34 \\
34
\end{tabular}
\end{tabular} & 2 \\
\hline 2 & 17 & 0 & \\
\hline
\end{tabular}
Therefore, $(435785667,131901878)=17$.
(ii) Solution: Since $15959989=3989 \times 4001$,
the two numbers 3989 and 4001 on the right are both prime, and they are clearly not factors of 7733, so $(15959989,7738)=1$.
|
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:
$$2 \times 2 \times 3=12$$
To find the greatest common divisor of several positive integers, first decompose these positive integers into prime factors, then take the common prime factors (the same prime factors are taken according to the number of times they are common) and multiply them.
|
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 above three numbers, we can see that the prime factors 2, 2, 3 (or $2^{2}, 3$) are common to all three numbers, and their product is the greatest common divisor of these three numbers:
$$2 \times 2 \times 3=12 \text { or } 2^{2} \times 3=12 \text {. }$$
For the sake of clarity, let's first discuss the greatest common divisor and the least common multiple, and then discuss the Fundamental Theorem of Arithmetic. To find the greatest common divisor of several numbers, first decompose these numbers into prime factors and write them in exponential form; then, among the common prime factors, take the powers with the smallest exponents and multiply them.
|
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 $b_{2}=1, b_{3}=5$, which means $2640=(5120)_{8}$.
|
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
$$\begin{array}{l}
1008=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7=2^{4} \times 3^{2} \times 7 \\
1260=2 \times 2 \times 3 \times 3 \times 5 \times 7=2^{2} \times 3^{2} \times 5 \times 7 \\
882=2 \times 3^{2} \times 7^{2} \\
1134=2 \times 3^{4} \times 7
\end{array}$$
Therefore, $(1008,1260,882,1134)=2 \times 3^{2} \times 7=126$.
|
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 $\left(12371^{56}+\right.$ $34)^{72 c} \equiv 1(\bmod 111)$. Therefore, we obtain
$$\left(12371^{56}+34\right)^{72 c+28} \equiv 70(\bmod 111)$$
|
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, $14^{3} \equiv 80(\bmod 111),(80)^{3} \equiv 68(\bmod 111) ，(68)(50) \equiv$ $70(\bmod 111)$, which gives us
$$(50)^{28} \equiv 70(\bmod 111)$$
By Lemma 3, we have $(50)^{*} \equiv 70^{2}(\bmod 111)$. We also have $70^{2} \equiv 16$ $(\bmod 111)$, so by (20) we get $12371^{56} \equiv 16(\bmod 111)$. By Lemma 3, we have $\left(12371^{56}+34\right)^{28} \equiv 50^{28}(\bmod 111)$. By (22), we get $\left(12371^{56}+34\right)^{28} \equiv 70(\bmod 111)$. Therefore, the remainder when $\left(12371^{56}+34\right)^{28}$ is divided by 111 is 70.
|
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
$$\begin{aligned}
\varphi(N) & =9450\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \\
& =\frac{2 \cdot 3^{3} \cdot 5^{2} \cdot 7 \cdot 2 \cdot 4 \cdot 6}{2 \cdot 3 \cdot 5 \cdot 7}=2160
\end{aligned}$$
(ii) Solution: Let the sum of all positive integers not greater than 9450 and coprime with 9450 be $S$. By Problem 5, we have
$$\begin{aligned}
S & =\frac{1}{2} \cdot 9450 \cdot \varphi(9450) \\
& =\frac{1}{2} \times 9450 \times 2160 \\
& =10206000
\end{aligned}$$
|
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 $1000=2^{3} \cdot 5^{3}$. Thus,
$$2^{3} \cdot 5^{3} \mid 2^{m} \cdot 989^{m}\left(1978^{n-m}-1\right)$$
Since $989^{m}$ and $1978^{n-m}-1$ are both odd, $2^{3} \mid 2^{m} . . m$ must be at least 3.
$$\text { Also, }\left(5^{3}, 2^{m} \cdot 989^{m}\right)=1 \text {, so } 5^{3} \mid\left(1978^{n \cdots m}-1\right) \text {, }$$
i.e.,
$$1978^{n-m} \equiv 1(\bmod 125)$$
The problem now is to find the smallest positive integer $n-m$ that satisfies the above congruence. At this point, taking $m=3, n+m=(n-m)+2 m$ will also be the smallest.
Since $\varphi(125)=5^{2} \cdot 4=100,(1978,125)=1$, by Theorem 1 we get
$$1978^{100} \equiv 1(\bmod 125)$$
We can prove that $(n-m) \mid 100$. Because otherwise:
$100=(n-m) q+r, q$ is an integer, $r$ is a positive integer,
and $0<r<n-m$, then
$$1978^{100}=1.978^{(n-m) q} \cdot 1978^{r} \equiv 1978^{r}(\bmod 125)$$
But
$$1978^{100} \equiv 1(\bmod 125)$$
so
$$1978^{r} \equiv 1(\bmod 125)$$
However, $r<n-m$, which contradicts the assumption that $n-m$ is the smallest positive integer that satisfies the congruence. Therefore, $(n-m) \mid 100$.
Since $125 | (1978^{n-m}-1)$, the last digit of $1978^{n-m}$ must be 1 or 6. It is easy to verify that the last digit of $1978^{n-m}$ is 6 only when $4 \mid(n-m)$. So $n-m$ is a multiple of 4, a divisor of 100, and can only be one of 4, 20, 100.
Since
$$\begin{aligned}
1978^{4} & =(125 \times 15+103)^{4} \equiv 103^{4}(\bmod 125) \\
103^{2} & =\left(3+4 \cdot 5^{2}\right)^{2} \equiv 3^{2}+2 \cdot 3 \cdot 4 \cdot 5^{2} \\
& \equiv 609 \equiv-16(\bmod 125) \\
103^{4} & \equiv(-16)^{2} \equiv 6(\bmod 125)
\end{aligned}$$
So
$$1978^{4} \equiv 1(\bmod 125)$$
And
$$1978^{20}=\left(1978^{4}\right)^{5} \equiv 6^{5} \equiv 1(\bmod 125)$$
Therefore, the smallest value of $n-m$ is 100. Now taking $m=3$, we have $n=103, n+m=106$.
|
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 in the previous section):
$$14^{2} \equiv 1, 14^{4} \equiv 1, 14^{7} \equiv 1, 14^{14} \equiv 1 \pmod{29}$$
Calculations yield:
$$\begin{array}{c}
14^{2} \equiv 22, \quad 14^{4} \equiv 22^{2} \equiv (-7)^{2} \equiv 22 \\
14^{7} \equiv (14)^{4}(14)^{2}(14) \equiv (20)(22)(14) \equiv 12 \\
14^{14} \equiv (12)^{2} \equiv -1 \pmod{29}
\end{array}$$
Thus, the conditions in (21) are indeed satisfied, and 14 must be a primitive root modulo 29.
