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 |
|---|---|---|---|---|---|---|---|---|---|
176*. Prove that in a right-angled triangle with integer sides, the length of at least one of the legs is divisible by 3.
$177^{\circ}$. Prove that if the sum $a^{2}+b^{2}$, where $a$ and $b$ are integers, is divisible by 3, then each of the numbers $a$ and $b$ is divisible by 3. | $\triangle$ From the divisibility $\left(a^{2}+b^{2}\right) \vdots 3$ it follows that the numbers $a$ and $b$ either both divide, or both do not divide by 3.
Consider the second possibility. Then each of the numbers $a^{2}$ and $b^{2}$, when divided by 3, gives a remainder of 1:
$$
a^{2}=3 k+1, \quad b^{2}=3 l+1
$$
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,111 |
## 6.2.
$180^{\circ}$. Prove that for any integer $a$, the difference $a^{5}-a$ is divisible by 5. | $\triangle$ Let's factor the given difference:
$$
a^{5}-a=a\left(a^{4}-1\right)=a\left(a^{2}-1\right)\left(a^{2}+1\right)=(a-1) a(a+1)\left(a^{2}+1\right)
$$
Now we can consider five cases:
$$
a=5 k, \quad a=5 k+1, \quad a=5 k+2, \quad a=5 k+3, \quad a=5 k+4(k \in \mathbb{Z})
$$
but this is quite lengthy. Let's loo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,112 |
185. Prove that the sum
$$
9 a^{5}-5 a^{3}-4 a
$$
is divisible by 15 for any integer $a$. | $\triangle$ It is necessary to prove that the given sum is divisible by 3 and by 5.
1) To prove the divisibility of the sum by 3, it is sufficient to prove that $5 a^{3}+4 a$ is divisible by 3 (since $9 a^{5}$ is divisible by 3). We have:
$$
5 a^{3}+4 a=\left(5 a^{3}-5 a\right)+(5 a+4 a)=5\left(a^{3}-a\right)+9 a
$$
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,113 |
189. An integer $a$ is not divisible by 5. Can the number $3 a^{4}+1$ be divisible by 5? | $\triangle$ From the statement of problem 183, it follows that in this case $a^{4}$ when divided by 5 can only leave a remainder of $1: a^{4}=5 k+1(k-$ an integer non-negative). Then
$$
3 a^{4}+1=3(5 k+1)+1=15 k+4
$$
The last sum is not divisible by 5 for any non-negative integer $k$.
Answer: cannot. | cannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,114 |
193*. Find all integer values of $x$ and $y$ for which at least one of the numbers
$$
x^{2}-2 x y+2 y^{2}, \quad x^{2}+2 x y+2 y^{2}
$$
is divisible by 5. | $$
\begin{gathered}
\left(x^{2}-2 x y+2 y^{2}\right)\left(x^{2}+2 x y+2 y^{2}\right)=\left(\left(x^{2}+2 y^{2}\right)-2 x y\right)\left(\left(x^{2}+2 y^{2}\right)+2 x y\right)= \\
=\left(x^{2}+2 y^{2}\right)^{2}-4 x^{2} y^{2}=x^{4}+4 x^{2} y^{2}+4 y^{4}-4 x^{2} y^{2}=x^{4}+4 y^{4}
\end{gathered}
$$
Since the sum $x^{4... | allintegersxy,simultaneouslydivisibleornotdivisible5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,115 |
195. Prove that the difference $26^{n}-7^{n}$ for any natural $n$ is divisible by 19. | $\triangle$ Based on the first divisibility, this difference is divisible by the difference of the bases of the powers, i.e., by $26-7=19$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,116 |
201. Prove that the sum $13^{49}+5^{49}$ is divisible by 18. | $\triangle$ Based on the second of the known divisibilities, this sum is divisible by the sum of the bases of the powers, i.e., by $13+5=18$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,118 |
207. Is it true that the difference $47^{100}-14^{100}$ is divisible by 61? | $\triangle$ Using the third of our divisibility rules: when the exponent is even, the difference of powers with this (same) exponent is divisible by the sum of the bases, i.e., by $47+14=61$.
Answer: correct. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,119 |
211. Is the difference $17^{15}-3^{15}$ divisible by 4? | $\triangle$ The difference of the bases of the powers $17-3=14$ is divisible by 2, but not by 4. This does not mean that the given difference $17^{15}-3^{15}$ is not divisible by 4. Let's factorize it using the formula (1:
$$
17^{15}-3^{15}=14 \cdot\left(17^{14}+17^{13} \cdot 3+17^{12} \cdot 3^{2}+\ldots+3^{14}\right)... | doesnotdivide | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,120 |
216. Prove that the number $2 \cdot 7^{n}+1$ is divisible by 3 for any natural $n$. | $\triangle$ Let's transform the given sum as follows:
$$
2 \cdot 7^{n}+1=2 \cdot 7^{n}+3-2=2\left(7^{n}-1\right)+3
$$
Since the difference $7^{n}-1$ is divisible by 6, and therefore by 3, the entire sum is divisible by 3.
Another solution: since the difference $7^{n}-1$ is divisible by 6, then
$$
7^{n}-1=6 k, \quad... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,121 |
218. Are there natural numbers $n$ for which the sum
$$
6^{2 n}+3^{n}+3^{n+2}
$$
is not divisible by 11?
219 ${ }^{\circ}$. Prove that if $p(x)$ is a polynomial with integer coefficients, and $a$ and $b$ are different integers, then the difference $p(a)-p(b)$ is divisible by $a-b$. | $\triangle$ Let's write the polynomial $p(x)$ of degree $n$ in canonical form:
$$
p(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}
$$
Here, by condition, all coefficients are integers. Then we get:
$$
\begin{gathered}
p(a)-p(b)=\left(a_{0} a^{n}+a_{1} a^{n-1}+\ldots+a_{n-1} a+a_{n}\right)-\left(a_{0} b^{n}+a_{1... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,122 |
222. Find all natural $n$ for which the sum $2^{n}+1$ is divisible by 3. | $\triangle$ Obviously, all odd $n$ will work, since in this case the sum $2^{n}+1$ is divisible by $2+1$, i.e., by 3. And what if $n$ is even?
