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23 Let $k$ and $m$ be positive integers. Find the minimum possible value of $\left|36^{k}-5^{m}\right|$.
|
23. Notice that
$$36^{k}-5^{m} \equiv \pm 1(\bmod 6)$$
and
$$36^{k}-5^{m} \equiv 1(\bmod 5)$$
Therefore, the positive integers that $36^{k}-5^{m}$ can take, in ascending order, are $1, 11, \cdots$.
If $36^{k}-5^{m}=1$, then $k>1$, so taking both sides modulo 8, we require
$$5^{m} \equiv-1(\bmod 8)$$
However, $5^{m} \equiv 1$ or $5(\bmod 8)$, which is a contradiction.
Additionally, when $k=1, m=2$, we have $36^{k}-5^{m}=11$. Therefore, the smallest positive integer that $36^{k}-5^{m}$ can take is 11.
Now consider the smallest positive integer that $5^{m}-36^{k}$ can take. From
$$\begin{array}{l}
5^{m}-36^{k} \equiv 4(\bmod 5) \\
5^{m}-36^{k} \equiv \pm 1(\bmod 6)
\end{array}$$
we know that
$$5^{m}-36^{k} \geqslant 19$$
In summary, the smallest possible value of $\left|36^{k}-5^{m}\right|$ is 11.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
30 Given that the three sides of $\triangle A B C$ are all integers, $\angle A=2 \angle B, \angle C>90^{\circ}$. Find the minimum perimeter of $\triangle A B C$.
|
30. Let the corresponding side lengths of $\triangle ABC$ be $a, b, c$. Draw the angle bisector of $\angle A$ intersecting $BC$ at point $D$, then
$$C D=\frac{a b}{b+c},$$
Using $\triangle A C D \backsim \triangle B C A$, we know
$$\begin{array}{c}
\frac{C D}{b}=\frac{b}{a} \\
a^{2}=b(b+c)
\end{array}$$
That is $\square$
And
Therefore
By
Let
$$\begin{array}{c}
\angle C>90^{\circ} \\
c^{2}>a^{2}+b^{2} \\
a^{2}=b(b+c) \\
(b, b+c)=d
\end{array}$$
Then $(b, c)=d$, and $d^{2} \mid a^{2}$, hence $d \mid a$.
To find the minimum value of $a+b+c$, we can set $d=1$, in which case both $b$ and $b+c$ are perfect squares. Let
$$b=m^{2}, b+c=n^{2}, m, n \in \mathbf{N}^{*}$$
Then $a=m n$. Using $a+b>c$ and $c^{2}>a^{2}+b^{2}$, we know
And $\square$
$$\begin{array}{c}
m n+m^{2}>n^{2}-m^{2} \\
\left(n^{2}-m^{2}\right)^{2}>(m n)^{2}+m^{4}
\end{array}$$
Thus
$$m>n-m$$
That is $\square$
$$n3 m^{2} n^{2},$$
That is $\square$
$$n^{2}>3 m^{2},$$
So $\square$
$$3 m^{2}<n^{2}<4 m^{2}$$
Thus, there is a perfect square between $3 m^{2}$ and $4 m^{2}$, which requires $m \geqslant 4$, at this time $n \geqslant 7$, hence
$$a+b+c \geqslant 4 \times 7+7^{2}=77$$
Clearly, $(a, b, c)=(28,16,33)$ satisfies the conditions, so the minimum perimeter of $\triangle A B C$ is 77.
|
77
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
40 Find the smallest positive integer $n$ such that the indeterminate equation
$$n=x^{3}-x^{2} y+y^{2}+x-y$$
has no positive integer solutions.
|
40. Let $F(x, y)=x^{3}-x^{2} y+y^{2}+x-y$, then $F(1,1)=1, F(1,2)=2$.
Therefore, when $n=1,2$, the equation has positive integer solutions.
Next, we prove that $F(x, y)=3$ has no positive integer solutions.
Consider the equation $F(x, y)=3$ as a quadratic equation in $y$
$$y^{2}-\left(x^{2}+1\right) y+x^{3}+x-3=0$$
If there exist positive integer solutions, then
$$\Delta=\left(x^{2}+1\right)^{2}-4\left(x^{3}+x-3\right)=x^{4}-4 x^{3}+2 x^{2}-4 x+13$$
must be a perfect square.
Notice that, when $x \geqslant 2$, we have $\Delta < (x^{2}-2 x-2)^{2}$, so when $x \geqslant 6$, $\Delta$ is not a perfect square. And when $x=1,2,3,4,5$, the corresponding $\Delta=8,-3,-8,29,168$ are not perfect squares, hence there are no positive integer solutions when $n=3$.
In summary, the smallest positive integer $n=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 1 There are 2012 lamps, numbered $1, 2, \cdots, 2012$, arranged in a row in a corridor, and initially, each lamp is on. A mischievous student performed the following 2012 operations: for $1 \leqslant k \leqslant 2012$, during the $k$-th operation, the student toggled the switch of all lamps whose numbers are multiples of $k$. Question: How many lamps are still on at the end?
|
Let $1 \leqslant n \leqslant 2012$, we examine the state of the $n$-th lamp. According to the problem, the switch of this lamp is pulled $d(n)$ times. An even number of pulls does not change the initial state of the lamp, while an odd number of pulls changes the state of the lamp from its initial state.
