problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
2. The Fundamental Theorem of Arithmetic states that every positive integer greater than 1 can be decomposed into a product of a finite number of prime numbers. If the order of the prime factors in the product is not considered, then the decomposition is unique, i.e., \( n = p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdots \cd... | We need to prove that $k=l$ and the prime numbers $p_{1}, p_{2}, \cdots, p_{k}$ are a permutation of $q_{1}, q_{2}, \cdots, q_{l}$.
From (1), we see that $p_{1}$ divides $q_{1} q_{2} \cdots q_{l}$, so $p_{1}$ divides $q_{1}$ or $q_{2} \cdots q_{l}$. If $p_{1} \mid q_{1}$, since $p_{1}$ and $q_{1}$ are both primes, then... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,304 |
Example 3 (29th Russian Mathematical Olympiad) Prove: from any 5 pairwise coprime three-digit numbers, it is possible to select 4 numbers that are coprime.
untranslated text retained the original line breaks and formatting. | Analyzing, first set the greatest common divisor of any 4 numbers, then use proof by contradiction to prove that at least one of the 5 common divisors is 1.
Proof: Let the 5 numbers be \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\).
If the proposition does not hold, then there exist
\[
\begin{array}{l}
\left(a_{1}, a_{2}, a_{3... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,305 |
Example 4 (15th Asia Pacific Mathematical Olympiad) Let $k, k \geqslant 14, p_{k}$ be the largest prime less than $k$. If $p_{k} \geqslant \frac{3 k}{4}$, and $n$ is a composite number, prove:
(1) If $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !.
(2) If $n>2 p_{k}$, then $n$ divides $(n-k)$ !. | Prove (1) When $n=2 p_{k}$,
Since $k>p_{k}$, then $p_{k}>2 p_{k}-k=n-k$, so, $p_{k} \nmid(n-k)!$, hence $n \nmid(n-k)!$.
(2) When $n>2 p_{k}$,
Since $n$ is a composite number, let $n=a b(2 \leqslant a \leqslant b)$, if $a \geqslant 3$.
( i ) $a \neq b$, then $n>2 p_{k} \geqslant \frac{3 k}{2}, b \leqslant \frac{n}{3}$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,306 |
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 $2,3,5,7,11,13,1... | 41040 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,307 |
3. Given the equation $x^{4}-p x^{3}+q=0$ has an integer root, find the prime numbers $p$ and $q$.
untranslated text:
已知方程 $x^{4}-p x^{3}+q=0$ 有一整数根, 求素数 $p 、 q$.
translated text:
Given the equation $x^{4}-p x^{3}+q=0$ has an integer root, find the prime numbers $p$ and $q$. | 3. Solution: Let the integer $x$ satisfy $q=x^{3}(p-x)$.
(1)
Then $x \mid q$.
Since $q$ is a prime, it can only be that $x= \pm 1, x= \pm q$.
When $x= \pm q$, from (1) we have
$q^{2} \mid 1$, which is impossible (since $q$ is a prime).
When $x=-1$, we have $p+q+1=0$, which is also impossible.
When $x=1$, from (1) we ge... | p=3, q=2 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,308 |
7. (2003 IMO Slovenia National Team Selection Test) Let $S$ be a finite non-empty set of positive integers greater than 1, and has the following property: there exists a number $s \in S$, such that for any positive integer $n$, either $(s, n)=1$ or $(s, n)=s$. Prove: there must exist two numbers $s, t \in S$ (not neces... | 7. Proof: Let $n$ be the smallest positive integer that is not coprime with any number in the set $S$ (such a number exists; in fact, since $1 \notin S$, the product of all numbers in $S$ is not coprime with any number in $S$).
Among all prime factors of $n$, each prime appears exactly once, otherwise, $n$ would not b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,309 |
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... | 436 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,312 |
Example 2 Let $p$ be a prime, $1 \leqslant k \leqslant p-1$, prove: $C_{p-1}^{k} \equiv(-1)^{k}(\bmod p)$. | $$\begin{array}{l}
A \equiv \frac{(p-1)(p-2) \cdots(p-k)}{k!}(\bmod p), \\
\text { so } k!\cdot A \equiv(p-1)(p-2) \cdots(p-k)(\bmod p) \text {, } \\
\equiv(-1) \cdot(-2) \cdots(-k) \quad(\bmod p) \\
\equiv(-1)^{k} k!\quad(\bmod p) . \\
\text { Also }(k!, p)=1,
\end{array}$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,313 |
$$\text { 9. } 3333^{8888}+8888^{3333} \equiv$$
$$\qquad (\bmod 7)$$. | 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 | 742,315 |
Example 2 (2003 Croatian Mathematical Olympiad) For all odd primes $p$ and positive integers $n(n \geqslant p)$, prove:
$$C_{n}^{p} \equiv\left[\frac{n}{p}\right](\bmod p)$$ | Analysis and Solution Note that $C_{n}^{p}=\frac{n(n-1) \cdots(n-p+1)}{p!}$, and $n, n-1, \cdots, n-p+1$ form a complete residue system modulo $p$, so there exists $a \in\{n, n-1, \cdots, n-p+1\}$ such that $p \mid a$, clearly $a=p \cdot\left[\frac{n}{p}\right]$. Let $C_{n}^{p} \equiv c(\bmod p)$.
Then $C_{n}^{p}=\fra... | C_{n}^{p} \equiv\left[\frac{n}{p}\right](\bmod p) | Number Theory | proof | Yes | Yes | number_theory | false | 742,317 |
Example 3 Let $n$ be a positive integer, integer $k$ be coprime with $n$, and $0<k<n$. Let $M=\{1,2, \cdots, n-1\}$, and color each number in $M$ with one of two colors, black or white, according to the following rules: (i) for each $i$ in $M, i$ and $n-i$ are the same color, (ii) for each $i$ in $M, i \neq k, i$ and $... | Analysis: From $(k, n)=1$, the numbers in $M$ can be reordered, and then it is only necessary to prove that adjacent numbers have the same color.
Proof: Since $(k, n)=1, 0, 1, 2, \cdots, n-1$ is a complete residue system modulo $n$,
so $k \cdot 0, k \cdot 1, k \cdot 2, \cdots, k(n-1)$ is also a complete residue system ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,318 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.