Let $n$ be even: $n=2 k(k \in N)$. Transform the given sum:
$$
2^{n}+1=2^{2 k}+1=\left(2^{2 k}-1\right)+(1+1)=\left(4^{k}-1\right)+2
$$
Since $4^{k}-1$ is divisible by 3, and... | foralloddn | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,123 |
225. Prove that the sum
$$
2^{n+2} \cdot 3^{n}+5 n-4
$$
is divisible by 25 for any natural number $n$. | $\triangle$ Transform the sum:
$$
\begin{gathered}
2^{n+2} \cdot 3^{n}+5 n-4=4 \cdot 2^{n} \cdot 3^{n}-4+5 n=4\left(6^{n}-1\right)+5 n= \\
=5\left(4 \cdot\left(6^{n-1}+6^{n-2}+\ldots+6+1\right)+n\right)= \\
=5\left(4 \cdot 6^{n-1}+4 \cdot 6^{n-2}+\ldots+4 \cdot 6+4-4 n+5 n\right)= \\
=5\left(4 \cdot\left(6^{n-1}-1\rig... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,124 |
230. Prove that the sum
$$
11^{n+2}+12^{2 n+1}
$$
is divisible by 133 for any natural number $n$. | $\triangle$ Let's transform this sum:
$$
11^{n+2}+12^{2 n+1}=11^{2} \cdot 11^{n}+12 \cdot 12^{2 n}=121 \cdot 11^{n}+12 \cdot 144^{n}
$$
Now, let's notice that
$$
121+12=133, \quad 144-11=133
$$
But how can we use this?
Let's perform the following transformations:
$$
\begin{aligned}
12 \cdot 144^{n}+121 \cdot 11^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,125 |
233*. Is the sum $19^{19}+69^{69}$ divisible by 44? | $\triangle$ Since $44=4 \cdot 11$, where the numbers 4 and 11 are coprime, it is sufficient to check whether this sum is divisible by 4 and 11.
Divisibility by 4 is simple: use the "plus-minus" trick, adding and subtracting 1. We get:
$$
19^{19}+69^{69}=\left(19^{19}+1\right)+\left(69^{69}-1\right)
$$
The number in ... | divisible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,126 |
239*. Find all natural $n$ for which the difference $2^{n}-1$ is divisible by 7. | $\triangle$ Let's perform an enumeration, assigning $n$ the values $1,2,3,4,5,6$ and 7. It turns out that the difference $2^{n}-1$ is divisible by 7 only when $n=3$ and $n=6$. It seems that this difference is divisible by 7 only when $n$ is divisible by 3. Let's test our hypothesis. For this, we will consider three cas... | n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,127 |
245*. Can the number $5^{n}-1$ be divisible by the number $4^{n}-1$ for the same natural $n$? | $\triangle$ The difference $4^{n}-1$ for any natural $n$ is divisible by 3, hence the difference $5^{n}-1$ is also divisible by 3. However, the latter is true only for even $n$ (check this!): $n=2k (n \in N)$. But for even $n$,
$$
4^{n}-1=4^{2k}-1
$$
is also divisible by 5, while the difference $5^{n}-1$ is never div... | cannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,128 |
251. Why is any six-digit number of the form $\overline{a b a b a b}$ divisible by 13? | $\triangle$ Obviously, this number is divisible by the two-digit number $\overline{a b}$. After division, we get:
$$
\overline{a b a b a b}=\overline{a b} \cdot 10101
$$
and the number 10101 is divisible by 13 (see the numerical equalities written above). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,129 |
254. Given two three-digit numbers, the sum of which is divisible by 37. These numbers are written one after the other. Is it true that the resulting six-digit number is necessarily divisible by 37? | $\triangle$ Let's denote the three-digit numbers by $a$ and $b$. Then their sum $a+b$ is divisible by 37.
The six-digit number takes the form $1000a + b$. We can transform it by isolating the term $a+b$:
$$
1000a + b = 999a + (a + b)
$$
In the resulting sum, not only the second term, but also the first term is divis... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,130 |
256. Prove that if a six-digit number is divisible by 7, or by 11, or by 13, or by 37, then the number obtained by moving the first digit to the end is also divisible by the same number. | $\triangle$ Let the original six-digit number be denoted as $\overline{a b c d e f}$. Then the new number obtained by rearranging the digits is $\overline{b c d e f a}$.
Let $\overline{b c d e f}=x$. Then the initial number is $10^{5} a+x$, and the new number is $10 x+a$. Why does the divisibility of the sum $10^{5} a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,131 |
258. Find all natural $n$ for which the number $10^{n}-1$ is divisible by 101. | $\triangle$ Let's transform this number:
$$
10^{n}-1=999 \ldots 9 \text { ( } n \text { nines). }
$$
For which number of nines is the resulting number divisible by 101? Checking shows that the numbers 9, 99, and 999 are not divisible by 101, but the number 9999 is divisible by 101, since 9999 is divisible by 1111, an... | n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,132 |
261*. Prove that if all three-digit numbers from 111 to 999 are written in any order, the resulting multi-digit number is divisible by 37. | $\triangle$ The total number of such three-digit numbers is $999-110=889$. Let's write the number itself in the form
$$
S=a_{1}+a_{2} \cdot 10^{3}+a_{3} \cdot 10^{6}+\ldots+a_{888} \cdot 10^{2661}+a_{889} \cdot 10^{2664}
$$
where $a_{1}, a_{2}, \ldots, a_{889}$ are the different three-digit numbers specified in the c... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,133 |
263. Prove that if the sum $6a+11b$ (where $a$ and $b$ are integers) is divisible by 31, then the sum $a+7b$ is also divisible by 31. | $\triangle$ Multiply the second of these sums by 6 to equalize the coefficients of $a$ in the sums. From the resulting sum $6 a+42 b$, subtract the original:
$$
(6 a+42 b)-(6 a+11 b)=31 b
$$
It turned out that the difference of these two numbers is divisible by 31. Since, in addition, the subtrahend is divisible by 3... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,134 |
268. What three digits should be appended to the number 456 so that the resulting six-digit number is divisible by 504? Find all solutions. | $\triangle$ Let's denote the unknown digits in sequential order as $x, y$, and $z$. Then the six-digit number takes the form $\overline{456 x y z}$.
Since $504=7 \cdot 8 \cdot 9$, it might seem that we need to use the divisibility rules for 8 and 9. However, since the digits $x, y$, and $z$ are adjacent, it is easier ... | 120or624 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,135 |
274. How many natural numbers divisible by 3 or 7 are there among the natural numbers from 1 to 500? | $\triangle$ First, let's calculate how many natural numbers from 1 to 500 are divisible by 3. Let's list all such numbers:
$$
3, \quad 6=3 \cdot 2, \quad 9=3 \cdot 3, \quad 12=3 \cdot 4, \ldots, \quad 498=3 \cdot 166
$$
From this, we can see that the number of such numbers is 166.
Now let's find how many natural num... | 214 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,136 |
### 8.1.5.
The tasks of this cycle are grouped around the statement of problem 281.