Using the properties of $d(n)$ and the previous discussion, since there are exactly 44 perfect squares among the numbers $1,2, \cdots, 2012$, it follows that finally, $2012-44=1968$ lamps are still on.
|
1968
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
11 Given a positive integer $n$, among its positive divisors, there is at least one positive integer ending in each of the digits $0,1,2, \cdots, 9$. Find the smallest $n$ that satisfies this condition.
|
11. The smallest $n$ that satisfies the condition is $270$.
In fact, from the condition, we know that $10 \mid n$, and we start the discussion from the factor of the last digit of $n$ being 9. If $9 \mid n$, then $90 \mid n$, and it can be directly verified that 90 and 180 are not multiples of any number ending in 7; if $19 \mid n$, then $190 \mid n$, and $n=190$ is not a multiple of any number ending in 7. However, 270 is a multiple of $10, 1, 2, 3, 54, 5, 6, 27, 18, 9$, which meets the conditions. Therefore, the smallest $n$ is 270.
|
270
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
12 Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the left $m$ digits of $M$ are multiples of $m$.
Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the number formed by the left $m$ digits of $M$ is a multiple of $m$.
|
12. Let $M=\overline{a_{1} a_{2} \cdots a_{9}}$ be a number that satisfies the given conditions. From the conditions, we know that $a_{5}=5$, and $a_{2}$, $a_{4}$, $a_{6}$, $a_{8}$ are a permutation of $2$, $4$, $6$, $8$. Consequently, $a_{1}$, $a_{3}$, $a_{7}$, $a_{9}$ are a permutation of $1$, $3$, $7$, $9$. Therefore,
$$a_{4}=2 \text { or } 6\left(\text { because } 4 \mid \overline{a_{3} a_{4}}\right),$$
Furthermore,
$$8 \mid \overline{a_{7} a_{8}}$$
Thus,
$$a_{8}=2,6$$
Hence,
$$\left(a_{4}, a_{8}\right)=(2,6),(6,2)$$
By analyzing these two cases further, we can find a number $M=381654729$ that satisfies the conditions.
|
381654729
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
17 Let $a, b, c, d$ all be prime numbers, and $a>3b>6c>12d, a^{2}-b^{2}+c^{2}-d^{2}=1749$. Find all possible values of $a^{2}+b^{2}+c^{2}+d^{2}$.
|
17. From $a^{2}-b^{2}+c^{2}-d^{2}=1749$ being odd, we know that one of $a, b, c, d$ must be even, indicating that $d=2$. Then,
from
$$\begin{array}{l}
a^{2}-b^{2}+c^{2}=1753 \\
a>3 b>6 c>12 d
\end{array}$$
we know $c \geqslant 5, b \geqslant 2 c+1, a \geqslant 3 b+1$, so
$$\begin{aligned}
a^{2}-b^{2}+c^{2} & \geqslant(3 b+1)^{2}-b^{2}+c^{2} \\
& =8 b^{2}+6 b+c^{2}+1 \\
& \geqslant 8(2 c+1)^{2}+6(2 c+1)+c^{2}+1 \\
& =33 c^{2}+44 c+15
\end{aligned}$$
Thus,
$$33 c^{2}+44 c+15 \leqslant 1753$$
Therefore, $c<7$, and combining $c \geqslant 5$ and $c$ being a prime number, we have $c=5$, leading to
$$a^{2}-b^{2}=1728=2^{6} \times 3^{3}$$
Using
$$b \geqslant 2 c+1=11, a \geqslant 3 b+1$$
we know
$$a-b \geqslant 2 b+1 \geqslant 23, a+b \geqslant 4 b+1 \geqslant 45$$
From $(a-b)(a+b)=2^{6} \times 3^{3}$ and $a, b$ both being odd primes, we have
$$(a-b, a+b)=(32,54)$$
Thus,
$$\begin{array}{c}
(a, b)=(43,11) \\
a^{2}+b^{2}+c^{2}+d^{2}=1749+2 \times\left(11^{2}+2^{2}\right)=1999
\end{array}$$
|
1999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
19 The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integers $m, n$, if $m \mid n, m<n$, then $a_{m} \mid a_{n}$, and $a_{m}<a_{n}$. Find the minimum possible value of $a_{2000}$.
|
19. From the conditions, when $m \mid n$ and $m < n$, we have $a_{n} \geqslant 2 a_{m}$. Therefore, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{4} \geqslant 2 a_{2} \geqslant 2^{2}$, similarly, $a_{8} \geqslant 2^{3}, a_{16} \geqslant 2^{4}, a_{30} \geqslant 2^{5}, a_{400} \geqslant 2^{6}, a_{2000} \geqslant 2^{7}$, which means $a_{2000} \geqslant 128$.
On the other hand, for any positive integer $n$, let the prime factorization of $n$ be
$$n=p_{1}^{q_{1}} p_{2}^{q_{2}} \cdots p_{k^{\prime}}^{q_{k}}$$
where $p_{1}<p_{2}<\cdots<p_{k}$ are prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are positive integers. Define
$$a_{n}=2^{a_{1}+a_{2}+\cdots+a_{k}}$$
Then the sequence $\left\{a_{n}\right\}$ satisfies the requirements of the problem, and
$$a_{2000}=2^{4+3} \leqslant 2^{7}$$
Therefore, the minimum value of $a_{2000}$ is 128.
|
128
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
22 Prove: For any positive integer $n$ and positive odd integer $m$, we have $\left(2^{m}-1,2^{n}+1\right)=1$.
|
22. Let $d=\left(2^{m}-1,2^{n}+1\right)$, then
hence
that is
$$\begin{array}{c}
d \mid 2^{m}-1 \\
d \mid\left(2^{m}\right)^{n}-1^{n} \\
d \mid 2^{n}-1
\end{array}$$
Additionally, $d \mid 2^{n}+1$, and since $m$ is odd, we have
$$2^{n}+1 \mid\left(2^{n}\right)^{m}+1^{m},$$
thus
$$d \mid 2^{m}+1$$
Comparing the two derived equations, we know $d \mid 2$, and since $2^{m}-1$ is odd, it follows that $d=1$.
|
1
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false
|
32 Given the pair of positive integers $(a, b)$ satisfies: the number $a^{a} \cdot b^{b}$ in decimal notation ends with exactly 98 zeros. Find the minimum value of $a b$.
|
32. Let the prime factorization of $a$ and $b$ have the powers of $2$ and $5$ as $\alpha_{1}$, $\beta_{1}$ and $\alpha_{2}$, $\beta_{2}$, respectively, then
$$\left\{\begin{array}{l}
a \cdot \alpha_{1} + b \cdot \alpha_{2} \geqslant 98 \\
a \cdot \beta_{1} + b \cdot \beta_{2} \geqslant 98
\end{array}\right.$$
and one of (1) and (2) must be an equality.