Further calculations yield:
$$\begin{aligned}
(14)^{28} & = (196)^{14} = (38416)^{7} \equiv (571)^{7} = (571)(326041)^{3} \\
& \equiv (571)(574)^{3} = (327754)(329476) \equiv (605)(645) \\
& = 390225 \equiv 1 \pmod{29^2}
\end{aligned}$$
Therefore, 14 is not a primitive root modulo \( 29^2 \). To find a primitive root of \( 29^2 \) from 14, we consider \( 14 + 29 = 43 \), since
$$43 \equiv 14 \pmod{29}$$
43 is still a primitive root modulo 29. However,
$$\begin{aligned}
(43)^{28} & = (3418801)^{7} \equiv (136)^{7} = (136)(18496)^{3} \\
& \equiv (136)(-6)^{3} = -29376 \equiv 59 \equiv 1 \pmod{29^2}
\end{aligned}$$
Thus, by Theorem 8, 43 is a primitive root modulo \( 29^2 \).
The method used in the above example is generally applicable. That is, if \( g \) is a primitive root modulo \( p \geqslant 3 \) and
$$g^{p-1} \equiv 1 \pmod{p^2}$$
then \( g + p \) must be a primitive root modulo \( p^2 \).
Since it is known that a primitive root modulo a prime \( p \) always exists, the above result implies that a primitive root modulo \( p^2 \) also always exists for an odd prime \( p \geqslant 3 \).
Surprisingly, for any odd prime power \( p^s, s \geqslant 2 \), condition (17) is also a necessary and sufficient condition for a primitive root \( g \) of \( p \) to remain a primitive root of \( p^s \). When \( g \) does not satisfy (17), taking \( g + p \) will yield a primitive root of \( p^s \). This is the result stated below.
|
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:
$$\begin{array}{l}
5040\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{4} \\
\quad=6 \sum(2 a)^{8}+60 \sum(a \pm b)^{8} \\
\quad+\sum(2 a \pm b \pm c)^{8}+6 \sum(a \pm b \pm c+d)^{8}
\end{array}$$
where the sum $\sum(2 a)^{8}$ represents the sum over each element in the set $\{a, b, c, d\}$, and the other sums have similar meanings. It is evident that the right-hand side of (31) consists of
$$6\binom{4}{1}+60\binom{4}{2}(2)+\binom{4}{1}\binom{3}{2}\left(2^{2}\right)+6\binom{4}{4}\left(2^{3}\right)=840$$
eighth powers of integers.
For any non-negative integer $n$, there exist integers $q \geqslant 0$ and $r$, $0 \leqslant r \leqslant 5039$, such that $n=5040 q+r$ holds.
First, consider $r$. Since $0 \leqslant r \leqslant 5039 < 3^{8}$, if $r$ can be expressed as the sum of eighth powers of non-negative integers, it can only be expressed as the sum of several $2^{8}$ and several $1^{8}$. Let
$$r=2^{8} k+1, k \geqslant 0, 0 \leqslant 1 \leqslant 2^{8}-1=255,$$
then it is clear that $k=\left[\frac{r}{2^{8}}\right] \leqslant\left[\frac{5039}{2^{8}}\right]=19$. $l=r-\left(2^{8}\right)(19) \leqslant 5039-(256)(19)=175.1$ is $r=(19) 2+1$ can be expressed as the sum of at most $19+175=194$ eighth powers of non-negative integers.
(2) If $0 \leqslant r<\left(2^{8}\right)(19)$. In this case, it is clear that $k \leqslant 18$ and $r=2^{8} k+l$ can be expressed as the sum of at most $k+l \leqslant 18+255-273$ eighth powers of non-negative integers. The fourth power of each $5040 q=5040 x_{1}^{+}+\cdots+5040 x_{i+1}^{+}$.
By Lagrange's theorem, each $x_{i}, 1 \leqslant i \leqslant g(4)$, can be expressed as the sum of four squares of non-negative integers $x_{i}=y_{i 1}^{2}+y_{i 2}^{2}+y_{13}^{2}+y_{i 4}^{2}$. Using the identity (31), for each $i (1 \leqslant i \leqslant g(4))$, the number $5040 x_{i}^{4}=5040\left(y_{i 1}^{2}+y_{i 2}^{2}+y_{i 3}^{2}+y_{i 4}^{2}\right)+$ can be expressed as the sum of 840 eighth powers of non-negative integers. Therefore, $5040 q$ can be expressed as the sum of $g(4) \times 840$ eighth powers of non-negative integers. Thus, $n=5040 q+r$ can be expressed as the sum of at most
$$g(4)(840)+273 \leqslant(50)(840)+273=42273$$
eighth powers of non-negative integers, completing the proof of Theorem 9.
|
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 by 6", and property $P_{3}$ is "an integer is divisible by 8". $A_{i}(i=1,2,3)$ is the subset of $S$ consisting of integers with property $P_{i}$. Note that
$$\begin{array}{l}
\left|A_{1}\right|=\left[\frac{1000}{5}\right]=200, \\
\left|A_{2}\right|=\left[\frac{1000}{6}\right]=166, \\
\left|A_{3}\right|=\left[\frac{1000}{8}\right]=125,
\end{array}$$
Since $\operatorname{LCM}\{5,6\}=30$, we have
$$\left|A_{1} \cap A_{2}\right|=\left[\frac{1000}{30}\right]=33 \text {, }$$
Since $\operatorname{LCM}\{5,8\}=40$, we have
$$\left|A_{1} \cap A_{3}\right|=\left[\frac{1000}{40}\right]=25$$
Since $\operatorname{LCM}\{6,8\}=24$, we have
$$\left|A_{2} \cap A_{3}\right|=\left[\frac{1000}{24}\right]=41$$
Since $\operatorname{LCM}\{5,6,8\}=120$, we have
$$\left|A_{1} \cap A_{2} \cap A_{3}\right|=\left[\frac{1000}{120}\right]=8$$
Thus, by Theorem 1, the number of integers in $S$ that are not divisible by 5, 6, or 8 is
$$\begin{aligned}
\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}\right|= & |S|-\left|A_{1}\right|-\left|A_{2}\right|-\left|A_{3}\right|+\left|A_{1} \cap A_{2}\right|+\left|A_{1} \cap A_{3}\right| \\
& +\left|A_{2} \cap A_{3}\right|-\left|A_{2} \cap A_{2} \cap A_{3}\right| \\
= & 1000-200-166-125+33+25+41-8 \\
= & 600
\end{aligned}$$
|
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
$$\begin{array}{c}
10^{5}-\left[\frac{10^{5}}{7}\right]-\left[\frac{10^{5}}{11}\right]-\left[\frac{10^{5}}{13}\right]+\left[\frac{10^{5}}{7 \times 11}\right] \\
+\left[\frac{10^{5}}{7 \times 13}\right]+\left[\frac{10^{5}}{11 \times 13}\right]-\left[\frac{10^{5}}{7 \times 11 \times 13}\right]=10^{5} \\
-14285-9090-7692+1298+1098+699-99=71929
\end{array}$$
|
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 groups, with 550 students joining the mathematics group, 460 students joining the Chinese group, and 350 students joining the foreign language group. Among them, 100 students participate in both the mathematics and foreign language groups, 120 students participate in both the mathematics and Chinese groups, and 140 students participate in all three groups. How many students participate in both the Chinese and foreign language groups?
