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
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
593*. Find some power of the number 2 with a natural exponent that ends in three identical digits. | Let's denote the number of thousands in the number $2^{k} (k \in \mathbb{N})$ by $x$, and the digit that appears three times at the end of this number by $y$. Then,
$$
2^{k}=1000 x+\overline{y y y}=1000 x+111 y
$$
From this, it follows that $2^{k}>1000$, and thus $k \geqslant 10$. We obtain that in this equation, $10... | 2^{39} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,215 |
595*. Does there exist a three-digit number whose cube ends in three sevens? | Let's denote this three-digit number as $x$. Then
$$
x^{3}=1000 a+777
$$
where $a$ is the number of thousands in $x^{3}$.
Since $x^{3}$ ends with the digit 7, the number $x$ ends with the digit 3: $x=10 k+3$, and the number $k$ is two-digit. We obtain:
$$
\begin{gathered}
(10 k+3)^{3}=1000 a+777, \quad 1000 k^{3}+9... | 753 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,216 |
607. The sum of a two-digit number and its reverse is a perfect square. Find all such numbers.
(Definition of the number reversed to the given one, see § 12, p. 12.4.) | $\triangle$ Let's add a two-digit number $\overline{a b}$ to its reverse:
$$
\overline{a b}+\overline{b a}=(10 a+b)+(10 b+a)=11(a+b)
$$
Since the number $11(a+b)$ is a perfect square, the sum $a+b$ is divisible by 11. However, since $a$ and $b$ are digits, it equals 11: $a+b=11$.
It is not difficult to go through al... | 29,38,47,56,65,74,83,92 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,217 |
610. A natural number is written using 10 sixes and several zeros. Can it be an exact power?
Can it be an exact power? | $\triangle$ The sum of the digits of this number is 60. Based on the divisibility rules for 3 and 9, the number is divisible by 3, but not by $3^{2}=9$. Therefore, it cannot be a perfect square, a perfect cube, or any perfect power with a natural exponent greater than 1.
Answer: cannot. | cannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,218 |
611. Can a natural number be a perfect square if the sum of its digits is: a) $20 ;$ b) $10 ?$ | $\triangle$ a) Let's divide the number 20 by 3 with a remainder: $20=3 \cdot 6+2$. The remainder is 2. Then the number itself, when divided by 3, also gives a remainder of 2 (see the statement of support problem 120 from § 5). However, the square of a natural number, when divided by 3, cannot give a remainder of 2 (see... | )cannot;b)can | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,219 |
613. A natural number ends with 316. Can it be a perfect cube?
614. Prove that a perfect square, when divided by 4, can only give a remainder of 0 or 1 (and, therefore, cannot give a remainder of 2 or 3). | $\triangle$ Indeed, if a natural number $a$ is even: $a=2 k(k \in N)$, then $a^{2}=4 k^{2}$, and if $a$ is odd: $a=2 k-1$, then
$$
a^{2}=(2 k-1)^{2}=4 k^{2}-4 k+1=4 k(k-1)+1
$$ | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,220 |
617. Prove that the product of four consecutive natural numbers, increased by 1, is a perfect square. | $\triangle$ Let the smallest of the four numbers be $n$. We get the expression
$$
n(n+1)(n+2)(n+3)+1
$$
Multiplying the factors in pairs: the first by the fourth, and the second by the third, we have:
$$
n(n+1)(n+2)(n+3)+1=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1
$$
If we now denote $n^{2}+3 n$ by $a$, we g... | (+1)^2 | Algebra | proof | Yes | Yes | olympiads | false | 45,221 |
620. Find the smallest natural number that, when multiplied by 2, becomes a perfect square, and when multiplied by 3, becomes a perfect cube. | $\triangle$ The desired number $x$ is divisible by 2 or 3. Since it is the smallest of all natural numbers with the given properties, its prime factorization consists only of twos and threes:
$$
x=2^{k} \cdot 3^{l} \quad(k \in N, l \in N)
$$
After multiplying the number $x$ by 2, the result is a perfect square, and a... | 72 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,222 |
625. A typesetter spilled part of the set - a five-digit number that is a perfect square, written with the digits $1,2,5,5$ and 6. Find all such five-digit numbers. | $\triangle$ Among the given digits, a perfect square can end in 1, 5, or 6. Let's consider all three cases.
1) Suppose the square ends in 1.
Since the second-to-last digit of a perfect square in this case must be even (see the statement of problem 566 from § 13), it is either 2 or 6. Let's form all possible five-digi... | 15625=125^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,223 |
629. Prove that the number
$$
1998^{2}+1998^{2} \cdot 1999^{2}+1999^{2}
$$
is a perfect square. | $\triangle$ Let's replace 1998 with $a$ and transform the resulting expression:
$$
\begin{gathered}
a^{2}+a^{2}(a+1)^{2}+(a+1)^{2}=a^{2}+a^{4}+2 a^{3}+a^{2}+a^{2}+2 a+1= \\
=a^{4}+2 a^{3}+3 a^{2}+2 a+1=\left(a^{2}+a+1\right)^{2}
\end{gathered}
$$
Therefore, the given number is the square of the number $1998^{2}+1998+... | 3994003 | Algebra | proof | Yes | Yes | olympiads | false | 45,224 |
633. Can the product $n(n+1)$ for any natural $n$ be a perfect power? | $\triangle$ Suppose that
$$
n(n+1)=a^{k}
$$
where $k$ is a natural number greater than $1$, and $a$ is an integer. We can limit ourselves to the case where $a$ is a natural number (think about why).
Two consecutive natural numbers $n$ and $n+1$ are coprime (see the statement of problem 454 from § 11), so the given e... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,225 |
635. What exact square is equal to the product of four consecutive odd numbers? | Let's denote the smallest of the odd numbers by $n$. Then
$$
n(n+2)(n+4)(n+6)=a^{2}
$$
where $a$ is an integer, and also odd. We transform the left side of this equation by multiplying the first factor by the fourth, and the second by the third:
$$
\left(n^{2}+6 n\right)\left(n^{2}+6 n+8\right)=a^{2}, \quad\left(n^{... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,226 |
636*. Find all five-digit numbers that are perfect squares and remain perfect squares when the first, the first two, or the first three digits are erased. | $\triangle$ Let the desired number be denoted as $\overline{a b c d e}$. Then each of the numbers
$$
\overline{d e}, \overline{c d e}, \overline{b c d e}, \overline{a b c d e}
$$
is a perfect square.