If (2) is an equality, i.e., $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$, then when $\beta_{1}$ and $\beta_{2}$ are both positive integers, the left side is a multiple of 5. When $\beta_{1}$ or $\beta_{2}$ is zero, the other must be greater than zero, in which case the left side is still a multiple of 5, leading to a contradiction. Therefore, (1) must be an equality.
From $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$, if $\alpha_{1}$ or $\alpha_{2}$ is zero, without loss of generality, let $\alpha_{2} = 0$, then $\alpha_{1} > 0$. In this case, $a \cdot \alpha_{1} = 98$. If $\alpha_{1} \geqslant 2$, then $4 \mid a$, which is a contradiction. Hence, $\alpha_{1} = 1$, and thus $a = 98$. Substituting $a = 98$ into (2), we know $\beta_{1} = 0$, so $b \cdot \beta_{2} > 98$. Combining $\alpha_{2} = 0$, we find that the minimum value of $b$ is 75.
If $\alpha_{1}$ and $\alpha_{2}$ are both positive integers, without loss of generality, let $\alpha_{1} \geqslant \alpha_{2}$. If $\alpha_{2} \geqslant 2$, then $4 \mid a$ and $4 \mid b$, leading to $4 \mid 98$, which is a contradiction. Hence, $\alpha_{2} = 1$. Further, if $\alpha_{1} = 1$, then $a + b = 98$, but $\frac{a}{2}$ and $\frac{b}{2}$ are both odd, so $\frac{a}{2} + \frac{b}{2}$ is even, which is a contradiction. Therefore, $\alpha_{1} > 1$. In this case, if $\beta_{1}$ and $\beta_{2}$ are both positive integers, then $5 \mid a$ and $5 \mid b$, which contradicts $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$. Hence, one of $\beta_{1}$ and $\beta_{2}$ must be zero. If $\beta_{1} = 0$, then from (2) we know $b \cdot \beta_{2} > 98$, in which case the number of trailing zeros in $b^b$ is greater than 98 (since, in this case, $10 \mid b$. When $\beta_{2} = 1$, $b \geqslant 100$, so $10^{100} \mid b^b$. When $\beta_{2} \geqslant 2$, $50 \mid b$, if $b > 50$, then $10^{100} \mid b^b$; if $b = 50$, then $a \cdot \alpha_{1} = 48$, in which case when $\alpha_{1} \geqslant 4$, $2^5 \mid a \cdot \alpha_{1}$, and when $\alpha_{1} \leqslant 3$, $2^4 \nmid a \cdot \alpha_{1}$, both leading to a contradiction, so the number of trailing zeros in $b^b$ is greater than 98).
Similarly, if $\beta_{2} = 0$, then $a \cdot \beta_{1} > 98$, and similarly, the number of trailing zeros in $a^a$ is greater than 98, which is a contradiction. In summary, the minimum value of $ab$ is 7350 (when $(a, b) = (98, 75)$ or $(75, 98)$).
|
7350
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
39 The numbers $1,2, \cdots, 33$ are written on the blackboard. Each time, it is allowed to perform the following operation: take any two numbers $x, y$ from the blackboard that satisfy $x \mid y$, remove them from the blackboard, and write the number $\frac{y}{x}$. Continue until there are no such two numbers on the blackboard. How many numbers will be left on the blackboard at least?
|
39. Consider the objective function $S=$ the product of all numbers on the blackboard.
Initially, $S=33!=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$. Each operation targets $x, y (x \mid y)$, where $y=kx$. Removing $x, y$ and replacing them with $k$ results in $S$ becoming $\frac{S}{xy} \cdot k = \frac{S}{x^2}$. This indicates that after each operation, the parity of the exponent of each prime factor in $S$ remains unchanged. Specifically, $2, 3, 5,$ and $11$ always divide the $S$ obtained after each operation. Since $2 \times 3 \times 5 \times 11 > 33$, at least two numbers must remain on the blackboard such that their product is a multiple of $2 \times 3 \times 5 \times 11$.
Furthermore, note that the primes $17, 19, 23, 29, 31$ have no multiples greater than themselves that are less than or equal to 33. Therefore, no operation can remove any of these numbers.
The above discussion shows that at least 7 numbers must remain on the blackboard.
The following example demonstrates that exactly 7 numbers can remain:
$$\begin{array}{l}
(32,16) \rightarrow 2, (30,15) \rightarrow 2, (28,14) \rightarrow 2, (26,13) \rightarrow 2, (24,12) \rightarrow 2 \\
(22,11) \rightarrow 2; (27,9) \rightarrow 3, (21,7) \rightarrow 3, (18,6) \rightarrow 3; (25,5) \rightarrow 5 \\
(20,4) \rightarrow 5; (8,2) \rightarrow 4 \\
(5,5) \rightarrow 1; (4,2) \rightarrow 2; (3,3) \rightarrow 1, (3,3) \rightarrow 1, (2,2) \rightarrow 1, (2,2) \rightarrow 1 \\
(2,2) \rightarrow 1
\end{array}$$
Thus, the blackboard is left with $10, 17, 19, 23, 29, 31, 33$ and 7 ones. The 7 ones can be removed by pairing them with 17 in 7 operations.