|
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, we have
$$\left|A_{1}\right|=550,\left|A_{2}\right|=460,\left|A_{3}\right|=350$$
We also use $A_{12}$ to represent the set of students who joined both the math and Chinese groups, $A_{13}$ to represent the set of students who joined both the math and English groups, and $A_{23}$ to represent the set of students who joined both the Chinese and English groups, then we have
$$\left|A_{13}\right|=100, \quad\left|A_{12}\right|=120$$
We use $A_{123}$ to represent the set of students who joined all three groups, then
$$\left|A_{123}\right|=140$$
Noting the initial explanation, by Theorem 1 of this chapter, we get
$$\begin{array}{c}
0=1200-\left|A_{1}\right|-\left|A_{2}\right|-\left|A_{3}\right|+\left|A_{12}\right| \\
+\left|A_{13}\right|+\left|A_{23}\right|-\left|A_{123}\right|,
\end{array}$$
Thus, we have
$$\begin{aligned}
\left|A_{23}\right| & =-1200+550+460+350-100-120+140 \\
& =80
\end{aligned}$$
That is, the number of students who joined both the Chinese and English groups is 80.
|
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,
$$\left(\frac{-286}{p}\right)=-\left(\frac{143}{p}\right)$$
Since $143=4 \times 35+3, p=3(\bmod 4)$, by Theorem 3 we have
$$\left(\frac{143}{p}\right)=-\left(\frac{p}{143}\right)$$
From $p=143 \times 29880+103$, we get
$$\left(\frac{p}{143}\right)=\left(\frac{103}{143}\right)$$
By Theorem 3 and $103=3(\bmod 4), 143 \equiv 3(\bmod 4)$, we have
$$\begin{array}{l}
\left(\frac{103}{143}\right)=-\left(\frac{143}{103}\right)=-\left(\frac{40}{103}\right)=-\left(\frac{2^{2} \times 2 \times 5}{103}\right) \\
\quad=-\left(\frac{2 \times 5}{103}\right)=-\left(\frac{2}{103}\right)\left(\frac{5}{103}\right)=-\left(\frac{5}{103}\right) \\
=-\left(\frac{103}{5}\right)=-\left(\frac{3}{5}\right)=1
\end{array}$$
Therefore,
$$\left(\frac{-286}{p}\right)=1$$
This means that equation (51) has a solution.
|
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 the modular inverse of $n$ modulo $p$. From the properties of the Legendre symbol, it is easy to see that
$$\begin{aligned}
\left(\frac{n(n+1)}{p}\right) & =\left(\frac{n\left(n+n r_{n}\right)}{p}\right)=\left(\frac{n^{2}\left(1+r_{n}\right)}{p}\right) \\
& =\left(\frac{n}{p}\right)^{2}\left(\frac{1+r_{n}}{p}\right)=\left(\frac{1+r_{n}}{p}\right)
\end{aligned}$$
We will prove that for $n \neq m(\bmod p), p \nmid n m$, it must also be true that
$$r_{n} \not\equiv r_{m}(\bmod p)$$
That is, different numbers in the reduced residue system modulo $p$ must correspond to different inverses. We use proof by contradiction. If $r_{n} \equiv r_{m}(\bmod p)$, multiplying both sides by $n m$ gives
$$n\left(m r_{m}\right) \equiv m\left(n r_{n}\right)(\bmod p)$$
Then, by the definition of the inverse, we get
$$n \equiv m(\bmod p)$$
This leads to a contradiction. This proves that when $n$ runs through $1,2, \ldots, p-1$, the corresponding inverse $r_{n}$ also runs through $1,2, \ldots, p-1(\bmod p)$, just in a different order. Also, note that from $p-1 \equiv-1(\bmod p)$, we immediately get
$$(p-1)^{2} \equiv(-1)^{2} \equiv 1(\bmod p)$$
So $r_{p-1}=p-1$. Therefore, when $n$ takes $1,2, \ldots, p-2$, the inverse $r_{n}$ of $n$ also takes $1,2, \ldots, p-2(\bmod p)$, just in a different order. Thus, we get
$$\begin{array}{l}
\left(\frac{1.2}{p}\right)+\left(\frac{2.3}{p}\right)+\ldots+\left(\frac{(p-2)(p-1)}{p}\right) \\
= \sum_{n=1}^{p-2}\left(\frac{1+r_{n}}{p}\right)=\sum_{r=1}^{p-2}\left(\frac{1+r}{p}\right) \\
=\sum_{r=1}^{p-1}\left(\frac{r}{p}\right)-\left(\frac{1}{p}\right)
\end{array}$$
Since in a reduced residue system modulo $p$, there are exactly $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ quadratic non-residues, we have $\sum_{r=1}^{p-1}\left(\frac{r}{p}\right)=0$. Therefore, the required sum is $-\left(\frac{1}{p}\right)=-1$.
|
-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
$$k \equiv 5(\bmod 23),$$
Thus, we get $a \geqslant 3 \times 29 \times 5+1=436$. When $a=436$, $2001 \mid(55+436 \times 32)$. Therefore, the smallest value is 436.
|
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 two cases.
Case one: $3^{a}-2^{b}=41$, taking modulo 3 on both sides, we know $b$ is even; taking modulo 4 on both sides, we know $a$ is even. Let $a=2 m, b=2 n$, then $\left(3^{m}-2^{n}\right)\left(3^{m}+2^{n}\right)=41$, which gives $\left(3^{m}-2^{n}, 3^{m}+\right.$ $\left.2^{n}\right)=(1,41)$, leading to
$$2 \times 3^{m}=\left(3^{m}-2^{n}\right)+\left(3^{m}+2^{n}\right)=42$$
i.e., $3^{m}=21$, a contradiction.
Case two: $2^{b}-3^{a}=41$, in this case $b \geqslant 3$, taking modulo 8 on both sides leads to a contradiction.