The numbers $\overline{d e}$ and $\overline{c d e}$ can only be perfect squares simultaneously if $\overline{d e}=25$... | 81225,34225,27225,15625,75625 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,227 |
638*. Find all exact powers of the number 2 that occur among numbers of the form $6 n+8(n=0,1,2, \ldots)$. | $\triangle$ Let's write the equality
$$
2^{k}=6 n+8
$$
For which natural $k$ can the power $2^{k}$ be represented as $6 n+8?$
Calculations show that the values $k=3,5,7$ work, while the values $k=2,4,6,8$ do not. This suggests that, perhaps, all odd values of $k$, starting from 3, satisfy the condition of the proble... | powers\of\2\with\any\odd\natural\exponent,\greater\than\1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,228 |
643. A triangle has integer side lengths $a, b$, and $c$, and the length of one of its altitudes is equal to the sum of the lengths of the other two altitudes. Prove that the number $a^{2}+b^{2}+c^{2}$ is a perfect square. | $\triangle$ Let's denote the lengths of the heights of the triangle by $h_{1}, h_{2}$, and $h_{3}$, where $h_{1}$ is the length of the greatest height. Using the formula for the area of a triangle, we get:
$$
\frac{1}{2} a h_{1}=\frac{1}{2} b h_{2}=\frac{1}{2} c h_{3}, \quad a h_{1}=b h_{2}=c h_{3}
$$
From this, we c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,229 |
645. Prove that the number $1000!+6$ is not a perfect power. | $\triangle$ The sum $1000!+6$ is divisible by 2, but not by $2^{2}=4$: the first term of the sum is divisible by 4, while the second is not. Therefore, the sum cannot be a perfect power with any natural exponent greater than 1. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,230 |
648. Find all natural $n$, for which the number $2^{n}+65$ is a perfect square. | $\triangle$ Let's consider two cases.
1) Suppose $n$ is even: $n=2 k(k \in N)$. Then
$$
2^{2 k}+65=a^{2},
$$
where $a$ is a natural number. We get:
$$
a^{2}-2^{2 k}=65, \quad\left(a+2^{k}\right)\left(a-2^{k}\right)=65
$$
When is the product of two natural numbers equal to 65? There are two possibilities:
$$
\left... | 10,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,231 |
650. Find all natural $k$ for which the number
$$
2^{k}+8 k+5
$$
is a perfect square. | $\triangle$ Checking shows that the value $k=1$ does not work, the value $k=2$ works, and the values $k=3,4,5,6$ do not work. It seems that no $k>2$ satisfies the condition of the problem. Let's prove our assumption.
Assume that there exists such a $k>2$ that the number $2^{k}+8 k+5$ is a perfect square. Since this nu... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,232 |
652*. Prove that among the numbers
$$
2^{n}+4^{k} \quad(n \in N, k \in N)
$$
there is an infinite set of perfect squares. | $\triangle$ Transform the given sum:
$$
2^{n}+4^{k}=2^{n}+2^{2 k} .
$$
Let $n>2 k$. We get:
$$
2^{n}+2^{2 k}=a^{2}, \quad 2^{2 k}\left(2^{n-2 k}+1\right)=a^{2},
$$
where $a$ is a natural number.
Since the right side of the last equation is a perfect square and the factor $2^{2 k}$ from the left side is a perfect s... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,233 |
656. Prove that all members of the infinite sequence
$$
\text { 16; 1156; 111556; 11115556; ... }
$$
are perfect squares. | $\triangle$ Let's check this statement for the first four terms of the sequence:
$$
16=4^{2}, \quad 1156=34^{2}, \quad 111556=334^{2}, \quad 11115556=3334^{2}
$$
The written equalities suggest that for any natural $n$, the equality holds:
$$
\overbrace{111 \ldots 1}^{n+1} \overbrace{55 \ldots .56}^{n}=\overbrace{333... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,234 |
660*. All digits of a certain four-digit number, which is a perfect square, can be decreased by the same natural number, not greater than the smallest of its digits, so that a four-digit number, also a perfect square, is obtained. Find all such numbers. | $\triangle$ Obviously, the bases of both squares are two-digit numbers.
Let the base of the second, smaller, square be denoted by $x$, and the first by $y$. Let $k$ be the natural number by which each digit of the first square is reduced. We get:
$$
\begin{aligned}
x^{2}=\overline{a b c d} & =1000 a+100 b+10 c+d \\
y... | 3136,4489 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,235 |
663. Find all natural $n$ for which the number $n^{2}+3 n$ is a perfect square. | $\triangle$ For a natural $n$, the inequality holds:
$$
(n+1)^{2} \leqslant n^{2}+3 n<(n+2)^{2}
$$
(check it!). But since the numbers $(n+1)^{2}$ and $(n+2)^{2}$ are the two closest perfect squares to each other, the number $n^{2}+3 n$ will be a perfect square only when the left inequality turns into an equality:
$$... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,236 |
667*. Are there such natural numbers $a$ and $b$ for which the numbers $a^{2}+b$ and $b^{2}+a$ are perfect squares? | $\triangle$ Let $a \geqslant b$. Apply the "squeezing" method for the number $a^{2}+b$ in the following form:
$$
a^{2}<a^{2}+b \leqslant a^{2}+a<(a+1)^{2} .
$$
Since the number $a^{2}+b$ is squeezed between two consecutive perfect squares and both inequalities are strict, this number cannot be a perfect square.
Answ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,237 |
670*. There is an infinite arithmetic progression, all members of which are natural numbers. One of them is a perfect square. Prove that the progression contains an infinite set of perfect squares. | $\triangle$ Let's denote the common difference of the arithmetic progression by $d$, and its term, which is a perfect square, by $a^{2} (a \in N)$. Then for any natural $n$ we get:
$$
(a+n d)^{2}=a^{2}+2 a n d+n^{2} d^{2}=a^{2}+d\left(2 a n+n^{2} d\right)
$$
From this, it follows that the perfect square $(a+n d)^{2}$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,238 |
672*. Prove that the sum of the squares of any 10 consecutive natural numbers is not a perfect square. | $\triangle$ Let the smallest of 10 consecutive natural numbers be denoted by $n-4$ (assuming that $n>4$). Then the largest of them is $n+5$. We obtain:
$$
(n-4)^{2}+(n-3)^{2}+(n-2)^{2}+(n-1)^{2}+n^{2}+(n+1)^{2}+(n+2)^{2}+(n+3)^{2}+
$$
$$
+(n+4)^{2}+(n+5)^{2}=10 n^{2}+10 n+85=5\left(2 n^{2}+2 n+17\right)=5\left(2\left... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,239 |
678. The number 43 is written in the septenary (base-7) numeral system. In which system is it written with the same digits but in reverse order? | $\triangle$ First, let's convert the number 43 to the decimal system:
$$
43_{7}=4 \cdot 7+3=28+3=31
$$
139
Now, let's determine the base $d$ of the system in which the equality $31=34_{d}$ holds. We will have:
$$
31=3 d+4, \quad 3 d=27, \quad d=9
$$
Answer: in base 9. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,241 |
685. For what base of the numeral system does the following rebus have a solution
$$
\begin{aligned}
& \text { KITO } \\
& + \text { KIOTO } \\
& \hline \text { TOKIO, }
\end{aligned}
$$
where identical letters represent identical digits, and different letters represent different digits? Find all solutions to the reb... | $\triangle$ From the last column, it is clear that the digit O is equal to zero.