In conclusion, at least 7 numbers remain.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
2 (1) Let $m, n$ be coprime positive integers, $m, n>1$. Let $a$ be an integer coprime to $m n$. Suppose the orders of $a$ modulo $m$ and modulo $n$ are $d_{1}, d_{2}$, respectively, then the order of $a$ modulo $m n$ is $\left[d_{1}, d_{2}\right]$;
(2) Find the order of 3 modulo $10^{4}$.
|
2. (1) Let the order of $a$ modulo $mn$ be $r$. From $a^{r} \equiv 1(\bmod m n)$, we can deduce $a^{r} \equiv 1(\bmod m)$ and $a^{r} \equiv 1(\bmod n)$. Therefore, $d_{1} \mid r$ and $d_{2} \mid r$, which implies $\left[d_{1}, d_{2}\right] \mid r$. On the other hand, from $a^{d_{1}} \equiv 1(\bmod m)$ and $a^{d_{2}} \equiv 1(\bmod n)$, we can infer $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod m)$ and $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod n)$. Since $(m, n)=1$, it follows that $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod m n)$, thus $r \mid \left[d_{1}, d_{2}\right]$. Combining both results, we conclude $r=\left[d_{1}, d_{2}\right]$.
(2) Direct calculation shows that the order of 3 modulo $2^{4}$ is 4. It is also easy to see that the order of 3 modulo 5 is 4, hence by part (1) of Example 5, the order of 3 modulo $5^{4}$ is $4 \times 5^{3}$. Therefore, by part (1) of this problem, the order of 3 modulo $10^{4}$ is $[4, 4 \times 5^{3}]=500$.
|
500
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 1 A positive integer, when added to 100, becomes a perfect square. If 168 is added to it, it becomes another perfect square. Find this number.
|
Let the number to be found be $x$. By the problem, there exist positive integers $y, z$, such that
$$\left\{\begin{array}{l}
x+100=y^{2} \\
x+168=z^{2}
\end{array}\right.$$
Eliminating $x$ from the above two equations, we get
$$z^{2}-y^{2}=68$$
Factoring the left side of this binary quadratic equation and standardizing the right side, we have
$$(z-y)(z+y)=2^{2} \times 17$$
Since $z-y$ and $z+y$ are both positive integers, and $z-y < z+y$, it follows from (1) and the unique factorization theorem (Unit 3 (5)) that we must have
$$\left\{\begin{array} { l }
{ z - y = 1 , } \\
{ z + y = 2 ^ { 2 } \times 1 7 ; }
\end{array} \left\{\begin{array}{l}
z-y=2, \\
z+y=2 \times 17 ;
\end{array} ;\left\{\begin{array}{l}
z-y=2^{2}, \\
z+y=17
\end{array}\right.\right.\right.$$
Solving these systems of linear equations one by one, we find $y=16, z=18$, hence $x=156$.
|
156
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
3. Let $m>n \geqslant 1$, find the minimum value of $m+n$ such that $: 1000 \mid 1978^{m}-1978^{n}$.
|
3. Solution: When $n \geqslant 3$, from $1000 \mid 1978^{m}-1978^{n}$, we get
$125 \mid 1978^{m-n}-1$, and since $1978=15 \times 125+103$,
we have $125 \mid 103^{m-n}-1$, so $25 \mid 103^{m-n}-1$.
Thus, $25 \mid 3^{m-n}-1$.
Let $m-n=l$, then: $25 \mid 3^{l}-1$ (obviously $l$ is even, otherwise it is easy to see that $3^{l}-1 \equiv \pm 3-1 \equiv 4$ or $2(\bmod 5)$, which is a contradiction), let $l=2a$,
then $25 \mid 9^{a}-1$, and $9^{a}-1=(10-1)^{a}-1$
$$\begin{array}{l}
\equiv C_{a}^{2} 10^{2}(-1)^{a-2}+C_{a}^{1} 10(-1)^{a-1}+(-1)^{a}-1(\bmod 25) \\
\equiv 10 a(-1)^{a-1}+(-1)^{a}-1(\bmod 25)
\end{array}$$
Obviously, $a$ is even.
So $9^{a}-1 \equiv 10 a(-1)(\bmod 25)$.
Thus, $5 \mid a$, and since $a$ is even, $10 \mid a$.
Let $a=10b$, then $l=20b$.
So $103^{t}-1=(100+3)^{20b}-1$
$$\begin{array}{l}
\equiv C_{20b}^{1} 100 \cdot 3^{20b-1}+3^{20b}-1(\bmod 125) \\
\equiv 9^{10b}-1(\bmod 125) \\
\equiv(10-1)^{10b}-1(\bmod 125) \\
\equiv C_{10b}^{2} \cdot 10^{2}-C_{10b}^{1} 10(\bmod 125) \\
\equiv-100b(\bmod 125)
\end{array}$$
So $5 \mid b$, thus $l=20b \geqslant 100$. Therefore, $m+n=l+2n \geqslant 106$.
When $m=103, n=3$, the above equality holds.
If $n \leqslant 2$, it is easy to see that $8 \times 1978^{m}-1978^{n} \Rightarrow 1000 \times 1978^{m}-1978^{n}$, so the minimum value of $m+n$ is 106.
|
106
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
14. Let $S$ be the set of all prime numbers $p$ that satisfy the following condition: the number of digits in the smallest repeating block of the decimal part of $\frac{1}{p}$ is a multiple of 3, i.e., for each $p \in S$, there exists the smallest positive integer $r=r(p)$, such that $\frac{1}{p}=0 . a_{1} a_{2} \cdots a_{3 r} a_{1} a_{2} \cdots a_{3 r} \cdots$
For each $p \in S$ and any integer $k \geqslant 1$, define $f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2 r(p)}$.