In summary, the smallest prime $p=41$.
|
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
$$\begin{aligned}
& \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}=\frac{(a p)(a p-1) \cdot \cdots \cdot((a-b) p+1)}{(b p)!}-\frac{a!}{b!(a-b)!} \\
= & \frac{a(a-1) \cdot \cdots \cdot(a-b+1) \prod_{k=a-b}^{a-1}(k p+1)(k p+2) \cdots \cdots \cdot(k p+(p-1))}{b!\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdots \cdots(k p+(p-1))} \\
& -\frac{a!}{b!(a-b)!} \\
= & \frac{a!}{b!(a-b)!} \cdot \frac{1}{\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdot \cdots \cdot(k p+(p-1))}\left\{\prod_{k=a-b}^{a-1}(k p+1)\right. \\
& \left.\cdot(k p+2) \cdot \cdots \cdot(k p+(p-1))-\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdot \cdots \cdot(k p+(p-1))\right\},
\end{aligned}$$
Thus, we only need to prove that $p^{3} \mid A$, where
$$\begin{aligned}
A= & \prod_{k=a-b}^{a-1}(k p+1) \cdot(k p+2) \cdot \cdots \cdot(k p+(p-1)) \\
& -\prod_{k=0}^{b-1}(k p+1) \cdot(k p+2) \cdot \cdots \cdot(k p+(p-1))
\end{aligned}$$
For this, we set
$$\begin{aligned}
f(x) & =(x+1)(x+2) \cdot \cdots \cdot(x+(p-1)) \\
& =x^{p-1}+\alpha_{p-2} x^{p-2}+\cdots+\alpha_{1} x+(p-1)!
\end{aligned}$$
Then, by the conclusion of Example 1 in Section 2.4, we know that $p^{2} \mid \alpha_{1}$. Therefore,
$$\begin{aligned}
A= & \prod_{k=a-b}^{a-1} f(k p)-\prod_{k=0}^{b-1} f(k p) \\
\equiv & ((p-1)!)^{b-1} \sum_{k=a-b}^{a-1} \alpha_{1} k p+((p-1)!)^{b} \\
& -((p-1)!)^{b-1} \sum_{k=0}^{b-1} \alpha_{1} k p-((p-1)!)^{b} \\
\equiv & 0\left(\bmod p^{3}\right) .
\end{aligned}$$
Therefore, the maximum integer $c=3$.
|
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 16 lies between 0 and $n$. Therefore, when $n \leqslant 14$, equation (9) cannot hold (since $1599 \equiv 15(\bmod 16)$).
Notice that $5^{4}+12 \times 3^{4}+1^{4}+1^{4}=1599$, so when $n=15$, equation (9) has an integer solution $(5, \underbrace{3, \cdots, 3}_{12 \uparrow}, 1,1)$.
In summary, the smallest positive integer $n=15$.
|
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 $a \geqslant 3 b+2, b \geqslant 2 c+1, c \geqslant 5$, hence
$$\begin{aligned}
1753 & \geqslant(3 b+2)^{2}-b^{2}+c^{2}=8 b^{2}+12 b+4-c^{2} \\
& \geqslant 8(2 c+1)^{2}+12 b+4=33 c^{2}+32 c+12 b+12 \\
& \geqslant 33 c^{2}+160+132+12
\end{aligned}$$
Therefore, $c^{2}3 b$, we know $a-b>2 b \geqslant 22$. Also, $a-b$ and $a+b$ are both even, and $a, b$ are both odd, so $a-b=32$, and $a+b=54$. Thus, $a=43, b=11$, and consequently, $a^{2}+b^{2}+c^{2}+d^{2}=1999$.
|
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 $B(n)$.
Prove: The function $f$ is a bounded function, and find the maximum value of $f$.
|
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 $x, y$,
$$[2 x]+[2 y] \geqslant[x]+[x+y]$$
In fact, let $x=m+\alpha, y=n+\beta$, where $m, n \in \mathbf{N}$, and $\alpha, \beta \in[0,1)$, then (2) is equivalent to proving
that is,
$$\begin{array}{c}
2 m+[2 \alpha]+2 n+[2 \beta] \geqslant m+(m+n)+[\alpha+\beta] \\
n+[2 \alpha]+[2 \beta] \geqslant[\alpha+\beta] .
\end{array}$$
Since $[\alpha+\beta] \leqslant 1$, and when $[\alpha+\beta]=1$, at least one of $\alpha, \beta$ is not less than $\frac{1}{2}$, thus in this case $[2 \alpha]+[2 \beta] \geqslant 1$. Therefore, (3) holds. The lemma is proved.
Returning to the original problem, we first prove: For any $n \in \mathbf{N}^{*}$, the number $\frac{(2 n)!\cdot 2000!}{n!(n+1000)!}$ is a positive integer. To do this, we only need to prove that for any prime $p$, the power of $p$ in the prime factorization of $(2 n)!\cdot 2000!$ is not less than the power of $p$ in the prime factorization of $n!(n+1000)!$. Using property 5, we only need to prove:
$$\sum_{k=1}^{+\infty}\left(\left[\frac{2 n}{p^{k}}\right]+\left[\frac{2000}{p^{k}}\right]\right)-\left(\left[\frac{n}{p^{k}}\right]+\left[\frac{n+1000}{p^{k}}\right]\right) \geqslant 0$$
If we let $x=\frac{n}{p^{k}}, y=\frac{1000}{p^{k}}$, using (2) we know that inequality (4) holds, so $\frac{(2 n)!\cdot 2000!}{n!(n+1000)!} \in \mathbf{N}^{*}$, i.e.,
$$\frac{A(n)}{B(n)} \cdot 2000!\in \mathbf{N}^{*}$$
The above discussion shows that for any $n \in \mathbf{N}^{*}$, we have $f(n) \mid 2000$!. Noting that $f(n)$ is 1 or a prime number, and by direct verification, 1999 is a prime number, so for any $n \in \mathbf{N}^{*}$, we have $f(n) \leqslant 1999$. Therefore, $f(n)$ is a bounded function.
Furthermore, when $n=999$, we know that
$$B(n)=1999 \cdot(999!)$$
At this time, $f(n)=1999$. Hence, the maximum value of $f(n)$ is 1999.
|
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. Find the maximum possible value of $n$.
|
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), (even, odd), and (even, even). Therefore, among $A, B, C, D$, there is one point such that the midpoint of the line segment connecting it to $P$ is a point in $S$, hence $n \leqslant 3$.
On the other hand, we prove: for any three points $A, B, C$ in $S$, there exists another point $P$ in $S$ such that the interiors of $A P, B P, C P$ do not contain any integer points.
Note that for integer points $X\left(x_{1}, y_{1}\right)$ and $Y\left(x_{2}, y_{2}\right)$ on the plane, the necessary and sufficient condition for the line segment $X Y$ to have no integer points in its interior is $\left(x_{1}-x_{2}, y_{1}-y_{2}\right)=1$. Therefore, we can assume $A\left(a_{1}, a_{2}\right), B\left(b_{1}, b_{2}\right), C(0,0)$.
To find a point $P(x, y)$ such that $A P, B P, C P$ have no integer points in their interiors, $x, y$ need to satisfy the following conditions:
$$\left\{\begin{array}{l}
x \equiv 1(\bmod y) \\
x-a_{1} \equiv 1\left(\bmod y-a_{2}\right) \\
x-b_{1} \equiv 1\left(\bmod y-b_{2}\right)
\end{array}\right.$$
Using the Chinese Remainder Theorem, if there exists $y \in \mathbf{Z}$ such that $y, y-a_{2}, y-b_{2}$ are pairwise coprime, then an $x$ satisfying (2) exists, thus finding a point $P$ that meets the requirements.