281 $\mathbf{1}^{\circ}$. Prove that for any odd $a$, the difference $a^{2}-1$ is divisible by 8. | $\triangle$ Since
$$
a^{2}-1=(a+1)(a-1)
$$
and the numbers $a+1$ and $a-1$ are even when $a$ is odd, the difference $a^{2}-1$ is divisible by 4. But why is it also divisible by 8?
Let $a=2 k+1$, where $k$ is an integer, and transform this difference:
$$
a^{2}-1=(2 k+1)^{2}-1=4 k^{2}+4 k+1-1=4 k^{2}+4 k=4 k(k+1)
$$
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,137 |
285. Prove that for any integer $n$ the number
$$
n^{5}-5 n^{3}+4 n
$$
is divisible by 120. | $\triangle$ Let's factorize the number 120 into prime factors:
$$
120=2^{3} \cdot 3 \cdot 5
$$
Moreover, let's factorize the expression $n^{5}-5 n^{3}+4 n$:
$$
\begin{gathered}
n^{5}-5 n^{3}+4 n=n\left(n^{4}-5 n^{2}+4\right)=n\left(n^{2}-4\right)\left(n^{2}-1\right)= \\
=(n-2)(n-1) n(n+1)(n+2)
\end{gathered}
$$
It ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 45,138 |
289. Find all integers $a$ for which the fraction
$$
\frac{a^{2}-21 a+17}{a}
$$
is an integer. | $\triangle$ Dividing the numerator of the fraction term by term by the denominator:
$$
\frac{a^{2}-21 a+17}{a}=a-21+\frac{17}{a}
$$
The representation of an algebraic fraction as the sum of a polynomial and a fraction with the same denominator, where the degree of the numerator is less than the degree of the denomina... | \1,\17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,139 |
291. Find all integers $n$ for which the fraction $\frac{n-2}{n+1}$ takes integer values. | $\triangle$ To isolate the integer part of the fraction in the numerator, we add and subtract 1 (the "plus-minus" technique):
$$
\frac{n-2}{n+1}=\frac{(n+1)-1-2}{n+1}=\frac{(n+1)-3}{n+1}=1-\frac{3}{n+1}
$$
The fraction $\frac{3}{n+1}$ must be equal to an integer. Let's consider all cases.
1) Suppose $n+1=1$. Then $n... | 2,0,-2,-4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,140 |
293. Find all integers $k$ for which the value of the fraction
$$
\frac{2 k^{2}+k-8}{k-1}
$$
is an integer. | $\triangle$ To isolate the integer part of the fraction, we will divide its numerator by the denominator using the "corner" method, similar to dividing a multi-digit natural number by another natural number. We get:
![](https://cdn.mathpix.com/cropped/2024_05_21_475fdfbb473422773b9eg-063.jpg?height=139&width=368&top_l... | 6,2,0,-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,141 |
295. Find all natural numbers $n$ for which the number $7 n+5$ is divisible by the number $3 n-1$. | $\triangle$ Here, the straightforward division of the dividend by the divisor using the "corner" method does not work, as the number $7 / 3$ is not an integer. Let's try an alternative approach.
If the fraction $\frac{7 n+5}{3 n-1}$ is a natural number, then the tripled fraction
$$
\frac{3(7 n+5)}{3 n-1}=\frac{21 n+1... | 1,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,142 |
297*. Find all real $x$ for which the fraction
$$
\frac{x^{2}+2 x-3}{x^{2}+1}
$$
takes integer values. | $\triangle$ Note that the number $x$, in general, is not an integer, so the methods used in previous problems are of little help.
Let
$$
\frac{x^{2}+2 x-3}{x^{2}+1}=a
$$
where the number $a$ is an integer. Simplify the expression:
$$
x^{2}+2 x-3=a x^{2}+a, \quad(a-1) x^{2}-2 x+(a+3)=0
$$
Consider two cases.
1) Le... | -3,-1,-\frac{1}{2},0,\frac{1}{3},1,2,\frac{-1\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,143 |
310. Prove that the sum of the cubes of any three consecutive integers is divisible by: a) $3 ;$ b) $9$. | Let's denote the average of three numbers by $n$. We will transform the given sum:
$$
\begin{gathered}
(n-1)^{3}+n^{3}+(n+1)^{3}=n^{3}-3 n^{2}+3 n-1+n^{3}+n^{3}+3 n^{2}+1= \\
=3 n^{3}+6 n=3 n\left(n^{2}+2\right)
\end{gathered}
$$
From this, it follows that this sum is divisible by 3. But why is it also divisible by 9... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,144 |
314*. Is it true that if the sum $a^{2}+b^{2}$ (where $a$ and $b$ are natural numbers) is divisible by 11, then the numbers $a$ and $b$ are divisible by 11? | $\triangle$ Let's look at the remainders that the square of a natural number can give when divided by 11. The results of the calculations can be presented in the following table.
| Remainder of $a$ divided by 11 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,145 |
319. Prove that the sum $n^{2}+3 n+4$ is not divisible by 49 for any integer $n$. | $\triangle$ The given sum is divisible by 7, for example, when $n=2$. But why is it never divisible by 49?
Assume the opposite: there exists an integer $n$ such that the number $n^{2}+3 n+4$ is divisible by 49. Then the following equality holds:
$$
n^{2}+3 n+4=49 k
$$
where $k$ is an integer. We will consider this e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,146 |
323. Find all two-digit numbers that are divisible by the product of their digits. | Let's denote a two-digit number as $\overline{a b}$. Then
$$
\overline{a b}: a b, \quad(10 a+b): a b
$$
Since the sum $10 a+b$ is divisible by $a$ and the first term $10 a$ is divisible by $a$, the second term $b$ is also divisible by $a: b=k a$, where $k$ is an integer non-negative number less than 10.
Similarly, t... | 11,12,24,36,15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,147 |
329. Construct an example of a polynomial $P(x)$ of the third degree with integer coefficients and the leading coefficient equal to 1, which is divisible by 3 for all integer values of $x$. | $\triangle$ For example, we can take
$$
P(x)=x(x+1)(x+2)
$$
Among three consecutive integers $x, x+1$, and $x+2$, one is divisible by 3. | P(x)=x(x+1)(x+2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,148 |
335. Find all pairs of natural numbers such that the sum of each with 1 is divisible by the other number. | Let's denote the required numbers by $a$ and $b$. According to the problem,
$$
(a+1) \div b, \quad (b+1) \div a.
$$
We will prove that the numbers $a$ and $b$ are coprime: $\text{GCD}(a, b) = 1$.