(1) Prove that the set $S$ contains infinitely many primes;
(2) For $k \geqslant 1$ and $p \in S$, find the maximum value of $f(k, p)$.
|
14. Prove: (1) The length of the smallest repeating cycle of $\frac{1}{p}$ is the smallest integer $d (d \geqslant 1)$ such that $10^{d}-1$ is divisible by $p$.
Let $q$ be a prime number, and $N_{q}=10^{2 q}+10^{q}+1$. Then $N_{q} \equiv 3(\bmod q)$. Let $p_{q}$ be a prime factor of $\frac{N_{q}}{3}$. Then $p_{q}$ cannot be divisible by 3, because $N_{q}$ is a factor of $10^{3 q}-1$. The repeating cycle length of $\frac{1}{p_{q}}$ is $3 q$, so the length of the smallest repeating cycle of $\frac{1}{p_{q}}$ is a divisor of $3 q$. If the smallest repeating cycle length is $q$, then from $10^{q} \equiv 1\left(\bmod p_{q}\right)$ we get $N_{q}=10^{2 q}+10^{q}+1 \left(\bmod p_{q}\right)$, which is a contradiction! If the smallest repeating cycle length is 3, there is only one case, i.e., $p_{q}$ is a factor of $10^{3}-1=3^{3} \times 37$, i.e., $p_{q}=37$. In this case, $N_{q}=3 \times 37 \equiv 3(\bmod 4)$, while $N_{q}=10^{2 q}+10^{q}+1 \equiv 1(\bmod 4)$, which is a contradiction! Thus, for each prime $q$, we can find a prime $p_{q}$ such that the length of the smallest repeating cycle of the decimal part of $\frac{1}{p_{q}}$ is $3 q$.
(2) Let the prime $p \in S$, and $3 r(p)$ be the length of the smallest repeating cycle of $\frac{1}{p}$. Then $p$ is a factor of $10^{3 r(p)}-1$ but not a factor of $10^{r(p)}-1$, so $p$ is a factor of $N_{r}=10^{2 r(p)}+10^{r(p)}+1$.
Let $\frac{1}{p}=0 . \overline{a_{1} a_{2} \cdots}, x_{j}=\frac{10^{j-1}}{p}, y_{j}=\left\{x_{j}\right\}=0 . \overline{a_{j} a_{j+1} \cdots}$, where $\{x\}$ denotes the fractional part of $x$, then $a_{j}<10 y_{j}$. Thus, $f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2 r(p)}<10\left(y_{k}+y_{k+r(p)}+y_{k+2 r(p)}\right)$. Since $x_{k}+x_{k+r(p)}+x_{k+2 r(p)}=\frac{10^{k-1} \cdot N_{r(p)}}{p}$ is an integer, $y_{k}+y_{k+r(p)}+y_{k+2 r(p)}$ is also an integer and is a number less than 3, so $y_{k}+y_{k+r(p)}+y_{k+2 r(p)} \leqslant 2$, thus $f(k, p)<20$, so $f(k, p) \leqslant 19$. Since $f(2,7)=4+8+7=19$, the maximum value sought is 19.
|
19
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false
|
Example 5 (2005 National High School Mathematics Competition Question) Define the function
$$f(k)=\left\{\begin{array}{l}
0, \text { if } k \text { is a perfect square } \\
{\left[\frac{1}{\{\sqrt{k}\}}\right], \text { if } k \text { is not a perfect square }}
\end{array} \text {, find } \sum_{k=1}^{240} f(k)\right. \text {. }$$
|
When $k$ is not a perfect square, $k$ must lie between two consecutive perfect squares. Divide $[1,240]$ into several intervals with perfect squares as boundaries, and sum $f(k)$ for each subinterval.
Solving $15^{2}<240<16^{2}$.
Since $f(k)=0$ when $k$ is a perfect square, we have
$$\begin{aligned}
\sum_{k=1}^{240} f(k) & =\sum_{k=1+1}^{2^{2}-1} f(k)+\sum_{k=2^{2}+1}^{3^{2}-1} f(k)+\cdots+\sum_{k=14^{2}+1}^{15^{2}-1} f(k)+\sum_{k=226}^{240} f(k) \\
& =\sum_{n=1}^{14} \sum_{k=n^{2}+1}^{(n+1)^{2}-1} f(k)+\sum_{k=226}^{240} f(k) .
\end{aligned}$$
When $n^{2}+1 \leqslant k \leqslant(n+1)^{2}-1$, $[\sqrt{k}]=n,\{k\}=\sqrt{k}-[k]=\sqrt{k}-n$.
Thus, $\left[\frac{1}{\{\sqrt{k}\}}\right]=\left[\frac{1}{\sqrt{k}-n}\right]=\left[\frac{\sqrt{k}+n}{k-n^{2}}\right]=\left[\frac{[\sqrt{k}+n]}{k-n^{2}}\right]=\left[\frac{2 n}{k-n^{2}}\right]$.