From the second-to-last column $2 \mathrm{~T}=$ I or $2 \mathrm{~T}=$ I $+d$, where $d$ is the base of the numeral system. But since in the third column the sum is equal to K, not I, only the second option remains.
We obtain the system o... | 4350_{7}+43050_{7}=50430_{7} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,242 |
690. Find the divisibility rule for 2 in a number system with any odd base. | Let's denote the base of the numeral system as $d$. Take any number in this system
$$
x=\overline{a_{1} a_{2} a_{3} \ldots a_{n-1} a_{n d}}=a_{1} \cdot d^{n-1}+a_{2} \cdot d^{n-2}+a_{3} \cdot d^{n-3}+\ldots+a_{n-1} \cdot d+a_{n}
$$
Add and subtract the sum $a_{1}+a_{2}+a_{3}+\ldots+a_{n}$ (the "plus-minus" trick).
W... | Aisdivisible2ifonlyifthesumofitsdigitsisdivisible2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,243 |
693. Two students, A and B, are hosting a math evening at school. A leaves the hall, and B asks the audience to name any natural number from 1 to 31. Then he calls five audience members to the stage and lines them up. A returns and, by looking at the five audience members on stage, immediately names the chosen number. ... | $\triangle$ Let's use the binary number system. Each natural number from 1 to 31 can be represented by a number containing five digits of this system; for example, the number 6 is represented as $00110_{2}$. We will denote the digit one in this system by a boy, and the digit zero by a girl, called to the stage. Then, f... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,244 |
695. We have a balance scale and weights of $1, 2, 4, 8, 16$, and 32 kg, one of each weight. Prove that with these weights, you can weigh any load from 1 to 63 kg (with an accuracy of 1 kg), by placing the weights only on one side of the scale. | $\triangle$ Let's see how this problem can be solved using number systems.
Suppose we need to weigh a load of $n$ kg, where $n$ is a natural number satisfying the inequality $1 \leqslant n \leqslant 63$. Let's write the number $n$ in binary. For this, no more than six digits are needed, since
$$
n<64=2^{6}=1000000_{2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,245 |
697. You have thought of a natural number between 1 and 500. I will try to guess it by asking you no more than 9 questions, to which you will truthfully answer only "yes" or "no." What should these questions be? | $\triangle$ Such a series of questions is possible.
1) Divide the number you have in mind by 2. Does it divide by $2$?
If the answer is "yes," I will write down the digit 0, and if "no," the digit 1.
2) Divide the quotient obtained from the first division by 2. Does it divide by 2?
If the answer is "yes," I will wr... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,246 |
700*. In what number system is the number $11111_{d}$ a perfect square? | $\triangle$ Let's represent this number as a sum, decomposing it by powers of $d$:
$$
11111_{d}=d^{4}+d^{3}+d^{2}+d+1
$$
We need to find a natural number $d>1$ such that the resulting sum is a square of a quadratic trinomial with argument $d$ and integer coefficients.
We will bound this sum between two squares:
$$
... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,247 |
702. Prove that any natural number not less than 2 can be represented as the sum of two coprime numbers. | $\triangle$ Let's represent a natural number $n \geqslant 2$ as a sum in the following way:
$$
n=(n-1)+1
$$
The natural numbers $n-1$ and 1 are coprime. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,248 |
705. The numbers 12 and 17 are written on the board. It is allowed to add to the board a number equal to the sum of any two of the already written numbers. Is it possible, by repeating this operation, to achieve that the number 2000 is written on the board?
706 ${ }^{\circ}$. Prove that any odd number can be represent... | $\triangle$ Let's take any odd number in the form $2 a+1$, where $a$ is an integer, and transform it as follows:
$$
2 a+1=\left(a^{2}+2 a+1\right)-a^{2}=(a+1)^{2}-a^{2}
$$
10 sec. 579 | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,249 |
709. Prove that any even number of the form $4a+2$, where $a$ is an integer, cannot be represented as the difference of two perfect squares. | $\triangle$ Suppose there exist such integers $x$ and $y$ that
$$
4 a+2=x^{2}-y^{2}, \quad(x+y)(x-y)=4 a+2
$$
Since the number $4 a+2$ is even, the numbers $x$ and $y$ must be of the same parity, and therefore the sum $x+y$ and the difference $x-y$ are even numbers. However, this means that the left side of the last ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,250 |
712. Let $a$ be any natural number. Find some representation of the number $a^{3}$ as the difference of two perfect squares. | $\triangle$ Let $x$ and $y$ be such integers that
$$
x^{2}-y^{2}=a^{3}, \quad(x+y)(x-y)=a^{3}
$$
For the last equality to hold, it is sufficient that the following two equalities hold:
$$
\left\{\begin{array}{l}
x+y=a^{2} \\
x-y=a .
\end{array}\right.
$$
This means that if the numbers $x$ and $y$ satisfy this syste... | ^{3}=[\frac{(+1)}{2}]^{2}-[\frac{(-1)}{2}]^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,251 |
719. Find all numbers of the form $222 \ldots 2$ that can be represented as the sum of two perfect squares. | $\triangle$ Let
$$
222 \ldots 2=a^{2}+b^{2}
$$
where $a$ and $b$ are integers. Then the numbers $a$ and $b$ can only be odd:
$$
a=2 k+1, b=2 l+1 \quad(k \in Z, l \in Z)
$$
Therefore, the sum $a^{2}+b^{2}$ when divided by 8 gives a remainder of 2:
$$
a^{2}+b^{2}=(2 k+1)^{2}+(2 l+1)^{2}=4 k(k+1)+4 l(l+1)+2
$$
On th... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,252 |
720. Prove that any natural number can be represented as the sum of two perfect squares if and only if the doubled number can also be represented in the same form. | $\triangle$ 1) First, we will prove that if
$$
n=a^{2}+b^{2}
$$
where $a$ and $b$ are integers, then the number $2n$ can also be represented in the same form. Multiply the given equation by 2 and transform the resulting sum:
$$
2 n=2 a^{2}+2 b^{2}=\left(a^{2}+2 a b+b^{2}\right)+\left(a^{2}-2 a b+b^{2}\right)=(a+b)^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,253 |
725. The number $11 \ldots 122 \ldots 2$ (written with 100 ones and 100 twos) can be expressed as the product of two consecutive natural numbers. | $\triangle$ First, let's try this for the numbers 12 and 1122:
$$
12=3 \cdot 4, \quad 1122=33 \cdot 34
$$
Apparently, in our case, one of the factors is written entirely with threes, and the other is one more than that.