Indeed, if $\text{GCD}(a, b) = d$, then
$$
(a+1) \div b, \quad b \div d \Rightarrow (a+1) \div d
$$
But the first term ... | 1,1;1,2;2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,149 |
338. Find all such triples of pairwise coprime natural numbers such that the sum of any two of them is divisible by the third. | $\triangle$ Let's denote these numbers by $a, b,$ and $c$. Then
$$
(a+b): c, \quad(a+c): b, \quad(b+c): a
$$
We will apply the same method of solution as in problem 335. Since the numbers $a, b,$ and $c$ are pairwise coprime, the system of three divisibilities is equivalent to one:
$$
(a+b+c): a b c
$$
(check this!... | 1,1,1;1,1,2;1,2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,150 |
343. A nine-digit number divisible by 37 was split into two parts and these parts were swapped. Is the resulting nine-digit number always divisible by 37? | Let the original number be
$$
m=\overline{a_{1} a_{2} \ldots a_{9}}
$$
and it is divided into parts
$$
x=\overline{a_{1} a_{2} \ldots a_{k}}, \quad y=\overline{a_{k+1} a_{k+2} \ldots a_{9}} \quad(1 \leqslant k \leqslant 9)
$$
Then
$$
m=x \cdot 10^{9-k}+y
$$
By the condition
$$
m=37 b, \quad x \cdot 10^{9-k}+y=37... | always | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,151 |
344. Prove that if $p$ is a prime number greater than 2, then the numerator of the fraction equal to the sum
divides by $p$.
$$
S=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p-1}
$$ | $$
\begin{gathered}
S=\left[1+\frac{1}{p-1}\right]+\left[\frac{1}{2}+\frac{1}{p-2}\right)+\left[\frac{1}{3}+\frac{1}{p-3}\right)+\ldots+\left(\frac{1}{\frac{1}{2}(p-1)}+\frac{1}{\frac{1}{2}(p+1)}\right)= \\
=\frac{p}{p-1}+\frac{p}{2 \cdot(p-2)}+\frac{p}{3 \cdot(p-3)}+\ldots+\frac{p}{\frac{1}{2}\left(p^{2}-1\right)}
\en... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,152 |
351*. Find all natural $n>1$, for which $(n-1)!$ is divisible by $n$. | $\triangle$ Consider two possibilities.
1) Let the number $n-$ be prime.
But then the number ( $n-1$ )! is coprime with $n$ and, consequently, cannot be divisible by $n$. Therefore, this case is ruled out.
2) Let the number $n-$ be composite
If in this case $n$ can be factored into two distinct factors $a$ and $b$,... | allcomposite,except4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,154 |
352*. Does there exist a two-digit number $\overline{a b}$ with different digits, which is divisible by the number $\overline{b a}$? | $\triangle$ Suppose such a two-digit number exists. Then
$$
\overline{a b}=k \cdot \overline{b a}
$$
where $k-$ is a natural number, $1<k \leqslant 9$.
Let's add the numbers $\overline{a b}$ and $\overline{b a}$. On one hand, their sum is
$$
\overline{a b}+\overline{b a}=k \cdot \overline{b a}+\overline{b a}=(k+1) ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,155 |
359. Natural numbers $a$ and $b$ are such that
$$
31 a=54 b
$$
Prove that the number $a+b$ is composite. | $\triangle$ Since the number $31a$ is divisible by 54 and the numbers 31 and 54 are coprime, then $a$ is divisible by $54: a=54n$, where $n \in N$. Then
$$
31 \cdot 54 \cdot n=54 b, \quad b=31 n
$$
From this
$$
a+b=54 n+31 n=85 n,
$$
therefore, the number $a+b$ is composite. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,156 |
361. A prime number is divided by 21 with a remainder. Find all values of the remainder that are composite numbers. | $\triangle$ Let's denote a prime number by $p$, the quotient by $q$, and the remainder by $r$. Then
$$
p=21 q+r \quad(0<r<21)
$$
Since $p$ is a prime number, from all composite values of $r$ less than 21, we need to select those that are coprime with 3 and 7, otherwise the sum $21 q+r$ would be divisible by 3 or 7. T... | 4,8,10,16,20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,157 |
364. Find all natural numbers $a$ for which the number $a^{2}-10 a+21$ is prime. | $\triangle$ Let's factor this quadratic trinomial into linear factors:
$$
a^{2}-10 a+21=(a-3)(a-7)
$$
From this, it is clear that the given number is, in general, composite. When is it prime? When one of the factors is 1 and the other is a prime number, or when one of them is -1 and the other is $-p$, where $p$ is a ... | 8,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,158 |
367. Find all natural $n$, for which the number $n^{4}+4$ is composite. | $\triangle$ Let's try to factor the expression $n^{4}+4$ into factors with integer coefficients. We are accustomed to the fact that the sum of squares does not factor into factors with integer coefficients. It turns out that sometimes it does. In this case, it is done using the "plus-minus" trick as follows:
$$
n^{4}+... | n\neq1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,159 |
375. Prove that the number $2^{17}+2^{5}-1$ is composite.
$376^{*}$. Prove that any number of the form
$$
a=101010 \ldots 101
$$
( $n$ zeros, $n+1$ ones, where $n>1$ ) is composite. | $\triangle$ Transform the number $a$, considering that it has $2 n+1$ digits, and therefore, the first one is in the $2 n$ place:
$$
\begin{gathered}
a=101010 \ldots 101=10^{2 n}+10^{2 n-2}+10^{2 n-4}+\ldots+10^{2}+1= \\
=\frac{1}{10^{2}-1}\left(10^{2}-1\right)\left(10^{2 n}+10^{2 n-2}+\ldots+10^{2}+1\right)=\frac{1}{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,160 |
381. The numbers $p$ and $p+15$ are prime. Find all such $p$. | $\triangle$ When $p=2$, we get that the number $p+15=17$ is prime.
Let $p>2$. Then $p$ is odd. Consequently, the number $p+15$ is even. Hence, this number is composite. Therefore, no prime $p>2$ works.
Answer: 2.
$382^{\circ}$. Prove that the numbers $p, p+2$, and $p+4$ are all prime only in the case when they form ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,161 |
391. Find three consecutive prime numbers (not necessarily differing by 1 or 2 from each other), the sum of whose squares is also a prime number. List all solutions. | $\triangle$ Let's denote these prime numbers in ascending order as $p, q$, and $r$. We will consider three cases.
1) Suppose $p=2$.
Then the numbers $q$ and $r$ are odd, which means the sum $4+q^{2}+r^{2}$ is even. Therefore, the case $p=2$ is impossible.
2) Suppose $p=3$.