So $\sum_{k=n^{2}+1}^{(n+1)^{2}-1} f(k)=\sum_{k=n^{2}+1}^{(n+1)^{2}-1}\left[\frac{2 n}{k-n^{2}}\right]=\sum_{i=1}^{2 n}\left[\frac{2 n}{i}\right]$,
$$\begin{aligned}
\sum_{k=226}^{240} f(k) & =\sum_{k=226}^{240}\left[\frac{2 \cdot 15}{k-15^{2}}\right]=\sum_{i=1}^{15}\left[\frac{30}{i}\right] . \\
\text { Therefore, } \sum_{k=1}^{240} f(k) & =\sum_{n=1}^{14} \sum_{i=1}^{2 n}\left[\frac{2 n}{i}\right]+\sum_{i=1}^{15}\left[\frac{30}{i}\right] \\
& =\sum_{n=1}^{14}\left(\sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+\sum_{i=n+1}^{2 n}\left[\frac{2 n}{i}\right]\right)+\sum_{i=1}^{15}\left[\frac{30}{i}\right] \\
& =\sum_{n=1}^{14}\left(\sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+n\right)+\sum_{i=1}^{15}\left[\frac{30}{i}\right] \\
& =\sum_{n=1}^{14} \sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+\sum_{i=1}^{15}\left[\frac{30}{i}\right]+\sum_{n=1}^{14} n \\
& =\sum_{n=1}^{15} \sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+105 .
\end{aligned}$$
Let $T_{n}=\sum_{i=1}^{n}\left[\frac{2 n}{i}\right], 1 \leqslant n \leqslant 15$.
$$\begin{array}{l}
\text { Then } T_{1}=\left[\frac{2}{1}\right]=2, T_{2}=\left[\frac{4}{1}\right]+\left[\frac{4}{2}\right]=4+2=6, \\
T_{3}=\left[\frac{6}{1}\right]+\left[\frac{6}{2}\right]+\left[\frac{6}{3}\right]=6+3+2=11, \\
T_{4}=\left[\frac{8}{1}\right]+\left[\frac{8}{2}\right]+\left[\frac{8}{3}\right]+\left[\frac{8}{4}\right]=8+4+2 \times 2=16, \\
T_{5}=\left[\frac{10}{1}\right]+\left[\frac{10}{2}\right]+\cdots+\left[\frac{10}{5}\right]=10+5+3+2 \times 2=22, \\
T_{6}=\left[\frac{12}{1}\right]+\left[\frac{12}{2}\right]+\cdots+\left[\frac{12}{6}\right]=12+6+4+3+2 \times 2=29, \\
T_{7}=\left[\frac{14}{1}\right]+\left[\frac{14}{2}\right]+\cdots+\left[\frac{14}{7}\right]=14+7+4+3+2 \times 3=34, \\
T_{8}=\left[\frac{16}{1}\right]+\left[\frac{16}{2}\right]+\cdots+\left[\frac{16}{8}\right]=16+8+5+4+3+2 \times 3=42, \\
T_{9}=\left[\frac{18}{1}\right]+\left[\frac{18}{2}\right]+\cdots+\left[\frac{18}{9}\right]=18+9+6+4+3 \times 2+2 \times 3=49, \\
T_{10}=\left[\frac{20}{1}\right]+\left[\frac{20}{2}\right]+\cdots+\left[\frac{20}{10}\right]=20+10+6+5+4+3+2 \times 4=56, \\
T_{11}=\left[\frac{22}{1}\right]+\left[\frac{22}{2}\right]+\cdots+\left[\frac{22}{11}\right]=22+11+7+5+4+3 \times 2+2 \times 4=63, \\
T_{12}=\left[\frac{24}{1}\right]+\left[\frac{24}{2}\right]+\cdots+\left[\frac{24}{12}\right]=24+12+8+6+4 \times 2+3 \times 2+2 \times 4=72, \\
T_{13}=\left[\frac{26}{1}\right]+\left[\frac{26}{2}\right]+\cdots+\left[\frac{26}{13}\right]=26+13+8+6+5+4+3 \times 2+2 \times 5=78, \\
T_{14}=\left[\frac{28}{1}\right]+\left[\frac{28}{2}\right]+\cdots+\left[\frac{28}{14}\right]=28+14+9+7+5+4 \times 2+3 \times 2+2 \times 5=87, \\
T_{15}=\left[\frac{30}{1}\right]+\left[\frac{30}{2}\right]+\cdots+\left[\frac{30}{15}\right]=30+15+10+7+6+5+4+3 \times 3+2 \times 5=96, \\
t_{2} \sum_{k=1}^{240} f(k)=\sum_{n=1}^{15} T_{n}+105 \\
=2+6+11+16+22+29+34+42+49+56+63+72+78+87+96+105 \\
=768
\end{array}$$
|
768
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
For non-negative integers $x$, the function $f(x)$ is defined as follows:
$$f(0)=0, f(x)=f\left(\left[\frac{x}{10}\right]\right)+\left[\lg \frac{10}{x-10\left[\frac{x-1}{10}\right]}\right]$$
What is the value of $x$ when $f(x)$ reaches its maximum in the range $0 \leqslant x \leqslant 2006$?
|
1. Solution: Let $x=10 p+q$, where $p$ and $q$ are integers and $0 \leqslant q \leqslant 9$.
Then $\left[\frac{x}{10}\right]=\left[p+\frac{q}{10}\right]=p$.
Thus, $\left[\frac{x-1}{10}\right]=\left[p+\frac{q-1}{10}\right]$ has a value of $p-1$ (when $q=0$) or $p$ (when $q \neq 0$), and $x-10\left[\frac{x-1}{10}\right]$ has a value of 10 (when $q=0$) or $q$ (when $q \neq 0$),
$\left[\lg \frac{10}{x-10\left[\frac{x-1}{10}\right]}\right]$ has a value of 1 (when $q=1$) or $0$ (when $q \neq 1$).
Therefore, $f(x)=f(p)+1$ (when $q=1$) or $f(p)$ (when $q \neq 1$).
Hence, the value of $f(x)$ is the number of digit 1s in the decimal representation of $x$, so when $0 \leqslant x \leqslant 2006$, the maximum value of $f(x)$ is 4, at which point $x=1111$.
|
1111
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
4. Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$, prove that: $k=5$.
|
4. Prove: If $a=b$, then $k=\frac{2 a^{2}}{a^{2}-1}=2+\frac{2}{a^{2}-1}$.
Then $\left(a^{2}-1\right) \mid 2$ and $a^{2}-1>0$.