Let's factorize the given number. But how do we get a factor written entirely with threes? We wil... | 33\ldots3\cdot33\ldots34 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,254 |
727. Prove that the number
$$
2 \cdot\left(6^{n}+1\right)
$$
cannot be represented as the product of two consecutive natural numbers for any natural $n$. | $\triangle$ The number $6^{n}+1$ ends with the digit 7, and therefore, the number $2 \cdot\left(6^{n}+1\right)$ ends with the digit 4. On the other hand, the product $k(k+1)$ for a natural number $k$ can only end with the digits 0, 2, or 6 (check this!). Therefore, the equation
$$
2 \cdot\left(6^{n}+1\right)=k(k+1)
$$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,255 |
729. Does there exist a natural number $n$ such that the number $3 n+1$ can be represented as the product of two natural numbers? | $\triangle$ The number $3n+1$ when divided by 3 gives a remainder of 1, while the number $k(k+1)$ for a natural number $k$ gives a remainder of 0 or 2 (prove it!). Therefore, such a natural number $n$ does not exist.
Answer: does not exist. | doesnotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,256 |
731*. Can the number $n^{2}+7 n+8$ for any natural $n$ be represented as the product of two consecutive natural numbers? | $\triangle$ Let's apply the "bracketing" method (see the solution to problem 663 in § 14). The following inequality holds:
$$
(n+2)(n+3)<n^{2}+7 n+8<(n+3)(n+4)
$$
(check this!).
Since the products $(n+2)(n+3)$ and $(n+3)(n+4)$ are the closest to each other products of two consecutive natural numbers, and the number ... | cannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,257 |
733. Prove that the square of any natural number greater than 1 can be represented as the sum of consecutive odd natural numbers. | $\triangle$ First, let's check this statement with examples:
$$
4=1+3, \quad 9=1+3+5, \quad 16=1+3+5+7, \quad 25=1+3+5+7+9
$$
It appears that the smallest of the odd numbers is always 1, and the number of addends in the case of the square $n^{2}$, where $n$ is a natural number, is equal to $n$. Let's verify our hypot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,258 |
736. Find all representations of the number 1 in the form of the sum
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$
where $x, y, z$ are natural numbers. | $\triangle$ Given:
$$
1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}, \quad x y z=y z+x z+x y
$$
Let $x \leqslant y \leqslant z$. Then among the three products $x y, x z$, and $y z$, the largest is $y z$. Using this, we obtain the inequality:
$$
x y z=x y+x z+y z \leqslant y z+y z+y z=3 y z
$$
Thus, $x y z \leqslant 3 y z$,... | 1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,259 |
739*. Can the number 456 be represented as the product of several natural numbers such that the sum of the squares of all these numbers is also equal to 456? | $\triangle$ The number 456 can be factored into prime factors: $456=2^{3} \cdot 3 \cdot 19$. Let's list all these prime factors: $2,2,2,3$ and 19.
Now let's write the equation
$$
456=2^{2}+2^{2}+2^{2}+3^{2}+19^{2}+x
$$
where $x$ is a natural number. We will solve for $x$:
$$
456=4+4+4+9+361+x, \quad 456=382+x, \qua... | 456=1\cdot1\cdot1\cdot\ldots\cdot1\cdot2\cdot2\cdot2\cdot3\cdot19(74factors,equalto1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,260 |
741*. Which years of the $XX$ century can be represented in the form $2^{n}-2^{k}$, where $n$ and $k$ are natural numbers? Find all such years. | $\triangle$ Let the desired year be denoted as $\overline{19 x y}$. Then
$$
\overline{19 x y}=2^{n}-2^{k}, \quad 2^{n}=\overline{19 x y}+2^{k}
$$
Therefore, $2^{n}>\overline{19 x y}$, from which it follows that $n>10$.
Consider several cases.
1) Suppose $n=11$. We get:
$$
\overline{19 x y}=2048-2^{k}
$$
The value... | 1984,1920 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,261 |
744*. Find all natural numbers that cannot be represented as the sum of three natural numbers, the product of which is a perfect square. | $\triangle$ First, we will look for all natural numbers that can be represented in such a form. Let's consider two cases.
1) Suppose the natural number $a$ is odd: $a=2k+1$, where $k$ is a natural number.
If $a$ is represented as the sum: $a=k+k+1$, then the product of these three addends is the square of a natural n... | 1,2,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,262 |
746. There are containers of two types: 130 kg and 160 kg. How many containers of the first type and how many of the second type are there if together they weigh 3 tons? List all solutions. | Let's denote the number of containers of the first type by $x$, and the second type by $y$. We get the equation
$$
130 x + 160 y = 3000, \quad 13 x + 16 y = 300
$$
Let's try to use divisibility by 13. For this, we represent $16 y$ as $13 y + 3 y$, and divide 300 by 13 with a remainder:
$$
13 x + 13 y + 3 y = 13 \cdo... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,263 |
752. Find the smallest natural number that is divisible by 28, and when divided by 15, gives a remainder of 4. | $\triangle$ The desired number is, on the one hand, $28 x$, and on the other $-15 y+4$, where $x$ and $y$ are natural numbers. We obtain the equation
$$
28 x=15 y+4
$$
We do not need to find all solutions to this equation in natural numbers, but only one solution - the one for which the values of $x$ and $y$ are mini... | 364 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,264 |
755. "Forty mice were walking, carrying forty groats,
Two poorer mice carried two groats each, Many mice - with no groats at all. The bigger ones carried seven each. And the rest carried four each.
How many mice walked without groats?" | $\triangle$ Let the number of mice that walked without groats be denoted by $x$, the number of large mice by $y$, and the number of those that carried four groats by $z$. We can set up a system of equations with the unknowns $x, y$, and $z$:
$$
\left\{\begin{array} { l }
{ 2 \cdot 2 + 7 y + 4 z = 4 0 , } \\
{ 2 + x +... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,265 |
757. Find all three-digit numbers that are 11 times the sum of their digits. | $\triangle$ Let the desired number be $\overline{a b c}$. We get:
$$
\overline{a b c}=11(a+b+c), \quad 100 a+10 b+c=11 a+11 b+11 c, \quad 89 a=b+10 c
$$
Let's try using divisibility by 10 (but we can also use 11):
$$
90 a-a=b+10 c, \quad 90 a-10 c=a+b .
$$
Therefore, the sum $a+b$ is divisible by 10, i.e., $a+b=10$... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,266 |
763. Find all solutions of the equation
$$
5 x-7 y=3
$$
in integers $x$ and $y$. | $\triangle$ Express from the equation the unknown whose coefficient is smaller in absolute value - in this case $x$:
$$
5 x=7 y+3, \quad x=\frac{7 y+3}{5}
$$
In the numerator of the obtained fraction, $7 y$ can be split into two terms, one of which is divisible by 5 for any integer $y$, and the coefficient of the oth... | 7-12,5-9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,267 |
766. Prove that the equation
$$
21 x-35 y=59
$$
has no solutions in integers $x$ and $y$. | $\triangle$ Indeed, for integer $x$ and $y$, the left side of the equation is divisible by 7, while the right side is not.