Take $y=m\left[a_{2}, b_{2}\right]+1, m \in \mathbf{Z}, m$ to be determined, then $\left(y, y-a_{2}\right)=\left(y, a_{2}\right)=\left(1, a_{2}\right)=1$, similarly $\left(y, y-b_{2}\right)=1$. Therefore, it is only necessary to take $m$ such that
$$\left(y-a_{2}, y-b_{2}\right)=1$$
which is equivalent to
$$\left(y-a_{2}, a_{2}-b_{2}\right)=1$$
To make (3) hold, set $a_{2}=a_{2}^{\prime} d, b_{2}=b_{2}^{\prime} d$, where $a_{2}^{\prime}, b_{2}^{\prime}, d \in \mathbf{Z}$, and $\left(a_{2}^{\prime}, b_{2}^{\prime}\right)=1$, then $y=m d a_{2}^{\prime} b_{2}^{\prime}+1$, and thus $y-a_{2}=d a_{2}^{\prime}\left(m b_{2}^{\prime}-1\right)+1$. Noting that when $\left(a_{2}^{\prime}, b_{2}^{\prime}\right)=1$, $\left(b_{2}^{\prime}, a_{2}^{\prime}-b_{2}^{\prime}\right)=1$, so there exists $m \in \mathbf{Z}$ such that $m b_{2}^{\prime} \equiv 1\left(\bmod a_{2}^{\prime}-b_{2}^{\prime}\right)$, such a determined $m$ satisfies: $d\left(a_{2}^{\prime}-b_{2}^{\prime}\right) \mid d a_{2}^{\prime}\left(m b_{2}^{\prime}-1\right)$, at this time
$$y-a_{2} \equiv 1\left(\bmod a_{2}-b_{2}\right),$$
Therefore, (3) holds.
In summary, the maximum value of $n$ that satisfies the conditions is 3.
|
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 $\underbrace{2+0}_{k \text { 个 }}$
$$\begin{aligned}
n & =\frac{10^{k-1}-1}{9} \times 10^{k+1}+\frac{2 \times\left(10^{k}-1\right)}{9} \times 10+5 \\
& =\frac{1}{9}\left(10^{2 k}-10^{k+1}+2 \times 10^{k+1}-20+45\right) \\
& =\frac{1}{9}\left(10^{2 k}+10 \times 10^{k}+25\right) \\
& =\frac{1}{9}\left(10^{k}+5\right)^{2}=(\underbrace{33 \cdots 35}_{k-1 \uparrow})^{2}
\end{aligned}$$
That is, $n$ is a perfect square.
By setting $k=3 t+2,3 t, 3 t+1, t \in \mathbf{N}$, and combining $S\left(1^{2}\right)=1$, we can see that for any $m \in\{0,1,2, \cdots, 2009\}$, if $m \equiv 1,4$ or $7(\bmod 9)$, then there exists $n \in \mathbf{N}^{*}$, such that $S\left(n^{2}\right)=m$.
Furthermore, for $k \in \mathbf{N}^{*}, n=\underbrace{99 \cdots 9}_{k \text { 个 }}$, we have
$$\begin{aligned}
n^{2} & =\left(10^{k}-1\right)^{2}=10^{2 k}-2 \times 10^{k}+1 \\
& =\underbrace{99 \cdots 9}_{k-1 \uparrow} \underbrace{00 \cdots 01}_{k-1 \uparrow},
\end{aligned}$$
Thus, $S\left(n^{2}\right)=9 k$. Therefore, when $m \equiv 0(\bmod 9)$, there also exists $n \in \mathbf{N}^{*}$, such that $S\left(n^{2}\right)=$ $m$.
The above discussion shows that the number of numbers that meet the condition is $\frac{4}{9} \times 2007+1=893$.
|
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,
$$\begin{aligned}
x & =1+\sum_{k=2}^{10^{6}} \frac{1}{\sqrt{k}}2 \sum_{k=1}^{10^{6}}(\sqrt{k+1}-\sqrt{k}) \\
& =2\left(\sqrt{10^{6}+1}-1\right)>1998 .
\end{aligned}$$
From this, we know that $[x]=1998$.
|
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 sequence of primes in ascending order. Suppose $m$ is a positive integer that meets the requirement. We first prove: if $p_{n} p_{n+1} \leqslant n$, then
$$p_{1} p_{2} \cdots p_{n} \leqslant m$$
In fact, if there are two different numbers $p_{u}, p_{v}$ in $p_{1}, p_{2}, \cdots, p_{n+1}$ that are not divisors of $m$, then $\left(p_{u} p_{v}, m\right)=1, p_{u} p_{v} \leqslant p_{n} p_{n+1} \leqslant m$, which contradicts the property of $m$. Therefore, at most one of $p_{1}, p_{2}, \cdots, p_{n+1}$ is not a divisor of $m$, so (5) holds.
Now, if there exists $m(>60)$ that meets the requirement, then when $m \geqslant 77=p_{4} p_{5}$, let $n$ be the largest positive integer such that $p_{1} p_{2} \cdots p_{n} \leqslant m$. Then, by (5), we know $n \geqslant 4$. Combining the conclusion from the previous problem, we have $p_{n+1} p_{n+2}5 \times 7$, so $2,3,5,7$ have at most one that is not a divisor of $m$ (see the proof of (5)), thus $m$ is a multiple of one of the numbers 105, 70, 42, 30, hence $m=70$. But in this case, $(33, m)=1$, which contradicts the property of $m$.
|
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 $a \in \mathbf{N}^{*}$, satisfying $(a-1) \mid 2008$, set $2008=(a-1) \cdot k$, in $x_{1}, x_{2}, \cdots, x_{2009}$ take $a$ to be $0, a-1$ to be $1, \cdots, a-1$ to be $k-1$, and let $x_{0}=k$, then we know that (1) holds.
Therefore, there are 8 positive integers $a$ that satisfy the condition.
|
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, then $2^{a}-1$ is also an odd composite number (this can be known through factorization). Let $2^{a-1}-1=a \times q, q$ be an odd positive integer, then
$$\begin{aligned}
2^{2^{a}-1-1}-1 & =2^{2\left(2^{a-1}-1\right)}-1 \\
& =2^{2 a q}-1 \\
& =\left(2^{a}\right)^{2 q}-1 \\
& \equiv 1^{2 q}-1 \\
& \equiv 0\left(\bmod 2^{a}-1\right)
\end{aligned}$$
Therefore $2^{a}-1$ is also a number that meets the requirement. By this recursion (combined with 341 meeting the requirement), it is known that there are infinitely many composite numbers that meet the condition.
|
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 the authenticity of each coin can be identified?
|
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 system of three linear equations for $b, c, d$ can determine the authenticity of $b, c, d$. Therefore, 3 weighings are sufficient.