The next prime numbers after 3 are 5 and 7... | 3,5,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,162 |
393. Prove that there exists an infinite set of composite numbers of the form $2^{n}-1$, where $n-$ is an odd natural number. | $\triangle$ Let $n$ be any composite number. Then $n=a b$, where $a$ and $b$ are natural numbers, $a>1, b>1$. It follows that the difference
$$
2^{n}-1=2^{a b}-1=\left(2^{a}\right)^{b}-1
$$
is divisible by $2^{a}-1$, and both the number $2^{a}-1$ and the quotient of this division are greater than 1. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,163 |
398. Prove that for any natural number $a$:
a) there exists a natural number $x$ such that the number $a x+1$ is composite;
b) there are infinitely many such $x$. | $\triangle$ a) Let's choose $x$ so that, for example, the equality
$$
a x+1=(a+1)^{2}
$$
holds. From this,
$$
a x+1=a^{2}+2 a+1, \quad x=a+2
$$
Since the number $(a+1)^{2}$ is composite, the statement is proven.
We can also find $x$ from the equalities:
$$
a x+1=(2 a+1)^{2}, \quad a x+1=(3 a+1)^{2}, \quad a x+1=(... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,164 |
401*. Let $p_{1}$ and $p_{2}$ be two consecutive odd prime numbers and $\frac{p_{1}+p_{2}}{2}=q$. Prove that the number $q$ is composite. | $\triangle$ Let $p_{1}<p_{2}$. Then
$$
p_{1}<\frac{p_{1}+p_{2}}{2}<p_{2}
$$
(check this!), i.e., $p_{1}<q<p_{2}$. Since $p_{1}$ and $p_{2}$ are the closest prime numbers to each other, the number $q$, which is between them, can only be composite. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,165 |
402*. Prove that the set of all prime numbers is infinite.
This is Euclid's theorem (4th century BC) from his famous work "Elements". It plays a crucial role in number theory. | $\triangle$ Suppose the opposite: the set of prime numbers is finite and consists of the numbers $2,3,5,7, \ldots, p$, where, consequently, $p$ is the largest prime number.
Consider the number
$$
a=2 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot p+1
$$
Let $q$ be any prime divisor of the number $a$.
Then $q \neq 2$, b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,166 |
404*. Prove that for any natural number $n$ there exist $n$ consecutive composite numbers.
This is also one of the theorems of number theory. If you think about it, we are faced with an amazing statement: after all, $n$ can be very large, for example, equal to $10^{20}$, and yet there will be $10^{20}$ composite numbe... | $\triangle$ Let's construct the following finite sequence:
$$
(n+1)!+2, \quad(n+1)!+3, \quad(n+1)!+4, \ldots, \quad(n+1)!+(n+1)
$$
Obviously, it has $n$ terms. The first term is divisible by 2, the second by 3, the third by 4, and so on, the last by $n+1$. Therefore, all terms of this sequence are composite numbers.
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,167 |
405. Find all natural numbers, when divided by 6, the quotient is the same as the remainder. | $\triangle$ Let the desired number be denoted by $a$, and the quotient and remainder simultaneously by $q$. Then
$$
a=6 q+q=7 q
$$
It might seem that the answer includes all natural numbers divisible by 7. However, this is not the case, since the remainder $q$ must satisfy the inequality $0<q<6$. By setting $q=1,2,3,... | 7,14,21,28,35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,168 |
408. From the set of numbers
$$
\{2 ; 3 ; 9 ; 27 ; 81 ; 243 ; 567\}
$$
numbers $a, b, q$ and $r$ are chosen such that equation (1) is satisfied. Find the dividend, divisor, quotient, and remainder. | $\triangle$ Let's start with the dividend. It is not equal to 2, 3, or 9, since it is the largest of the four numbers. It is also not equal to 27, 81, or 243, as dividing each of these numbers by all preceding numbers results in a remainder of 0 or 1 (check this!), and the numbers 0 and 1 are not among the elements of ... | 567,243,2,81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,169 |
420. When dividing a two-digit number by 6, the remainder is equal to the digit in the tens place, and when dividing the same number by 10, the quotient is 3 and the remainder is the digit in the units place of the dividend. Find all such two-digit numbers. | $\triangle$ Let the desired number be denoted by $\overline{x y}$. Then
$$
\left\{\begin{array}{l}
\overline{x y}=6 q+x \\
\overline{x y}=30+y
\end{array}\right.
$$
From the second equation, we get:
$$
10 x+y=30+y, \quad x=3
$$
Transform the first equation:
$$
30+y=6 q+3, \quad 27+y=6 q
$$
From this, the digit $y... | 33,39 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,171 |
422. The quotient of dividing a three-digit number by the sum of its digits is 13, and the remainder is 15. Find all such three-digit numbers. | Let's denote the desired number as $\overline{x y z}$. We obtain:
$$
100 x + 10 y + z = 13(x + y + z) + 15, \quad 87 x = 3 y + 12 x + 15, \quad 29 x = y + 4 z + 5
$$
Since
$$
y + 4 z + 5 \leqslant 9 + 4 \cdot 9 + 5 = 50
$$
then $29 x \leqslant 50$, from which $x = 1$.
Now let's solve the equation
$$
29 = y + 4 z ... | 106,145,184 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,172 |
425*. A two-digit number was divided by a number written with the same digits but in reverse order, and the quotient and remainder were equal. Find all such two-digit numbers. | $\triangle$ Let the desired number be denoted by $\overline{x y}$. Then
$$
10 x+y=(10 y+x) r+r, \quad 10 x+y=(10 y+x+1) r
$$
where $r-$ is the remainder (and the quotient at the same time). Obviously, $1 \leqslant r < 10$. But this is impossible.
Answer: 52. | 52 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,173 |
427. If the numbers 826 and 4373 are divided by the same natural number, the remainders will be 7 and 8, respectively. Find all values of the divisor. | $\triangle$ Let's write the corresponding equalities:
$$
826=b q_{1}+7, \quad 4373=b q_{2}+8
$$
where $b-$ is the unknown divisor, $q_{1}$ and $q_{2}$ are the incomplete quotients. Then
$$
b q_{1}=819, \quad b q_{2}=4365
$$
Let's factorize the numbers 819 and 4365 into prime factors:
$$
819=3^{2} \cdot 7 \cdot 13,... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,174 |
431. When dividing a natural number $a$ by 2, the remainder is 1, and when dividing by 3, the remainder is 2. What remainder will be obtained when dividing $a$ by 6? | $\triangle$ Let
$$
a=6 q+r
$$
where the remainder $r$ satisfies the inequality $0 \leqslant r \leqslant 5$. We will consider all possible values of $r$.
The case $r=0$ is impossible, otherwise the number $a$ would be divisible by 2 and 3.
The case $r=1$ is also impossible, because then $a$ when divided by 3 would l... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,175 |
438*. What remainder does $46^{925}$ give when divided by $21?$ | $\triangle$ The solution method we applied in the last problems is too cumbersome here due to the large base of the exponent and the large divisor. Let's try to modify it.