In general, for the equation
$$
a z+b y=c,
$$
where $a, b, c$ are given integers, there are only two possible cases: either it has no solutions in integers $x$ and $y$ (when $c$ is not divisible... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,268 |
767. Solve the equation in natural numbers $x$ and $y$
$$
65 x - 43 y = 2
$$ | $\triangle$ First, let's solve this equation in integers using a familiar method. We will provide the corresponding formulas without derivation (you can verify them by substitution into the equation):
$$
x=4-43 t, \quad y=6-65 t
$$
Now, let's find all integer $t$ for which $x$ and $y$ are positive. To do this, we sol... | 4-43,\quad6-65,where=0,-1,-2,\ldots | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,269 |
778*. Find the smallest natural number that can be represented in exactly two ways as $3 x+4 y$, where $x$ and $y$ are natural numbers. | $\triangle$ Let
$$
3 x+4 y=n
$$
where $n-$ is a natural number. Let's solve this equation first in integers:
$$
\begin{array}{cl}
3 x=n-4 y, & x=\frac{n-4 y}{3}=-y+\frac{n-y}{3} \\
\frac{n-y}{3}=t \in Z ; \quad n-y=3 t, \quad y=n-3 t ; \quad x=-(n-3 t)+t=4 t-n
\end{array}
$$
Next, let's find all $t$ for which $x$ a... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,270 |
## 18.1.
First, let's consider equations that do not contain terms with squares of unknowns. Therefore, such an equation contains a single term of the second degree with the product of unknowns \(xy\).
\(780^{\circ}\). Prove that the product of two integers \(x\) and \(y\) equals their sum only in two cases: \(x=y=0\... | $\triangle$ We need to solve the equation
$$
x y=x+y
$$
in integers $x$ and $y$.
Let's consider two methods to solve the problem.
1) Express $y$ in terms of $x$ from the equation:
$$
x y-y=x, \quad y(x-1)=x, \quad y=\frac{x}{x-1}
$$
11 Law. 579
161
(Check that the equation does not lose solutions when dividing ... | 02 | Algebra | proof | Yes | Yes | olympiads | false | 45,271 |
785. Solve the equation in integers
$$
3 x y+y=7 x+3
$$ | $\triangle$ Express $y$ in terms of $x$ from the equation:
$$
y=\frac{7 x+3}{3 x+1}
$$
To separate the integer part of the resulting fraction, multiply this equation by 3:
$$
3 y=\frac{21 x+9}{3 x+1}=\frac{(21 x+7)+2}{3 x+1}=7+\frac{2}{3 x+1}
$$
From this, $3 x+1$ can only take the values $-2, -1, 1$, and 2. We wil... | (0;3),(-1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,272 |
793. Solve the equation in integers
$$
2 x^{2}-2 x y+9 x+y=2
$$ | $\triangle$ Express from the equation the unknown that enters it only in the first degree - in this case $y:$
$$
2 x^{2}+9 x-2=2 x y-y, \quad y=\frac{2 x^{2}+9 x-2}{2 x-1}
$$
For the last fraction, let's separate the integer part - either using the rule of polynomial division by "corner" or by breaking down the numer... | (1;9),(2;8),(0;2),(-1;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,273 |
795. Does the equation
$$
x^{2}+2 x y=2002
$$
have a solution in integers? | $\triangle$ Let's represent the equation in the following form:
$$
(x+y)^{2}-y^{2}=2002
$$
Then the numbers $x+y$ and $y$ have the same parity. But the difference of squares of two integers with the same parity is divisible by 4, while 2002 is not.
Answer: no solutions. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,274 |
799. Solve the equation in integers
$$
x^{2}-y^{2}=1997
$$ | $\triangle$ Let's write the equation in the form:
$$
(x+y)(x-y)=1997
$$
The number 1997 is prime, so there are only four possible cases:
1) $\left\{\begin{array}{l}x+y=1997, \\ x-y=1 ;\end{array}\right.$
2) $\left\{\begin{array}{l}x+y=1, \\ x-y=1997 ;\end{array}\right.$
3) $\left\{\begin{array}{l}x+y=-1997, \\ x-y=-... | (999;998),(999;-998),(-999;-998),(-999;998) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,275 |
804. Solve the equation in integers
$$
x^{2}-x y+y^{2}=x+y
$$ | $\triangle$ The methods we used to solve problems 799-802 do not achieve the goal here. Let's try to gather all terms of the equation on the left side and arrange them by the degree of $x$:
$$
x^{2}-(y+1) x+\left(y^{2}-y\right)=0
$$
We will consider the last equation as a quadratic equation in terms of $x$. Its discr... | (1;2),(2;2),(0;0),(1;0),(0;1),(2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,276 |
806. Prove that the equation
$$
3 x^{2}=16 y^{2}+8 y+5
$$
has no solutions in integers. | $\triangle$ Let's complete the sum $16 y^{2}+8 y$ on the right side of the equation to a square of a sum:
$$
3 x^{2}=\left(16 y^{2}+8 y+1\right)+4, \quad 3 x^{2}=(4 y+1)^{2}+4
$$
From this, it is clear that the sum $4 y+1$ is not divisible by 3. Then $(4 y+1)^{2}$, when divided by 3, gives a remainder of 1 (see the s... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,277 |
808. Solve the equation in integers
$$
(x+y)^{2}=x-y
$$ | $\triangle$ It is hardly expedient to expand the expression $(x+y)^{2}$; instead, this expression should be utilized.
Let $x+y=t$, where $t$ is some integer. Then $x-y=t^{2}$.