Next, we prove that two weighings cannot guarantee the determination of the authenticity of each coin.
Notice that if two coins, for example, $a, b$, either both appear or both do not appear in each weighing, then when $a, b$ are one genuine and one counterfeit, swapping the authenticity of $a, b$ does not affect the weighing results, so their authenticity cannot be determined.
If there is a weighing in which at most two coins, for example, $a, b$, appear, then in the other weighing, $c, d$ can only have exactly one on the scale (otherwise, swapping the odd/even nature of $c, d$ does not affect the result), at this point, one coin does not appear in both weighings, and changing its authenticity does not affect the weighing results, thus its authenticity cannot be determined. Therefore, at least 3 coins must be on the scale in each weighing, which means there must be two coins that appear in both weighings, leading to a contradiction.
In summary, at least 3 weighings are required.
|
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 numbers among these 27 sum numbers?
|
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 sum of the 3 numbers in the first row of Figure 1, i.e., $a_{1}+a_{2}+a_{3}$ is even, while the sums of the other 5 rows in Figure 1 are all odd. In this case, summing the numbers in Figure 1 by rows and columns, we get
$$\begin{aligned}
& \left(a_{1}+a_{2}+a_{3}\right)+\left(a_{4}+a_{5}+a_{6}\right)+\left(a_{7}+a_{8}+a_{9}\right) \\
= & \left(a_{1}+a_{4}+a_{7}\right)+\left(a_{2}+a_{5}+a_{8}\right)+\left(a_{3}+a_{6}+a_{9}\right)
\end{aligned}$$
However, the left side of this equation is the sum of two odd numbers and one even number, while the right side is the sum of 3 odd numbers, leading to a contradiction where the left side is even and the right side is odd.
Therefore, among these 27 sums, there can be at most 24 odd numbers.
The example below (as shown in Figure 2) demonstrates that there exists a way to fill the numbers such that 24 of the 27 sums can be odd. In the tables of Figure 2, 0 represents an even number, and 1 represents an odd number, from left to right, representing the top, middle, and bottom layers of the unit cubes.
Thus, among these 27 sums, the maximum number of odd numbers is 24.
|
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
$$\begin{aligned}
a_{n} & =10^{2 n}+10^{2(n-1)}+\cdots+1 \\
& =\frac{10^{2(n+1)}-1}{10^{2}-1} \\
& =\frac{\left(10^{n+1}-1\right)\left(10^{n+1}+1\right)}{99} .
\end{aligned}$$
Notice that, $99<10^{n+1}-1,99<10^{n+1}+1$, and $a_{n}$ is a positive integer, so $a_{n}$ is a composite number (because the terms $10^{n+1}-1$ and $10^{n+1}+1$ in the numerator cannot be reduced to 1 by 99).
|
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
$$\begin{array}{c}
m \leqslant n<15 \\
1599 \equiv 15(\bmod 16)
\end{array}$$
This is a contradiction, so
$$n \geqslant 15.$$
Furthermore, when $n=15$, it is required that
$$x_{1}^{4} \equiv x_{2}^{4} \equiv \cdots \equiv x_{n}^{4} \equiv 1(\bmod 16)$$
This means that $x_{1}, x_{2}, \cdots, x_{n}$ must all be odd numbers, which points us in the right direction to find suitable numbers. In fact, among $x_{1}, x_{2}, \cdots, x_{15}$, if one number is 5, twelve are 3, and the other two are 1, then
$$\begin{aligned}
& x_{1}^{4}+x_{2}^{4}+\cdots+x_{15}^{4} \\
= & 5^{4}+12 \times 3^{4}+2 \\
= & 625+972+2 \\
= & 1599
\end{aligned}$$
Therefore, the minimum value of $n$ is 15.
|
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 $q \equiv 1 \pmod{3}$, then
$$7p + q = 14 + q \equiv 0 \pmod{3}$$
This is a contradiction;
If $q \equiv 2 \pmod{3}$, then
$$pq + 11 = 2q + 11 \equiv 4 + 11 \equiv 0 \pmod{3}$$
This is also a contradiction. Therefore, $q = 3$. In this case,
$$7p + q = 17, \quad pq + 11 = 17$$
Both are prime numbers, so
$$\left(p^2 + q^p\right)\left(q^2 + p^q\right) = \left(2^2 + 3^2\right)\left(3^2 + 2^3\right) = 221$$
Case two: Suppose $q$ is even, then $q = 2$. Similarly, we find that $p = 3$. In this case,
$$\left(p^2 + q^p\right)\left(q^2 + p^q\right) = \left(3^2 + 2^3\right)\left(2^2 + 3^2\right) = 221$$
In conclusion, the desired value is 221.
|
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 system modulo 3), which is a contradiction.
If $5 \nmid d$, then one of $p_{1}, p_{1}+d, \cdots, p_{1}+4 d$ is a multiple of 5, which can only be $p_{1}=$ 5, in this case the common difference $d$ is a multiple of 6.
And $5,11,17,23,29$ is a sequence of 5 primes in arithmetic progression, so, $p_{5}$ is at least 29.
|
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 of $k$ is 2.
|
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 a multiple of 12.
|
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 $\left\{F_{n}\right\}$
$$\begin{array}{l}
\equiv 1+1+2+3+5+(-3)+2+(-1)+1+0 \\
=11 \equiv 0(\bmod 11)
\end{array}$$
The proposition is proved. $\square$
(2) $\left\{F_{n}(\bmod 12)\right\}: 1,1,2,3,5,-4,1,-3,-2,-5,5,0,1, 1, \cdots$ is a purely periodic sequence with a period of 12. Direct verification shows that the smallest positive integer $k$ satisfying the condition is $k=36$.
Note: If $k$ is the smallest positive integer satisfying (2), and $n$ is any positive integer satisfying (2), then $k \mid n$ (this conclusion is left for the reader to prove). Therefore, after finding that $n=36$ (where the sum of the numbers in each period of $\left\{F_{n}(\bmod 12)\right\}$ is $\equiv 4(\bmod 12)$), it is only necessary to verify that the positive divisors of 36 do not satisfy the condition to conclude that 36 is the smallest positive integer satisfying the condition.
|
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, $a_{i}$ represents the number of matches in the $i$-th box. In this way, the parity (odd or even) of the sum of the three answers is the same as the parity of $a_{1}$ (the other boxes each appear exactly twice in the three questions). Therefore, after 3 questions, the parity of $a_{1}$ can be determined.
Next, prove that "at least 3 questions are needed." If there are only two questions, and $a_{1}$ appears in both questions, then there must be $a_{i}$ and $a_{j}$ in the two questions such that $a_{i}$ appears only in the first question, and $a_{j}$ appears only in the second question. By changing the parity of $a_{1}$, $a_{i}$, and $a_{j}$ simultaneously, the answers to each question remain the same, thus the parity of $a_{1}$ cannot be determined. If $a_{1}$ does not appear in both questions, when $a_{1}$ does not appear at all, change the parity of $a_{1}$; when $a_{1}$ appears only once, change the parity of $a_{1}$ and $a_{i}$ (where $a_{i}$ is a box that appears with $a_{1}$ in the question), then the answers to the two questions remain the same, and the parity of $a_{1}$ cannot be determined.