Divide 46 by 21 with a remainder and transform the exponent:
$$
46^{925}=(21 \cdot 2+4)(21 \cdot 2+4) \ldots(21 \cdot 2+4)
$$
If we multiply 925 ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,176 |
445. Find all pairs of natural numbers whose sum is 288, and the greatest common divisor is 36. | $\triangle$ Let the required numbers be $a$ and $b$. Since their greatest common divisor is 36, then
$$
a=36 k, \quad b=36 n
$$
where $k$ and $n$ are coprime natural numbers. Then
$$
a+b=288, \quad 36 k+36 n=288, \quad k+n=8
$$
What are $k$ and $n$ from the last equation? They are either 7 and 1, or 5 and 3. From h... | 252,36;180,108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,177 |
447. Find the greatest common divisor of the numbers 111111 and 111111111. | $\triangle$ Obviously, each of the given numbers is divisible by 111, i.e., the number 111 is their common divisor. But will 111 be their greatest common divisor?
To answer this question, we will divide the given numbers by 111 and determine whether the resulting quotients are coprime numbers. We will have:
$$
111111... | 111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,178 |
449. Find the greatest common divisor of all nine-digit numbers, in the representation of which each of the digits $1,2,3, \ldots, 9$ appears exactly once. | $\triangle$ Let's denote this greatest common divisor by $d$.
From all nine-digit numbers of the specified form, we will take only two - 123456798 and 123456789.
Since these numbers are divisible by $d$, their difference, which is 9, is also divisible by $d$: $9: d$. Therefore, $d=1, d=3$ or $d=9$.
Which of these ca... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,179 |
452. Natural numbers $a$ and $b$, where $a>b$, are coprime. Find all values of the greatest common divisor of the numbers $a+b$ and $a-b$. | $\triangle$ Let
$$
\text { GCD }(a+b, a-b)=d
$$
Then
$$
(a+b): d, \quad(a-b): d
$$
Therefore, the sum and difference of the numbers $a+b$ and $a-b$ are divisible by $d$. The sum of these numbers is $2 a$, and the difference is $-2 b$. We get:
$$
2 a: d, \quad 2 b: d .
$$
But the numbers $a$ and $b$ are coprime by... | 1or2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,180 |
453. Let $a$ and $b$ be natural numbers, $a > b$, and the numbers $a+b$ and $a-b$ are coprime. Find all possible values of the greatest common divisor of the numbers $a$ and $b$.
454. Prove that any two consecutive natural numbers are coprime. | Let's denote two consecutive natural numbers by $n$ and $n+1$, and their greatest common divisor by $d$. Then
\[
\begin{gathered}
(n+1): d, \\
n: d .
\end{gathered}
\]
Subtracting these numbers, we get: $1: d$. Therefore, $d=1$.
455. Prove that the greatest common divisor of any two consecutive even natural numbers ... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,181 | |
459. Find all values of the greatest common divisor of the numbers $8 a+3$ and $5 a+2$, where $a$ is a natural number. | $\triangle$ Let's denote the greatest common divisor of these numbers by $d$. Then
$$
(8 a+3): d, \quad(5 a+2): d
$$
Multiply the sum $8 a+3$ by 5, and the sum $5 a+2$ by 8. We get:
$$
(40 a+15): d, \quad(40 a+16): d
$$
But two consecutive natural numbers $40 a+15$ and $40 a+16$ are coprime (see the statement of pr... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,182 |
461. For which number and for which natural values of $a$ is the fraction
$$
\frac{2 a+5}{3 a+4}
$$
reducible?
Find all solutions. | $\triangle$ Let
$$
\text { GCD }(2 a+5,3 a+4)=d \text {. }
$$
Therefore,
$$
(2 a+5): d, \quad(3 a+4): d
$$
Now multiply the number $2 a+5$ by 3, and the number $3 a+4$ by 2. Then
$$
(6 a+15): d, \quad(6 a+8): d
$$
Subtract the obtained numbers $6 a+15$ and $6 a+8$:
$$
7: d \text {. }
$$
Hence, $d=1$ or $d=7$. T... | =7k+1\quad(k=0,1,2,\ldots) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,183 |
466. For what number and for which natural $n$ is the fraction $\frac{n^{2}+1}{n+1}$ reducible? Find all solutions. | $\triangle$ Let
$$
\text { GCD }\left(n^{2}+1, n+1\right)=d \text {. }
$$
Since $n+1$ is divisible by $d$, then the product
$$
n(n+1)=n^{2}+n
$$
is also divisible by $d$. Now we get:
$$
\begin{aligned}
& \left(n^{2}+n\right): d \\
& -\left(n^{2}+1\right): d \\
& (n-1): d
\end{aligned}
$$
Therefore, the numbers $n... | 2foralloddn | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,184 |
471. Prove that for any natural $n$ the numbers $n^{5}+4 n^{3}+3 n$ and $n^{4}+3 n^{2}+1$ are coprime. | $\triangle$ Let's introduce the notation:
$$
\text { GCD }\left(n^{5}+4 n^{3}+3 n, n^{4}+3 n^{2}+1\right)=d
$$
Suppose that $d>1$.
We divide the number $n^{5}+4 n^{3}+3 n$ by the number $n^{4}+3 n^{2}+1$ with a remainder using the polynomial division rule "by corner" (see § 8, point 8.1.6, solution of problem 293):
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,185 |
472. For which number and for which natural $a$ are the fractions reducible:
a) $\frac{a^{4}+a^{3}-a^{2}+a+1}{a^{2}-1}$;
b) $\frac{a^{4}+6 a^{3}+15 a^{2}+18 a+8}{a^{4}+6 a^{3}+13 a^{2}+12 a+3}$?