We obtain the system of equations:
$$
\left\{\begin{array}{l}
x+y=t \\
x-y=t^{2}
\end{array}\right.
$$
By adding and subtracting the equatio... | [\frac{(1+)}{2};\frac{(1-)}{2}),\in\mathrm{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,278 |
810. Prove that the equation
$$
x^{2}-y^{2}=a^{5}
$$
has a solution in integers $x$ and $y$ for any integer $a$. | $\triangle$ For this equality to hold, it is sufficient that the following equalities are satisfied:
$$
\left\{\begin{array}{l}
x+y=a^{4} \\
x-y=a
\end{array}\right.
$$
This means that if $x$ and $y$ satisfy the system of equations written here, then they also satisfy the original equation (but, of course, not vice v... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,279 |
812. Solve in integers $x, y$ and $z$ the equation
$$
5 x^{2} + y^{2} + 3 z^{2} - 2 y z = 30
$$ | $\triangle$ Transform the equation by completing the square for the difference $y^{2}-2 y z$:
$$
5 x^{2}+\left(y^{2}-2 y z+z^{2}\right)+2 z^{2}=30, \quad 5 x^{2}+(y-z)^{2}+2 z^{2}=30
$$
From this, it is clear that the unknowns can only take values with small absolute values. Let's determine the constraints for $x$:
... | (1;5;0),(1;-5;0),(-1;5;0),(-1;-5;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,280 |
814. Does the equation
$$
x^{3}+y^{3}=9 z+5
$$
have solutions in integers $x, y$ and $z ?$ | $\triangle$ The cube of an integer, when divided by 9, can only leave a remainder of 0, 1, or 8 (we have already encountered this in solving some problems in § 8). Consequently, the sum of two perfect cubes, when divided by 9, cannot leave a remainder of 5. In other words, the given equation does not hold for any integ... | doesnothave | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,281 |
816. Solve in integers $x, y, z$ and $t$ the system of equations
$$
\left\{\begin{array}{l}
x y+z t=1 \\
x z+y t=1 \\
x t+y z=1
\end{array}\right.
$$ | $\triangle$ Subtract, for example, the second equation from the first in the system:
$$
x(y-z)+t(z-y)=0, \quad(y-z)(x-t)=0
$$
Then $y-z=0$ or $x-t=0$. Let's consider both cases.
1) Let $y-z=0$, i.e., $y=z$.
The system of equations takes the following form:
$$
\left\{\begin{array}{l}
x z+z t=1 \\
x z+z t=1 \\
x t+z... | (0;1;1;1),(1;0;1;1),(1;1;0;1),(1;1;1;0),(0;-1;-1;-1),(-1;0;-1;-1),(-1;-1;0;-1),(-1;-1;-1;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,282 |
821. Find all natural numbers $p$ and $q$ for which the quadratic equation
$$
x^{2}-p q x+p+q=0
$$
has only integer roots. | Let's denote the integer roots of this equation as $x_{1}$ and $x_{2}$. Then, by Vieta's theorem:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=p q \\
x_{1} x_{2}=p+q
\end{array}\right.
$$
We have obtained a peculiar system of equations: the sum of two unknowns out of four equals the product of the other two, while the prod... | p=1,q=5;p=5,q=1;p=q=2;p=2,q=3;p=3,q=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,283 |
823. Prove that the equation
$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+\ldots+x_{10}^{3}=600
$$
has an infinite number of solutions in integers $x_{1}, x_{2}, x_{3}, \ldots, x_{10}$. | $\triangle$ Let's try to guess some of the unknowns for now. Let's assume
$$
x_{1}=x_{2}=\ldots=x_{6}=4
$$
We get:
$$
\begin{gathered}
64+64+\ldots+64+x_{7}^{3}+x_{8}^{3}+x_{9}^{3}+x_{10}^{3}=600 \\
64 \cdot 6+x_{7}^{3}+x_{8}^{3}+x_{9}^{3}+x_{10}^{3}=600, \quad x_{7}^{3}+x_{8}^{3}+x_{9}^{3}+x_{10}^{3}=600-384=216
\e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,284 |
828. Solve the equation in integers
$$
3^{x}-5=4 y . \quad .
$$ | $\triangle$ First of all, $x$ must be non-negative, otherwise the power $3^{x}$ will turn into a fraction. Then the question of the problem reduces to the following: for which whole non-negative $x$ is the difference $3^{x}-5$ divisible by 4?
By trial and error, we establish that $3^{x} - 5$ is divisible by 4 for $x=0... | (2n;\frac{1}{4}(3^{2n}-5)) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,285 |
833. Find all two-digit numbers that are equal to triple the product of their digits. | $\triangle$ Let the desired number be denoted by $\overline{a b}$. Then
$$
\overline{a b}=3 a b, \quad 10 a+b=3 a b
$$
We have obtained a quadratic equation with two unknowns of the type considered in § 18, point 18.1, with the difference that this equation needs to be solved in integers from 0 to 9.
Express $b$ in ... | 15,24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,286 |
835. A three-digit number is equal to the cube of its unit digit. Find all such numbers. | Let's denote a three-digit number as $\overline{x y z}$. Then
$$
\overline{x y z}=z^{3}, \quad 10 \cdot \overline{x y}+z=z^{3}
$$
In which cases does the cube of the digit $z$ end in $z$? This is possible when $z=0,1,4,5$, 6, and 9. However, the values $z=0, z=1$, and $z=4$ are obviously not valid.
If $z=5$, then
$... | 125,216,729 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,287 |
841*. A natural number is equal to the cube of the number of its thousands. Find all such numbers. | $\triangle$ Let's denote the desired number as $1000 x+y$, where $x-$ is the number of its thousands, and $y<1000$. Estimate the number $k$ from above:
$$
k=\frac{y}{x}<\frac{1000}{31}<33
$$
However, the sum $1000+k$ must be a square of a natural number. Only $k=24$ fits:
$$
1000+k=1024=32^{2}
$$
Therefore, $x=32$.... | 32768 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,288 |
844. Solve the equation in natural numbers
$$
x^{2}+x y+y^{2}=x+y+9
$$ | $\triangle$ Let's gather all terms on the left side and multiply the resulting equation by 2, highlighting three squares of sums in the left part:
$$
\begin{gathered}
2 x^{2}+2 x y+2 y^{2}-2 x-2 y-18=0 \\
\left(x^{2}+2 x y+y^{2}\right)+\left(x^{2}-2 x+1\right)+\left(y^{2}-2 y+1\right)-2-18=0
\end{gathered}
$$
$$
(x+y... | (3;1),(1;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,289 |
847. Prove that the equation
$$
x^{2}+y^{2}=61^{3}
$$
has at least one solution in natural numbers. | $\triangle$ Let
$$
x=61 a, \quad y=61 b
$$
where $a$ and $b$ are natural numbers. Substitute these values of $x$ and $y$ into the equation:
$$
61^{2} a^{2}+61^{2} b^{2}=61^{3}, \quad a^{2}+b^{2}=61
$$
The last equation has a solution in natural numbers: $a=6, b=5$. Then the original equation has a solution:
$$
x=6... | 366,305 | Number Theory | proof | Yes | Yes | olympiads | false | 45,290 |
850. Solve the equation in integers
$$
\sqrt{x}+\sqrt{y}=\sqrt{98}
$$ | $\triangle$ From the equation, it is clear that
$$
0 \leqslant x \leqslant 98, \quad 0 \leqslant y \leqslant 98
$$
Let's represent it in the form $\sqrt{y}=\sqrt{98}-\sqrt{x}$ and square the last equation:
$$
y=98+x-2 \sqrt{98 x}, \quad y=98+x-14 \sqrt{2 x}
$$
## From this
$$
2 x=4 a^{2}, \quad x=2 a^{2},
$$
wher... | (0;98),(2;72),(8;50),(18;32),(32;18),(50;8),(72;2),(98;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,291 |
854. Solve the equation in natural numbers
$$
x y^{2}-x y-y^{2}+y=94
$$ | $\triangle$ Let's factorize the left side of the equation:
$$
x y^{2}-x y-y^{2}+y=y(x y-x-y+1)=y(y-1)(x-1) .
$$
Moreover, the number 94 can be factored into prime factors: $94=2 \cdot 47$.
