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 |
|---|---|---|---|---|---|---|---|---|---|
97. The number $A$ is written in the decimal system using 666 threes, and the number $B$ is written using 666 sixes. From which digits does the product $A \cdot B$ consist? | 97. The desired number will be equal to the product $A_{1} \cdot B_{1}$, where $A_{1}$ consists of 666 digits 9, and $B_{1}$ consists of 666 digits 2. But $A_{1}$ is one less than the number $10^{6666}$, which is expressed by a one followed by 666 zeros. Therefore, multiplying the number $B_{1}$ by this number is the s... | 22\ldots2177\ldots78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,368 |
98. Find the quotient and remainder of dividing the number $A$, which is written using 1001 sevens, by the number 1001. | 98. The number 777777 is divisible by 1001 and gives a quotient of 777. Therefore, the number $\underbrace{777 \ldots 70000}_{996 \text { times }}$ gives a quotient of
$777000777000 \ldots 77709010$.
,700 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,369 |
99. Find the smallest square that starts with six twos. | 99. Since the number 222222 is not a perfect square, the decimal representation of the unknown number has the form $222222 a_{7} a_{8} \ldots a_{n}$, where $a_{7}, a_{8}, \ldots, a_{n}$ are some unknown digits.
First, assume that the number of digits $n$ of the unknown number is even: $n=2 k$. We will now extract the ... | 222222674025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,370 |
100. Do there exist such integers $m$ and $n$ that $m^{2}=n^{2}+1954$? | 100. From the equality $m^{2}=n^{2}+1954$ it follows that the numbers $m^{2}$ and $n^{2}$ are of the same parity (i.e., both even or both odd); therefore, the same can be said about the numbers $m$ and $n$. But in this case, the number $1954=m^{2}-n^{2}=(m+n)(m-n)$ must necessarily be divisible by 4 (since the numbers ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,371 |
101. Append three digits to 523 ... so that the resulting six-digit number is divisible by 7, 8, and 9. | 101. The desired six-digit number, starting with the digits 523 and divisible without remainder by $7 \cdot 8 \cdot 9=504$, can be represented as $523000+X$, where $X$ is a three-digit number. Direct division gives $523000=504 \cdot 1037+352$, i.e., 523000 gives a remainder of 352 when divided by 504. Since the sum of ... | 523152523656 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,372 |
102. Find a four-digit number that, when divided by 131, gives a remainder of 112, and when divided by 132, gives a remainder of 98. | 102. Let $N$ be the desired number. According to the problem, we have:
$$
N=131 k+112=132 l+98
$$
where $k$ and $l$ are positive integers. Since $N$ is a four-digit number, it is clear that:
$$
l=\frac{N-98}{132}<\frac{10000-98}{132} \leqslant 75
$$
Further, we have:
$$
131 k+112=132 l+98 ; \quad 131(k-l)=l-14
$$
... | 1946 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,373 |
103. a) Prove that the sum of all n-digit numbers $(n>2)$ is $494 \underbrace{99 \ldots 9}_{(n-3) \text { times }} 5 \underbrace{00 \ldots 0}_{(n-2) \text { times }}$ (thus, the sum of all three-digit numbers is 494550, and the sum of all six-digit numbers is 494999550000).
b) Find the sum of all four-digit even numbe... | 103. a) The $2n$-digit number given in the condition of the problem can be expressed as follows:
$$
\begin{gathered}
4 \cdot 10^{2n-1} + 9 \cdot 10^{2n-2} + 4 \cdot 10^{2n-3} + 9(10^{2n-4} + 10^{2n-5} + \cdots \\
\ldots + 10^n) + 5 \cdot 10^{n-1} + 5 \cdot 10^{n-2} = 4 \cdot 10^{2n-1} + 9 \cdot 10^{2n-2} + \\
+ 4 \cdo... | 1769580 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,374 |
104. How many and which digits will be needed to write down all integers from 1 to 100000000 inclusive? | 104. Let's first consider all integers from 0 to 99999999; for those numbers that have fewer than eight digits, we will pad them with leading zeros to make them eight-digit numbers. We will thus have 100000000 eight-digit numbers, which will require, obviously, 800000000 digits for their representation. In this case, e... | 68888897 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,375 |
105. All integers are written in a row, starting from one. Determine which digit stands at the $206788-\mathrm{th}$ place.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 105. There are nine single-digit numbers, $99-9=90$ two-digit numbers, $999-99=900$ three-digit numbers, and generally $9 \cdot 10^{n-1}$ $n$-digit numbers.