473 $3^{\circ}$. Let $a$ and $b$ be natural numbers, where $a > b$, and the number $a$ is divided by the number $b$ with a r... | $\triangle$ Let's introduce the notation:
$$
\text { GCD }(a, b)=d_{1}, \quad \text { GCD }(b, r)=d_{2} .
$$
Since the numbers $a$ and $b$ are divisible by $d_{1}$, from the initial equation we get that the remainder $r$ is also divisible by $d_{1}$. Therefore, $d_{1}$ is a common divisor of the numbers $b$ and $r$. ... | 168 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,186 |
477*. What is the greatest common divisor of all numbers $7^{n+2}+8^{2 n+1}(n \in N)$? | $\triangle$ Let's denote the greatest common divisor of all such numbers by $d$. Replacing $n$ with $n+1$ in the given sum, we get the sum $7^{n+3} + 8^{2n+3}$, which must also be divisible by $d$. Additionally, let's multiply the sum $7^{n+2} + 8^{2n+1}$ by 7:
$$
\left(7^{n+2} + 8^{2n+1}\right) \cdot 7 = 7^{n+3} + 7 ... | 57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,187 |
479*. For what least natural $n$ are each of the fractions
irreducible
$$
\frac{5}{n+6}, \quad \frac{6}{n+7}, \frac{7}{n+8}, \ldots, \quad \frac{24}{n+25}
$$ | $\triangle$ Transform the given fractions as follows:
$$
\frac{5}{n+6}=\frac{5}{(n+1)+5}, \quad \frac{6}{n+7}=\frac{6}{(n+1)+6}, \ldots, \frac{24}{n+25}=\frac{24}{(n+1)+24}
$$
From this, it is clear that each of these fractions is irreducible if and only if the number $n+1$ is coprime with each of the numbers $5,6,7,... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,188 |
485. A father and son were walking one behind the other on a snow-covered road. The father's step length is $-80 \mathrm{~cm}$, the son's $-60 \mathrm{~cm}$. Their steps coincided 601 times, including at the beginning and the end of the journey. What distance did they cover? | $\triangle$ Let's find the distance that the father and son walked from one coincidence of steps to the next. It is equal to the least common multiple of the numbers 80 and 60, i.e., $240 \mathrm{~cm}=2.4$ m. Therefore, the entire distance they walked is
$$
600 \cdot 2.4 \mathrm{M}=1440 \mathrm{~m}=1 \mathrm{Km} 440 \... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,189 |
487. A customer wanted to buy all the eggs the seller had and asked how many eggs she had. She replied that she didn't remember, but she knew that if the eggs were divided into groups of $2, 3, 4, 5$ or 6, there would always be one egg left over. What is the smallest number of eggs the seller could have had? | $\triangle$ Let's temporarily set aside one egg. The number of remaining eggs is equal to the least common multiple of the numbers $2,3,4,5$ and 6, i.e., 60. We need to add one more egg to this number.
Answer: 61. | 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,190 |
492. Find the least common multiple of the natural numbers $n$ and $n+3$. | $\triangle$ The answer depends on the greatest common divisor of the numbers $n$ and $n+3$. It is 1 if $n$ is not divisible by 3, and 3 if $n$ is divisible by 3 (see problem 457).
Answer: $n(n+3)$, if $n$ is not divisible by 3, and $\frac{1}{3} n(n+3)$, if $n$ is divisible by 3. | n(n+3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,191 |
495. Find all pairs of natural numbers if their sum is 60, and the least common multiple is 72. | $\triangle$ Let the required numbers be $a$ and $b$.
The number 72 is divisible by each of them. Since
$$
72=8 \cdot 9=2^{3} \cdot 3^{2}
$$
the prime factorization of numbers $a$ and $b$ contains only twos and threes. Moreover, in the factorization of one of the numbers $a$ and $b$, the number 2 enters in the third ... | 2436 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,192 |
499. Find all pairs of natural numbers whose greatest common divisor is 24, and the least common multiple is 360. | $\triangle$ Let the required numbers be $a$ and $b$.
We factorize the numbers 24 and 360 into prime factors:
$$
24=2^{3} \cdot 3, \quad 360=2^{3} \cdot 3^{2} \cdot 5
$$
Therefore, the prime factorizations of the numbers $a$ and $b$ contain only twos, threes, and fives, and $2^{3}=8$ is a factor of each of the number... | 24360;72120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,193 |
502. Find the smallest fraction, which when divided by each of the fractions $\frac{21}{25}$ and $\frac{14}{15}$, results in a natural number.
The conditions of this problem are unique: here two fractions are mentioned, one of which divides the other exactly. | $\triangle$ Let the required fraction be $\frac{a}{b}$, where $a$ and $b$ are natural coprime numbers. Divide it by each of the given fractions.
$$
\frac{a}{b}: \frac{21}{25}=\frac{25 a}{21 b}, \quad \frac{a}{b}: \frac{14}{15}=\frac{15 a}{14 b}
$$
Since the obtained fractions are equal to natural numbers, $25 a$ is d... | \frac{42}{5} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,194 |
505*. Prove that for any two natural numbers $a$ and $b$, the product of their greatest common divisor and least common multiple is equal to the product of the numbers themselves:
$$
\text { GCD }(a, b) \cdot \operatorname{LCM}(a, b)=a b
$$ | $\triangle$ Let's factorize the numbers $a$ and $b$ into prime factors:
$$
a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{n}^{\alpha_{n}}, \quad b=p_{1}^{\beta_{1}} \cdot p_{2}^{\beta_{2}} \cdot \ldots \cdot p_{n}^{\beta_{n}}
$$
where all $p_{k}(k=1,2, \ldots, n)$ are distinct prime numbers, and ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,195 |
515. Find all natural numbers that decrease by 14 times when the last digit is erased. | $\triangle$ Let the desired number be $10 x + y$, where $x$ is the number of tens in the number, and $y$ is its last digit. Then
$$
10 x + y = 14 x, \quad y = 4 x
$$
Since $y$ is a digit, $x$ is also a digit, and does not exceed 2. Setting $x=1,2$, we find that $y=4,8$ respectively.
Answer: $14,28$. | 14,28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,196 |
518*. Does there exist a natural number that, by erasing its first digit, decreases: a) by $57 ;$ b) by 58 times | $\triangle$ a) Let the desired number be denoted as $10^{k} \cdot a + x$, where $k$ is a natural number, $a$ is the digit to be crossed out, and $x$ is the number formed by the last $k$ digits. We obtain:
$$
10^{k} \cdot a + x = 57 x, \quad 10^{k} \cdot a = 56 x
$$
Since the right-hand side of the last equation is di... | )7125;b)doesnotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,197 |
526. A six-digit number ends with the digit 7. If this digit is moved to the beginning of the number, the number increases by 5 times. What is this number? | The problem can be solved in two ways.
1) We will solve it as a problem of restoring the record.
where the five-digit numbers denoted by asterisks in the first multiplier and in the produc... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,198 |
532. A six-digit number starts with the digit 2. If this digit is moved to the last place, the number increases by 3 times. What is this number? | $\triangle$ Let's solve the problem in two ways.
1) Restore the record:
where the five-digit numbers denoted by asterisks in the first multiplier and in the product coincide. We sequentia... | 285714 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,199 |
538. Does there exist a natural number that, when its first digit is moved to the end, increases by 5, 6, or 8 times? | $\triangle$ The first digit of the original number can only be 1, because otherwise, when multiplying it by 5, 6, or 8, the resulting number would have one more digit than the original.