Let's think about which divisors of the number 94 differ by 1? This is only possible when $y-1=1, y=2$.
It remains to find $x$:... | (48,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,292 |
858. Find the smallest pair of natural numbers $x$ and $y$ that satisfy the equation
$$
5 x^{2}=3 y^{5}
$$ | $\triangle$ From the equation, it is clear that $x$ is divisible by 3, and $y$ is divisible by 5:
$$
x=3 x_{1}, \quad y=5 y_{1},
$$
where $x_{1}$ and $y_{1}$ are natural numbers. Then
$$
5 \cdot 3^{2} x_{1}^{2}=3 \cdot 5^{5} y_{1}^{5}, \quad 3 x_{1}^{2}=5^{4} y_{1}^{5}
$$
From this, it follows that $x_{1}$ is divis... | 675,15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,293 |
860. Prove that the equation
$$
y^{3}=x^{2}+x
$$
has no solutions in natural numbers. | $\triangle$ Let's write the equation in the form:
$$
y^{3}=x(x+1)
$$
Two consecutive natural numbers $x$ and $x+1$ are coprime (see the supporting problem 454 from § 11), so their product can only be a perfect cube if each of these numbers is a perfect cube.
$$
x+1=a^{3}, \quad x=b^{3}
$$
where $a$ and $b$ are natu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,294 |
863. The product of three prime numbers is 5 times their sum. Find all such triples of numbers. | $\triangle$ Let's denote these prime numbers as $x, y$, and $z$. Then
$$
x y z=5(x+y+z)
$$
One of the factors on the left side of the equation is equal to 5, for example, $x=5$:
$$
5 y z=5(5+y+z), \quad y z=y+z+5
$$
Express $z$ in terms of $y$ from the last equation and separate the integer part of the resulting fr... | 2,5,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,295 |
865*. Solve the equation in natural numbers $x, y$ and $z$
$$
x y + y z + z x = 2(x + y + z).
$$ | $\triangle$ Let's gather all the terms of the equation on the left side and rewrite it as:
$$
x(y-2)+y(z-2)+z(x-2)=0 .
$$
From the last equation, it follows that the inequalities
$$
x>2, \quad y>2, \quad z>2
$$
cannot all be true simultaneously, otherwise the left side of the equation would become positive. Therefo... | (1;2;4),(1;4;2),(2;1;4),(2;4;1),(2;2;2),(4;1;2),(4;2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,296 |
867. Solve the equation in natural numbers
$$
x+\frac{y}{z}=\frac{x^{2} y}{z}
$$ | $\triangle$ Transform the equation:
$$
x z+y=x^{2} y, \quad x z=y\left(x^{2}-1\right)
$$
From here, the product $y\left(x^{2}-1\right)$ is divisible by $x$. But the numbers $x$ and $x^{2}-1$ are coprime (think about how to prove this, using the solutions of problems 454 and 466 from § 11 as examples). Then $y$ is div... | (x;kx;k(x^{2}-1)),wherek\inN,x\inN,x>1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,297 |
872. Solve the equation in natural numbers
$$
\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{1998}
$$ | $\triangle$ Let's raise this equation to the third power:
$$
x+y+3 \sqrt[3]{x^{2} y}+3 \sqrt[3]{x y^{2}}=1998, \quad x+y+3 \sqrt[3]{x y}(\sqrt[3]{x}+\sqrt[3]{y})=1998
$$
What's next? Next, we replace the sum in the parentheses with $\sqrt[3]{1998}$:
$$
x+y+3 \sqrt[3]{1998 x y}=1998
$$
Factorize the number 1998 into... | (74;592),(592;74) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,299 |
875. a) Prove that the equation
$$
x^{3}+y^{3}=z^{5}
$$
has a solution in natural numbers.
b) Prove that it has an infinite number of solutions in natural numbers. | $\triangle$ a) Let's substitute in the equation
$$
x=y=2^{k}, \quad z=2^{l} \quad(k \in \mathbb{N}, l \in \mathbb{N})
$$
We get:
$$
2^{3 k}+2^{3 k}=2^{5 l}, \quad 2^{3 k+1}=2^{5 l} .
$$
From here, the exponents must be equal:
$$
3 k+1=5 l
$$
This last equation has solutions in natural numbers, for example, $k=3$,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,300 |
877*. Does the equation
$$
x^{2}+y^{3}+z^{4}=10^{25}
$$
have a solution in natural numbers? | $\triangle$ Let's use the equality
$$
1^{2}+2^{3}+1^{4}=10
$$
Multiply it by $10^{24}$. Why exactly by $10^{24}$? Firstly, because $10 \cdot 10^{24}=10^{25}$, and secondly, because 24 is divisible by 2, 3, and 4. By comparing the resulting numerical equality
$$
10^{24}+2^{3} \cdot 10^{24}+10^{24}=10^{25}
$$
with th... | 10^{12},2\cdot10^{8},10^{6} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,301 |
879. Solve the equation in natural numbers
$$
x!+12=y^{2}
$$ | $\triangle$ For $x>4$ the number $x!$ ends with the digit 0, which means the number $x!+12$ ends with the digit 2. However, the square of a natural number cannot end with the digit 2, so there are no solutions to the equation for $x>4$.
It remains to check the values $x=1,2,3,$ and 4. A solution is obtained only for $... | (4,6) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,302 |
881*. Solve the equation in natural numbers
$$
x!+y!=z!
$$ | $\triangle$ Consider two cases.
1) Let $x<y$. Then the equation can be reduced by $x$!:
$$
1+(x+1)(x+2) \ldots y=(x+1)(x+2) \ldots z
$$
From here, 1 is divisible by $x+1$. But this is impossible, since the sum $x+1$ is at least 2.