Single-digit numbers will occupy nine places in the sequence we have written, two-digit numbers $90 \cdot 2=180$ places, three-digit numbers $900 \cdot 3=2700$ pl... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,376 |
106. Will the fraction $0.1234567891011121314 \ldots$ which is obtained by writing all integers in sequence after the zero, be periodic? | 106. Suppose the fraction $0.1234 \ldots$ is periodic, $n$ is the number of digits in the period, and $k$ is the number of digits before the period. Consider the number $N = 10^{m}$, where $m$ is some integer not less than $n+k$; this is a one followed by $m$ zeros. In forming our fraction, we write down all integers i... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,377 |
107. Each of the integers from one to a billion inclusive is replaced by the sum of the digits of the number; single-digit numbers, of course, do not change, while the others decrease. Then each of the newly obtained numbers is again replaced by the sum of its digits, and so on until a billion single-digit numbers are ... | 107. It is a good fact to know that every positive integer $N$ gives the same remainder when divided by 9 as the sum of its digits. (This follows from the fact that the digit $a_{k}$ in the $(k+1)$-th position from the end in the decimal representation of $N$ represents the term $a_{k} \cdot 10^{k}$ in the expansion of... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,378 |
108. Can an integer, whose decimal representation contains
a) some number of sixes and some number of zeros;
b) the digits $1,2,3,4,5,6,7,8$, and 9 (each exactly once), with the digit 5 in the last place,
be a perfect square? | 108. a) It cannot. If the decimal representation of the number $N=n^{2}$ ends with a zero, then it necessarily ends with an even number of zeros, and the number $N_{1}$, obtained from $N$ by removing these zeros at the end, is also a perfect square; therefore, it is sufficient to prove that a number, which is written o... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,379 |
109. On cards, all five-digit numbers from 11111 to 99999 inclusive are written (the number of such numbers is obviously 88,889). Then these cards were laid out in one chain in a random order. Prove that the resulting 444,445-digit number ($444,445 = 88,889 \times 5$) is not a power of two. | 109. To prove the required statement, it is sufficient to check that the 444445-digit number $A$, obtained by writing down 88889 five-digit numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{88889}$ one after another, is divisible by some number other than 2. But $A$ is divisible by 11111. In the case of
; in this case, $a_{0}$ is the number of zeros among the digits of $X$, $a_{1}$ is the number of ones, $a_{2}$ is the number of twos, and so on. Therefore, the sum of all the digit... | 6210001000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,381 |
112. Let $A$ be some natural number. Prove that there are infinitely many natural numbers $N$, written with the digits $1,2, \ldots, 9$ (without zeros!), such that the numbers $N$ and $A N$ have the same sum of digits 1 .[^2] | 112. Let the number $A=\overline{a_{n} a_{n-1} \ldots a_{2} a_{1}}$ be p-digit. We can, of course, assume that $a_{1} \neq 0$, because zeros at the end of the representation of $A$ can simply be discarded - this does not change the sum of the digits of the number $A N$ for any $N$. Now consider the number $N=10^{m}-1=9... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,382 |
116. The distance from point $A$ to point $B$ is 899 km. Along the road connecting $A$ and $B$, there are kilometer markers indicating the distances to $A$ and to $B$ - the inscriptions on them look like this:
$$
0|999,1| 998,2|997, \ldots, 999| 0
$$
How many of these markers have only two different digits? | 116. It is clear that if the numbers on the pole are $\overline{\lambda y z} \mid \overline{x_{1} y_{1} z_{i}}$ (where $x$, $\boldsymbol{y}, \ldots$ are digits), then $\overline{x_{1} y_{1} z_{1}}=999-x y z$ and, therefore, $z_{1}=9-z, y_{1}=$ $=9-y, x_{1}=9-x$. (If $x=9$ or $x=y=9$, then the digits $x_{1}=$ $=0$, or $... | 40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,386 |
118. The highway, along which buses run 24/7, crosses a railway. Two trains pass along the road every hour, reaching the highway precisely at $n$ o'clock and at $n$ o'clock 38 minutes, where $n$ takes all values from 0 to 23; each time a train passes, the barrier, which blocks the highway, is lowered for 5 minutes. It ... | 118. Let's assume that the barrier is closed from 0 min. to 5 min. and from 38 min. to 43 min. every hour. If the specified 5-minute intervals start, for example, 1 minute before the train approaches the barrier, we will simply "set the clock back by 1 minute" (and after solving the problem, to correct the bus movement... | 20,15,12,7\frac{1}{2},5\frac{5}{11} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,388 |
119. What is the greatest possible value of the ratio of a three-digit number to the sum of its digits? | 119. Let \( N = \overline{a b} c \), where \( a, b, c \) are the digits of the number; it is clear that for "round" numbers \( N = 100, 200, \ldots, 900 \) we have \( \frac{N}{a+b+c} = 100 \). Furthermore, if the number \( N \) is not "round," then \( b+c > 0 \) and \( a+b+c \geqslant a+1 \), and since the leading digi... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,389 |
120. Cross out 100 digits from the number
$$
12345678910111213 \ldots 979899100
$$
so that the resulting number is
a) as large as possible;
b) as small as possible. | 120. In this number, there are 192 digits; in the number obtained from it by erasing 100 digits - 92 digits.
a) The first digits of the number we are interested in should be as large as possible. We can make the number start with 5 nines, borrowing these nines from the numbers $9, 19, 29, 37$, and 49; in this case, we... | 999997585960616263646566676869707172737475767778798081828384858687888 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,390 |
121. From all the digits from 1 to 9, form three three-digit numbers whose product is
a) the smallest; b) the largest. | 121. a) The first digits of the three sought numbers must be the smallest; therefore, these numbers in decimal notation have the form:
$$
\overrightarrow{1 A a}, \overrightarrow{2 B b}, \overrightarrow{3 C c},
$$
where the symbol $\overline{x y z}$ denotes the number written with the digits $x, y, z$.