Let's assume that when the first digit of the number is moved to the end, the number increases by 5 times. We get the following recor... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,201 |
540. Prove that there does not exist a natural number which, when its first digit is moved to the end, increases by 4 times. | $\triangle$ Obviously, the rearranged digit $a$ does not exceed 2 and is even, so $a=2$.
Let the original number be denoted as $2 \cdot 10^{n} + x$, where $n$ is a natural number, and $x$ is the number formed by the digits following the digit 2. The number $2 \cdot 10^{n} + x$ is $(n+1)$-digit; then the number $x$ is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,202 |
543. Does there exist a six-digit number that increases by 6 times when the first three digits, without changing their order, are moved to the end of the number? | $\triangle$ Let the desired number be denoted as $\overline{a b c d e f}$. Then
$$
\overline{a b c d e f} \cdot 6=\overline{d e f a b c}
$$
Let
$$
\overline{a b c}=x, \quad \overline{d e f}=y
$$
The equation becomes
$$
(1000 x+y) \cdot 6=1000 y+x
$$
Simplify it:
$$
6000 x+6 y=1000 y+x, \quad 5999 x=994 y, \quad ... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,203 |
548. A three-digit number has its digits arranged in ascending order, and the second digit is as many times greater than the first as the third digit is greater than the second. The sum of this number and its reverse is a prime number. Find all such numbers.
549. ${ }^{\circ}$. Prove that the difference between a natu... | $\triangle$ A natural number when divided by 9 gives the same remainder as the sum of its digits (see the statement of the reference problem 120 from § 5). But a natural number and its reverse have the same sum of digits, so the difference between such numbers is divisible by 9.
550 ${ }^{\circ}$. Let $x$ be an $n$-di... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,204 |
552. Find all four-digit numbers that are 9 times greater than their reversals. | $\triangle$ Let the desired number be denoted as $\overline{a b c d}$. Then
$$
\overline{a b c d}=9 \cdot \overline{d c b a}
$$
Two digits, $d$ and $a$, can be found immediately. Since the product $9 \cdot \overline{d c b a}$ is a four-digit number, the digit $d$ can only be 1. Since the product $9a$ ends with the di... | 9801 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,205 |
560. How many zeros are at the end of the product of all natural numbers from 1 to 60? | $\triangle$ Since zeros in the product of natural numbers are formed by multiplying twos by fives, it is desirable to know how many twos and fives are in the prime factorization of the given product. However, since there are more twos than fives in this factorization, it is sufficient to count the number of fives.
Fir... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,206 |
562. For what least natural $n$ does the product of all natural numbers from 1 to $n$ end with exactly 499 zeros?
$8^{*}$
115
563. a) Prove that if a natural number ends in the digit 5, then its square ends in 25.
b) Prove that to square such a number, it is sufficient to multiply the number of its tens by the numb... | $\triangle$ a) Let's take a natural number of the form $10 n+5$, where $n-$ is the number of its tens. We will square it:
$$
(10 n+5)^{2}=100 n^{2}+100 n+25=100 n(n+1)+25
$$
Since the product $n(n+1)$ is the number of hundreds in the square, the last two digits of the square form the number 25.
b) Now let's notice t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,207 |
565. Which natural numbers squared give the numbers:
a) 9025;
b) 13225;
c) 24025?
566º. Prove that if the last digit of the square of a natural number is odd, then its second-to-last digit is even. | $\triangle$ Let's take a natural number of the form $10 a+b$, where $a$ is the number of its tens, and $b$ is the last digit. We will square this number:
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2}
$$
The number of tens in this square is composed of the number of tens in the number $20 a b$, which is an even number $2 a b... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,208 |
568. Can the square of a natural number end in:
a) 31;
b) 61;
c) 39;
d) 69;
e) 65
f) 26?
569 ${ }^{\circ}$. Prove that the fifth power of a natural number ends in the same digit as the number itself. | $\triangle$ Let the original natural number be denoted by $a$. The numbers $a$ and $a^{5}$ end with the same digit if and only if the difference $a^{5}-a$ is divisible by 10.
$$
\left(a^{5}-a\right) \vdots 10
$$
116
It is obvious that this difference is divisible by 2: the numbers $a$ and $a^{5}$ have the same parit... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,209 |
573. Do there exist natural numbers $a, b$, and $c$ such that the decimal representation of the sum
$$
a^{60}+b^{60}+c^{60}
$$
ends in the digit 9?
574. Prove that the square of a two-digit number $\overline{a b}(b \neq 0)$ ends in $\overline{a b}$ only when $\overline{a b}$ is 25 or 76. | $\triangle$ Numbers $x=\overline{a b}$ and $x^{2}=\overline{a b^{2}}$ end with the same two digits if and only if their difference is divisible by 100:
$$
\left(x^{2}-x\right): 100, \quad x(x-1) \vdots 100
$$
Two consecutive natural numbers $x-1$ and $x$ are coprime (see the supporting problem 454 from § 11), so thei... | 25or76 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,210 |
582. The number $n^{2}+2 n(n \in N)$ ends with the digit 4. Find all possible values of its second-to-last digit. | $\triangle$ Add 1 to both sides of the equality
$$
n^{2}+2 n=10 a+4
$$
where $a-$ is the number of tens in the number $n^{2}+2 n$, to get:
$$
n^{2}+2 n+1=10 a+5, \quad(n+1)^{2}=10 a+5
$$
From this, it is clear that the number $n+1$ ends in 5, and therefore, its square ends in 25 (see the statement of problem 563). ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,211 |
585. Find the last non-zero digit of the product of all natural numbers from 1 to 40. | $\triangle$ First, let's find the last non-zero digit of the product $1 \cdot 2 \cdot 3 \cdot \ldots \cdot 10$. For this, we will remove the factors 10, 2, and 5 from the number. The remaining product is
$$
3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9
$$
which, as is not difficult to calculate, ends with the digit 8.
F... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,212 |
587. Find the last two digits of the power: a) $7^{38} ; \quad$ b) $7^{7^{7}}$. | $\triangle$ Let's solve a more general problem: what values can the last two digits of the power $7^{n}$ take for natural $n$? By assigning $n$ the values $1,2,3,4,5,6, \ldots$ in sequence, we will each time find the last two digits of this power - until repetitions occur. (Think about why repetitions will eventually o... | )49;b)43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,213 |
590. The last two digits of the decimal representation of the square of some natural number are equal and different from zero. What are these digits? Find all solutions. | $\triangle$ The square of a natural number can end in the digits $0,1,4,5,6$ or 9. The digit 0 is not suitable in this case. Then the square can end in
$$
11,44,55,66,99
$$
However, cases where the square ends in 11, 55, or 99 are impossible, as if the last digit of the square of a natural number is odd, then the sec... | 44 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,214 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.