So $a^{2}-1=1$ or 2, but this contradicts $a \in \mathbf{N}_{+}$.
Therefore, $a \neq b$, without loss of generality, let $a>b$.
When $b=1$, $k=\frac{a^{2}+1}{a-1}=a+1+\frac{2}{a-1}$, then $(a-1) \mid 2$ and $a \geqslant b+1=2$, so $a=2$ or 3, in either case, $k=5$.
Without loss of generality, assume $(a, b)$ is the ordered pair that minimizes $a+b$ among all ordered pairs satisfying $\frac{a^{2}+b^{2}}{a b-1}=k$.
From $\frac{a^{2}+b^{2}}{a b-1}=k$, we get $a^{2}-k b \cdot a+b^{2}+k=0$.
Viewing (1) as a quadratic equation in $a$, by Vieta's formulas, the roots are $a$ and $(k b-a)$, and $k b-a = \frac{b^{2}+k}{a} > 0$, and since $k b-a \in \mathbf{Z}$, we have $k b-a \in \mathbf{N}_{+}$, and $\frac{(k b-a)^{2}+b^{2}}{(k b-a) b-1}=k$. Thus, by the definition of $(a, b)$, we have $a+b \leqslant \frac{b^{2}+k}{a}+b\left(\frac{b^{2}+k}{a}=k b-a\right)$, so $a^{2}-b^{2} \leqslant k=\frac{a^{2}+b^{2}}{a b-1}$.
Therefore, $a^{2}-b^{2}a-b \geqslant 1$. That is, $(a-1)(b-1)b=2$, a contradiction. In conclusion, $k=5$.
|
5
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false
|
6. In the Cartesian coordinate system, the number of integer points $(x, y)$ that satisfy $(|x|-1)^{2}+(|y|-1)^{2}<2$ is $\qquad$ .
|
6. 16 Since $(|x|-1)^{2} \leqslant(|x|-1)^{2}+(|y|-1)^{2}<2$.
Therefore, $(|x|-1)^{2}<2, -1 \leqslant|x|-1 \leqslant 1$, which means $0 \leqslant|x| \leqslant 2$, similarly $0 \leqslant|y| \leqslant 2$.
Upon inspection, $(x, y)=(-1, \pm 1),(-1,0),(1,0),(0, \pm 1),(-1, \pm 2),(1, \pm 2),(-2, \pm 1),(2, \pm 1)$ are the 16 integer solutions that satisfy the original inequality.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 3 In the Cartesian coordinate system, the number of integer points that satisfy $(1) y \geqslant 3 x$; (2) $y \geqslant \frac{1}{3} x ;(3) x+y \leqslant 100$ is how many?
|
As shown in Figure 7-5, the region enclosed by the lines $y=3x$, $y=\frac{1}{3}x$, and $x+y=100$ forms a triangular region. The three vertices of this triangle are $O(0,0)$, $A(75,25)$, and $B(25,75)$. $\square$
Next, we calculate the number of integer points $N$ on the boundary and inside $\triangle OAC$. For a grid point $M(m, n)$ on the boundary or inside $\triangle OAC$, when $0 \leqslant m < 75$, $0 \leqslant n \leqslant \frac{1}{3} m$. When $75 \leqslant m \leqslant 100$, $0 \leqslant n \leqslant 100-m$. Therefore,
$$\begin{aligned}
N= & \sum_{m=0}^{74}\left(\left[\frac{1}{3} m\right]+1\right)+\sum_{m=75}^{100}(100-m+1) \\
& =\sum_{k=0}^{24}\left(\left[\frac{1}{3} \cdot 3 k\right]+1+\left[\frac{1}{3} \cdot(3 k+1)\right]+1+\left[\frac{1}{3} \cdot(3 k+2)\right]+1\right)+\sum_{m=75}^{100}(100-m+1) \\
& =\sum_{k=0}^{24}(3 k+3)+\sum_{m=75}^{100}(100-m+1) \\
& =975+351=1326
\end{aligned}$$
Similarly, the number of integer points on the boundary and inside $\triangle OBD$ is also $N=1326$.
The number of integer points on the line segment $OA$ is $25+1=26$.
Similarly, there are 26 integer points on the line segment $OB$.
Let $L$ be the number of integer points inside $\triangle OAB$. Then $L+2N-26 \times 2$ represents the number of integer points on the boundary and inside $\triangle OCD$, which numerically equals $\sum_{x=1}^{100}([100-x]+1)=\sum_{x=0}^{100}(101-x)=5151$.
Therefore, $L+2 \times 1326-2 \times 26=5151$.
Thus, the number of grid points $L=2551$.
|
2551
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
5. In $1 \sim 1000$, the number of pairs $(x, y)$ that make $\frac{x^{2}+y^{2}}{7}$ an integer is $\qquad$ pairs.
|
5. 10011 From $\frac{x^{2}+y^{2}}{7} \in \mathbf{Z}$ we know $7 \mid x^{2}+y^{2}$.
When $7 \mid x$ and $7 \mid y$, it is obvious that $7 \mid x^{2}+y^{2}$.
When $7 \nmid x$ or $7 \nmid y$, from $7 \mid x^{2}+y^{2}$ we know that $7 \nmid x$ and $7 \nmid y$. By Fermat's Little Theorem, $x^{6} \equiv 1(\bmod 7)$, which means $\left(x^{2}\right)^{3} \equiv 1(\bmod 7)$. Also, $x^{2} \equiv -y^{2}(\bmod 7)$, so $\left(-y^{2}\right)^{3} \equiv 1(\bmod 7)$, which means $-y^{6} \equiv 1(\bmod 7)$. By Fermat's Little Theorem, $y^{6} \equiv 1(\bmod 7)$, which contradicts $-y^{6} \equiv 1(\bmod 7)$.