2) Let $x=y$. We get:
$$
x!+x!=z!, \quad 2 x!=z!
$$
This equation can also be reduc... | (1;1;2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,303 |
883. Solve the equation in natural numbers
$$
1+x+x^{2}+x^{3}=2^{y}
$$ | $\triangle$ Obviously, $x$ is odd. Factorize the left side of the equation
$$
(1+x)\left(1+x^{2}\right)=2^{y}
$$
Then each of the numbers $1+x$ and $1+x^{2}$ must be a power of two. But for odd $x$
$$
\text { GCD }\left(1+x, 1+x^{2}\right)=2
$$
(think about why; see § 11, point 11.1.1). Since $1+x^{2} \geqslant 1+x... | (1;2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,304 |
885. Solve in prime numbers $x, y$ and $z$ the equation
$$
x^{y}+1=z
$$ | $\triangle$ The number $z$ is greater than 2, because if $z=2$, then $x=1$, which is impossible. Therefore, $z$ is odd, and consequently, the number $x$ is even. But $x$ is a prime number, so $x=2$. We get the equation:
$$
2^{y}+1=z
$$
Is the number $y$ even or odd? If $y$ is odd, then the sum $2^{y}+1$ is divisible ... | (2;2;5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,305 |
887. Solve the equation in natural numbers
$$
2^{x}-15=y^{2}
$$ | $\triangle$ Consider two cases.
1) Let $x$ be odd: $x=2 k+1(k \in N)$.
Let's see what remainder $2^{x}$ gives when divided by 3.
$$
2^{x}=2^{2 k+1}=2 \cdot 2^{2 k}=2 \cdot 4^{k}=2\left(\left(4^{k}-1\right)+1\right)=2\left(4^{k}-1\right)+2
$$
Since the difference $4^{k}-1$ is divisible by 3 (§ 7, first divisibility)... | (6;7),(4;1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,306 |
891. Solve the equation in natural numbers
$$
3^{x}-2^{y}=1
$$ | $\triangle$ First, let's determine whether the numbers $x$ and $y$ are even or odd.
If we write the equation as:
$$
3^{x}=2^{y}+1
$$
it is not hard to guess that the number $y$ is odd: only when $y$ is odd does the sum $2^{y}+1$ divide by the sum of the bases of the powers, i.e., by 3 (see § 7, second divisibility).... | (1;1),(2;3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,307 |
893. Prove that the equation
$$
x^{2}+x y+y^{2}=x^{2} y^{2}
$$
has no solutions in natural numbers. | $\triangle$ Let $x \leqslant y$. Then the largest term in the left-hand side of the equation is $y^{2}$. We obtain:
$$
x^{2} y^{2}=x^{2}+x y+y^{2} \leqslant y^{2}+y^{2}+y^{2}=3 y^{2}.
$$
Therefore,
$$
x^{2} y^{2} \leqslant 3 y^{2}, \quad x^{2} \leqslant 3
$$
from which $x=1$. But when $x=1$, the original equation b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,308 |
894. Find three natural numbers whose sum equals their product. List all solutions. | $\triangle$ Let the required numbers be denoted by $x, y$, and $z$. We obtain the equation:
$$
x+y+z=x y z
$$
For natural numbers $x, y$, and $z$, the right-hand side of this equation is generally greater than the left-hand side; therefore, the unknown numbers should be small. We will impose restrictions on these num... | 1,2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,309 |
900. Prove that the equation
$$
y^{2}=x^{2}+x+1
$$
has no solutions in natural numbers.
This is a problem by L. Euler, recently found in his notebook. | $\triangle$ Apply the "bracketing" method (see the solution to problem 663 in § 14).
For natural $x$, the inequality holds:
$$
x^{2}<x^{2}+x+1<x^{2}+2 x+1
$$
From this,
$$
x^{2}<y^{2}<(x+1)^{2}, \quad x<y<x+1
$$
But the last inequality is impossible for natural $x$ and $y$: $x$ and $x+1$ are the closest natural nu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,310 |
902*. Solve the equation in natural numbers
$$
y^{2}+y=x^{4}+x^{3}+x^{2}+x
$$ | $\triangle$ Multiply the equation by 4 and add 1 to both sides of the resulting equation so that its left side becomes a perfect square.
$$
(2 y+1)^{2}=4 x^{4}+4 x^{3}+4 x^{2}+4 x+1
$$
The right side of the last equation is, for almost all natural $x$, between $\left(2 x^{2}+x\right)^{2}$ and $\left(2 x^{2}+x+1\right... | (2,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,311 |
904. Prove that the equation
$$
x^{2}+y^{2}=5\left(z^{2}+t^{2}\right)
$$
has an infinite set of solutions in natural numbers $x, y, z$ and $t$. | $\triangle$ Let's first find one solution to this equation. For example, $x=3$, $y=1, z=1, t=1$ works, since
$$
3^{2}+1^{2}=5\left(1^{2}+1^{2}\right)
$$
But then the equation has the following infinite set of solutions in natural numbers:
$$
x=3 n, \quad y=n, \quad z=n, \quad t=n
$$
where $n-$ is any natural number... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,312 |
905. Solve the equation
$$
x^{3}=3 y^{3}+9 z^{3}
$$
in non-negative integers $x, y$ and $z$. | $\triangle$ From the equation, it is clear that the number $x$ is divisible by 3: $x=3 x_{1}$, where $x_{1}$ is a non-negative integer. We obtain:
$$
27 x_{1}^{3}=3 y^{3}+9 z^{3}, \quad 9 x_{1}^{3}=y^{3}+3 z^{3}
$$
From the last equation, it follows that $y$ is divisible by 3: $y=3 y_{1}$, where $y_{1}$ is a non-nega... | (0;0;0) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,313 |
908*. Does the equation
$$
x^{2}+y^{2}+z^{2}=2 x y z
$$
have a solution in non-negative integers? | $\triangle$ By the condition, the sum $x^{2}+y^{2}+z^{2}$ is divisible by 2. Therefore, among the numbers $x, y$, and $z$, either one is even and two are odd, or all of them are even. Let's consider both possibilities.
1) Suppose one of the numbers $x, y, z$ is even, and two are odd.
But in this case, the right-hand ... | (0,0,0) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,314 |
910*. Find some solution to the equation
$$
x^{2}-119 y^{2}=1
$$
in natural numbers. | $\triangle$ The identity
$$
n^{2}-(n+1)(n-1)=1
$$
may raise the question: what does this identity have to do with anything? Who knows... The author just felt like writing it.
Answer: for example, $x=120, y=11$. | 120,11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,315 |
915*. Solve the equation in natural numbers
$$
x^{2}+y^{2}=z^{2}
$$
The formulas for the unknowns $x, y$, and $z$ that we will obtain have been known in mathematics for over 3000 years (they first appeared in Ancient Babylon). These formulas are useful to know, and the method of solving the problem is instructive.
E... | $\triangle$ Let's limit ourselves to the case when the numbers $x, y$, and $z$ are pairwise coprime: if two of these three numbers have a greatest common divisor $d$ greater than 1, then for the equation (1) to have a solution in natural numbers, it is necessary that the third unknown also be divisible by $d$, and then... | 2ab,\quad^2-b^2,\quad^2+b^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,316 |
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