In summary, at least 3 questions are needed.
|
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 $\square$
$$\begin{array}{c}
a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}} \\
1 \equiv \underbrace{1+\cdots+1}_{2001 \uparrow}(\bmod a-1), \\
a-1 \mid 2000 .
\end{array}$$
On the other hand, if $a>1$ and $a-1 \mid 2000$, then we can take $x_{1}, x_{2}, \cdots, x_{2001}$ to be $a$ numbers as $0$, $a-1$ numbers as $1$, $a-1$ numbers as $2$, ..., $a-1$ numbers as $k-1$, where $k=\frac{2000}{a-1}$, and take $x_{0}=k$, then we have $a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}}$.
Therefore, $a$ is a number that satisfies the condition if and only if $a>1$ and $a-1 \mid 2000$, and there are 20 such $a$.
|
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
$$\begin{aligned}
& 5 x^{13} + 13 x^{5} + 9 a x \\
\equiv & 13 x + 9 a x \\
\equiv & (3 + (-1) \times 3) x \\
\equiv & 0 \pmod{5}
\end{aligned}$$
$$\begin{aligned}
& 5 x^{13} + 13 x^{5} + 9 a x \\
\equiv & 5 x + 9 a x \\
\equiv & (5 + 9 \times (-2)) x \\
\equiv & 0 \pmod{13}
\end{aligned}$$
Therefore,
$$65 \mid 5 x^{13} + 13 x^{5} + 9 a x$$
In summary, the smallest positive integer $a=63$.
|
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 the condition, and when $b=0$, from $a^{2}+b^{2} \equiv 0(\bmod 37)$ we know $a=0$. Hence, the number of pairs $(a, b)$ that satisfy the condition is $2 \times 36+1=73$ (pairs).
|
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,
$$x^{2} \geqslant 1000, x^{2}-x+1 \leqslant 1000$$
So,
$$32 \leqslant x<33$$
Hence,
$$x=32,$$
Then,
$$y=768$$
Therefore,
$$n=32768$$
|
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 $m=0,1,2,3,4(\bmod 5)$, correspondingly
$$m^{3} \equiv 0,1,3,2,4(\bmod 5)$$
So, from $m^{3} \equiv 1(\bmod 5)$ we know $m \equiv 1(\bmod 5)$, we can set $m=5 k+1$, at this time
$$m^{3}=(5 k+1)^{3}=125 k^{3}+75 k^{2}+15 k+1 \equiv 111(\bmod 125)$$
Hence
Thus
That is, $\square$
Therefore
This requires
Hence
$$\begin{array}{c}
75 k^{2}+15 k \equiv 110(\bmod 125) \\
15 k^{2}+3 k \equiv 22(\bmod 25) \\
15 k^{2}+3 k+3 \equiv 0(\bmod 25) \\
5 k^{2}+k+1 \equiv 0(\bmod 25) \\
5 k^{2}+k+1 \equiv 0(\bmod 5) \\
5 \mid k+1
\end{array}$$
We can set $k+1=5 l$, then
$$\begin{aligned}
5 k^{2}+k+1 & =5 \times(5 l-1)^{2}+5 l \\
& =125 l^{2}-50 l+5(l+1) \\
& \equiv 0(\bmod 25) \\
& 5 \mid l+1
\end{aligned}$$
We can set $l+1=5 r$, therefore
$$\begin{aligned}
n & =2 m=10 k+2=10(5 l-1)+2 \\
& =50 l-8=50(5 r-1)-8 \\
& =250 r-58
\end{aligned}$$
Combining that $n$ is a positive integer, we know
$$n \geqslant 250-58=192$$
Also, $192^{3}=7077888$ meets the requirement, hence the smallest positive integer satisfying the condition is 192.
|
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
$$84+r \equiv 0(\bmod 5)$$
In this case,
$$n<3 \times 42=126$$
If $r \equiv 2(\bmod 5)$, then
$$4 \times 42+r \equiv 0(\bmod 5)$$
In this case,
$$n<5 \times 42=210$$
If $r \equiv 3(\bmod 5)$, then
$$42+r \equiv 0(\bmod 5)$$
In this case,
$$n<2 \times 42=84$$
If $r \equiv 4(\bmod 5)$, then
$$3 \times 42+r \equiv 0(\bmod 5)$$
In this case,
$$n<4 \times 42=168$$
If $r \equiv 0(\bmod 5)$, then
$$r=5,$$
In this case, since 5, 47, 89, 131, and 173 are all prime numbers, $n$ is at most 215.
In summary, the largest positive integer sought is 215.
|
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 first condition requires $k \equiv 1(\bmod 11)$, set $k=11 m+1$,
The second condition requires
$$210(11 m+1) \equiv-1(\bmod 13)$$
Solving this, we get
$$m \equiv 4(\bmod 13)$$
Therefore, let $k=45$, then the 21 numbers $9440,9441, \cdots, 9460$ are not coprime with 30030, so taking $n=9440$ is sufficient.
|
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 this way, some points are not labeled, and some points are labeled with more than one number. Question: What is the smallest number among the numbers labeled on the point that is labeled with 2000?
|
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
$$2000=2^{4} \times 5^{3},$$
so $2^{4}\left|n, 5^{3}\right| n+1$; or $5^{3}\left|n, 2^{4}\right| n+1$; or one of $n$ and $n+1$ is a multiple of 2000. Solving these, the smallest $n$ is found to be $624, 1375, 1999$, with the smallest number satisfying (1) being 624. Therefore, the smallest number marked on the point labeled 2000 is 624.
|
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 red. Question: According to this rule, what is the maximum number of points that can be colored red on the circle? Prove your conclusion.
|
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 $2^n a \not\equiv 2^m a \pmod{800}$ holds, $2^n \equiv 2^m \pmod{800}$ must hold. Therefore, the maximum number of distinct elements in the sequence $a, 2a, 2^2a, \cdots$ under modulo 800 can be achieved when $a=1$. Thus, we only need to find the number of distinct elements in the sequence $1, 2, 2^2, \cdots$ under modulo 800.
Since $800 = 2^5 \times 5^2$, and for $n \geq 5$ we have
$$2^n \equiv 0 \pmod{2^5}$$
Additionally, $\{2^n \pmod{25}\}$ is: $2, 4, 8, 16, 7, 14, 3, 6, 12, -1, -2, -4, -8, -16, -7, -14, -3, -6, -12, 1, \cdots$, so $\{2^n \pmod{25}\}$ contains exactly 20 distinct elements. Combining this with $\{2^n \pmod{2^5}\}$ being $2, 4, 8, 16, 0, 0, \cdots$, we get that $\{2^n \pmod{800}\}$ contains $20 + 4 = 24$ (distinct) numbers.