In summary, $\frac{x^{2}+y^{2}}{7} \in \mathbf{Z}(x, y \in \mathbf{Z}) \Leftrightarrow 7|x, 7| y$.
Therefore, the number of integer pairs $(x, y)$ is $C_{142}^{2}=10011$.
|
10011
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
6. Let the planar region $T=\{(x, y) \mid x>0, y>0, x y \leqslant 48\}$, then the number of lattice points within $T$ is $\qquad$ .
|
$$\begin{array}{l}
\text { 6. } 202 \text { Let } T_{1}=\left\{(x, y) \mid 0<x \leqslant \sqrt{48}, 0<y \leqslant \frac{n}{x}\right\}, \\
T_{2}=\left\{(x, y) \mid 0<y \leqslant \sqrt{48}, 0<x \leqslant \frac{n}{y}\right\} .
\end{array}$$
Then $T=T_{1} \cup T_{2}, T_{1} \cap T_{2}=\{(x, y) \mid 0<x \leqslant \sqrt{48}, 0<y<\sqrt{48}\}$, using $\|x\|$ to denote the number of lattice points in set $x$, thus according to the principle of inclusion-exclusion, we have
$$\begin{aligned}
\|T\| & =\left\|T_{1} \cup T_{2}\right\| \\
& =\left\|T_{1}\right\|+\left\|T_{2}\right\|-\left\|T_{1} \cap T_{2}\right\| \\
& =\sum_{0<x \leqslant \sqrt{48}}\left[\frac{48}{x}\right]+\sum_{0<y \leqslant \sqrt{48}}\left[\frac{48}{y}\right]-[\sqrt{48}]^{2} \\
& =2 \sum_{0<k \leqslant 6}\left[\frac{48}{k}\right]-36=119 \times 2-36=202
\end{aligned}$$
|
202
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 7 How many positive integer factors does 20! have?
|
Analyze writing 20! in its standard factorization form $n=\beta_{1}^{a_{1}} \beta_{2}^{a_{2}} \cdots \beta_{k}^{q_{k}}$, and then using $r(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots$ $\left(\alpha_{k}+1\right)$ to calculate the result.
Solution Since the prime numbers less than 20 are $2,3,5,7,11,13,17,19$, in the standard factorization of 20!, the highest power of 2 $=\left[\frac{20}{2}\right]+\left[\frac{20}{4}\right]+\left[\frac{20}{8}\right]+\left[\frac{20}{16}\right]=18$,
the highest power of 3 $=\left[\frac{20}{3}\right]+\left[\frac{20}{9}\right]=8$,
the highest power of 5 $=\left[\frac{20}{5}\right]=4$,
the highest powers of $7,11,13,17,19$ are $2,1,1,1,1$ respectively.
Thus, the standard factorization of 20! is
$$20!=2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19$$
Therefore, the number of positive divisors $r(20!)$ of 20! is:
$$\begin{aligned}
r(20!) & =(18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1) \\
& =19 \cdot 9 \cdot 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \\
& =41040
\end{aligned}$$
Thus, the number of positive divisors of $20!$ is 41040.
|
41040
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 1 (2001 Irish Mathematical Olympiad) Find the smallest positive integer $a$ such that there exists a positive odd integer $n$ satisfying $2001 \mid$
$$55^{n}+a \cdot 32^{n}$$
|
Analysis Using the properties of congruence, noting that $2001=23 \times 87$, we can find that $a \equiv 1(\bmod 87)$ and $a \equiv-1$ $(\bmod 23)$. Thus, we can obtain the smallest value of $a$ that meets the requirements.
Solution Since $2001=87 \times 23$. By the problem, there exists a positive odd number $n$ such that
$87 \mid 55^{n}+a \cdot 32^{n}$,
and $23 \mid 55^{n}+a \cdot 32^{n}$.
From (1), we have: $0 \equiv 55^{n}+a \cdot 32^{n} \quad(\bmod 87)$
$$\begin{array}{l}
\equiv(-32)^{n}+a \cdot 32^{n} \quad(\bmod 87) \\
\equiv 32^{n}(a-1) \quad(\bmod 87)
\end{array}$$
Since $\left(32^{n}, 87\right)=1$, it follows that $a-1 \equiv 0(\bmod 87)$
From (2), we have: $0 \equiv 55^{n}+a \cdot 32^{n}(\bmod 23)$
$$\equiv 32^{n}+a \cdot 32^{n} \quad(\bmod 23)$$
Since $\left(32^{n}, 23\right)=1$, it follows that $a+1 \equiv 0(\bmod 23)$,
hence $a \equiv-1(\bmod 23)$
From (3), let $a=87 k+1(k \in \mathbf{N})$ and substitute into (4) to get
$$23|87 k+2 \Rightarrow 23| 18 k+2 \Rightarrow 23|-5 k+25 \Rightarrow 23| k-5$$
Since $k \geqslant 0$, it follows that $k \geqslant 5$.
Thus, the smallest value of $a$ is $87 \times 5+1=436$.
|
436
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
$$\text { 9. } 3333^{8888}+8888^{3333} \equiv$$
$$\qquad (\bmod 7)$$.
|
9.0 Hint: $3333 \equiv 1(\bmod 7)$, so $3333^{8888} \equiv 1(\bmod 7)$. $8888 \equiv 5(\bmod 7)$, so $8888^{3} \equiv 5^{3}(\bmod 7) \equiv-1(\bmod 7)$. Therefore, $8888^{3333} \equiv-1(\bmod 7)$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
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