Therefore, there are at most 24 points on the circle that are colored red.
|
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
$$\left\{\begin{array}{l}
x=282-19 t, \\
y=2+7 t .
\end{array} \text { ( } t \text { is an integer }\right)$$
Combining $x>0, y>0$ and $t$ being an integer, we can solve to get $0 \leqslant t \leqslant 14$.
Therefore, equation (1) has 15 sets of positive integer solutions.
|
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
$$\begin{array}{l}
1 \leqslant y \leqslant 1004,1 \leqslant y \leqslant 1002,1 \leqslant y \leqslant 1001 \\
1 \leqslant y \leqslant 999, \cdots, 1 \leqslant y \leqslant 2
\end{array}$$
Since when $y$ and $z$ are determined, the value of $x$ is uniquely determined, the number of positive integer solutions to (1) is
$$\begin{aligned}
& (1004+1002)+(1001+999)+\cdots+(5+3)+2 \\
= & 2006+2000+\cdots+8+2 \\
= & \frac{1}{2}(2006+2) \times 335=336340
\end{aligned}$$
In summary, there are 336340 sets of positive integer solutions.
|
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 in the output.)
Example 12 Let $m, n$ be positive integers, and $n>1$. Find the minimum value of $\left|2^{m}-5^{n}\right|$.
|
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
$$2^{m}-5^{n}=1 \text{ or } 2^{m}-5^{n}=-1 \text{.}$$
If $2^{m}-5^{n}=1$, then $m \geqslant 3$. Taking both sides modulo 8, we require
$$5^{n} \equiv 7(\bmod 8)$$
But for any positive integer $n$, $5^{n} \equiv 1$ or $5(\bmod 8)$, which is a contradiction. Therefore, $2^{m}-5^{n}=1$ does not hold.
If $2^{m}-5^{n}=-1$, then by $n>1$, we know $m \geqslant 3$. Taking both sides modulo 8, we get
$$5^{n} \equiv 1(\bmod 8)$$
This implies that $n$ is even. Let $n=2x$, where $x$ is a positive integer, then
$$2^{m}=\left(5^{x}-1\right)\left(5^{x}+1\right)$$
Since $5^{x}-1$ and $5^{x}+1$ are two consecutive even numbers, this requires
$$5^{x}-1=2,5^{x}+1=4$$
which is impossible.
Therefore, the minimum value of $\left|2^{m}-5^{n}\right|$ is 3.
|
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.,
$$x+y+(y+7)=67$$
This gives
$$x+2 y=60 .$$
Combining with
$$x \leqslant 2 y$$
We get
$$4 y \geqslant 60 \text{, i.e., } y \geqslant 15,$$
Thus,
$$x=60-2 y \leqslant 60-30=30,$$
Therefore, Jia is at most 30 years old (Note: Jia, Yi, and Bing being 30 years, 15 years, and 22 years old, respectively, meet the requirements).
|
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, the number of qualifying arrays is $327+493+658+819=2297$ (groups).
|
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 integer solutions, $c$ should be a composite number.
By direct calculation, it is known that the smallest $c$ that has exactly 3 sets of positive integer solutions is 10, and they are
$$(x, y)=(4,2),(2,3),(1,11)$$
The smallest positive integer
$$c=10$$
|
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 English, assuming it refers to 7 rectangles. If it refers to a single rectangle, the correct translation would be "A rectangle can be divided into $n$ identical squares...". Please clarify if needed.
|
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), thus $\frac{n+76}{n}=\left(\frac{x}{y}\right)^{2}$ is the square of a rational number. Therefore, we can set
$$n=k a^{2}, n+76=k b^{2}$$
where $k$, $a$, and $b$ are all positive integers. Consequently,
that is
$$\begin{array}{c}
k\left(b^{2}-a^{2}\right)=76 \\
k(b-a)(b+a)=76=2^{2} \times 19
\end{array}$$
Noting that $b-a$ and $b+a$ have the same parity, we have
$$(k, b-a, b+a)=(1,2,38),(4,1,19)$$
which gives
$$(k, a, b)=(1,18,20),(4,9,10)$$
The resulting $n$ is 324 in both cases. Therefore, $n=324$.
|
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,
$$\begin{aligned}
y^{2}-x^{2} & =84 \\
(y-x)(y+x) & =2^{2} \times 3 \times 7
\end{aligned}$$
Since $y-x$ and $y+x$ have the same parity, we have
$$(y-x, y+x)=(2,42),(6,14)$$
Solving these, we get
$$(x, y)=(20,22),(4,10)$$
However,
$$x^{2}+z^{2}=65$$
Thus, it can only be
$$(x, y)=(4,10)$$
In this case,
$$z=7$$
Therefore,
$$x^{2}+y^{2}+z^{2}=165$$
|
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
$$\begin{array}{l}
\max \left\{\alpha_{1}, \alpha_{2}\right\}=3, \max \left\{\alpha_{2}, \alpha_{3}\right\}=4, \max \left\{\alpha_{3}, \alpha_{1}\right\}=4 \\
\max \left\{\beta_{1}, \beta_{2}\right\}=\max \left\{\beta_{2}, \beta_{3}\right\}=\max \left\{\beta_{3}, \beta_{1}\right\}=3
\end{array}$$
Note that, the array $(a, b, c)$ is determined only when the non-negative integer arrays $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ and $\left(\beta_{1}, \beta_{2}, \beta_{3}\right)$ are both determined. From the conditions above, we know that $\alpha_{3}=4$, and at least one of $\alpha_{1}$ or $\alpha_{2}$ is 3, indicating that $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ has 7 possible combinations; at least two of $\beta_{1}$, $\beta_{2}$, and $\beta_{3}$ are equal to 3, so $\left(\beta_{1}, \beta_{2}, \beta_{3}\right)$ has 10 possible combinations.
In summary, the number of ordered arrays $(a, b, c)$ that satisfy the conditions is $7 \times 10 = 70$ (groups).
|
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
$$\begin{aligned}
\left(y^{2}-2\right)^{2} & =y^{4}-4 y^{2}+4 \\
& 3 \text{ holds, and } y^{4}-2 y^{2}-4 y+70 \text{ is a perfect square, so it can only be or }
$$\begin{array}{c}
y^{4}-2 y^{2}-4 y+70=\left(y^{2}-1\right)^{2} \\
y^{4}-2 y^{2}-4 y+70=\left(y^{2}\right)^{2}
\end{array}$$
The former has no positive integer solutions, while the latter has a unique positive integer solution \( y=5 \). In this case, we require \((x-12)^{2}=5^{4}\), and the positive integer \( x=37 \). Therefore, \( x^{2}+y^{4}=37^{2}+5^{4}=1994 \).
|
1994
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false
|
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