We will prove ... | 147\cdot258\cdot369 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,391 |
122. The sum of several consecutive positive integers is 1000. Find these numbers. | 122. We are given that $m+(m+1)+\ldots+(m+k)=1000$. By the formula for the sum of terms of an arithmetic progression, we have:
$$
\frac{2 m+k}{2} \cdot(k+1)=1000
$$
Or
$$
(2 m+k)(k+1)=2000
$$
Since
$$
(2 m+k)-(k+1)=2 m-1
$$
is odd, one of the last two factors is even, and the other is odd. Moreover, it is clear t... | =1000,k=0;=108,k=4;=28,k=24;=55,k=15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,392 |
123. a) Prove that any number that is not a power of two can be represented as the sum of at least two consecutive positive integers, while such a representation is impossible for powers of two.
b) Prove that any composite odd number can be represented as the sum of at least two consecutive odd numbers, while no prime... | 123. a) Let our number $N$ differ from any power of 2. Then the following equality holds:
$$
N=2^{k}(2 l+1)
$$
where $2^{k}$ is the highest power of 2 that divides $N$ ( $k$ can be zero), and $2 l+1$ is the smallest odd divisor of the number $N$. Further, we have:
$$
\begin{array}{r}
\left(2^{k}-l\right)+\left(2^{k}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,393 |
124. Prove that the product of four consecutive numbers plus one gives a perfect square. | 124. Let's denote these numbers as $n, n+1, n+2, n+3$. The sum of their product and one is given by
$$
\begin{aligned}
& n(n+1)(n+2)(n+3)+1=[n(n+3)][(n+1)(n+2)]+1= \\
& =\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1= \\
& =\left(n^{2}+3 n\right)^{2}+2\left(n^{2}+3 n\right)+1=\left(n^{2}+3 n+1\right)^{2}
\end{aligne... | (n^{2}+3n+1)^{2} | Algebra | proof | Yes | Yes | olympiads | false | 29,394 |
125. There are $4 n$ positive numbers such that from any four pairwise distinct numbers, a geometric proportion can be formed. Prove that among these numbers, there will be $n$ identical ones. | 125. We will prove that these numbers take no more than four different values. Suppose this is false and among our $4 n$ numbers there exist five numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, all pairwise distinct. Let us assume that $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$.
Consider the numbers $a_{1}, a_{2}, a_{3}$, and $a_{4... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,395 |
126. Given 27 weights, the weights of which are equal to $1^{2}, 2^{2}, 3^{2}, \ldots, 27^{2}$ respectively. Divide these weights into three groups of equal weight. | 126. Let's first distribute nine weights, whose weights are equal to $n^{2},(n+1)^{2},(n+2)^{2}, \ldots,(n+8)^{2}$, into three groups as follows:
1st group $n^{2},(n+5)^{2},(n+7)^{2}$
$$
n^{2}+(n+5)^{2}+(n+7)^{2}=3 n^{2}+24 n+74
$$
2nd group $(n+1)^{2},(n+3)^{2},(n+8)^{2}$ :
$$
(n+1)^{2}+(n+3)^{2}+(n+8)^{2}=3 n^{2}... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,396 |
127. There are 13 weights, each weighing a whole number of grams. It is known that any 12 of them can be distributed on the pans of a balance, six on each pan, so that equilibrium is achieved. Prove that all the weights have the same weight. | 127. From the condition of the problem, it follows first of all that all weights are either even or odd. Indeed, from the fact that any 12 weights can be divided into two groups of equal weight, it follows that the total weight of any 12 weights is even. In this case, the total weight of 12 weights will remain even eve... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,397 |
128*. Four (arbitrary) numbers $a, b, c, d$ are written down in a row. Below these numbers, a new set of four numbers $a_{1}=a b, b_{1}=b c, c_{1}=c d$, $d_{1}=d a$ is written; below them, the numbers $a_{2}=a_{1} b_{1}, b_{2}=b_{1} c_{1}$, $c_{2}=c_{1} d_{1}, d_{2}=d_{1} a_{1}$, and so on. Prove that either all the re... | 128. First of all, note that if for some natural numbers $k$ and $l \neq k$ the quadruples $a_{k}, b_{k}, c_{k}, d_{k}$ and $a_{l}, b_{l}, c_{l}, d_{l}$ obtained in this way are the same, then either all numbers $a_{k}=a_{l}, b_{k}=b_{l}, c_{k}=$ $=c_{l}, d_{k}=d_{l}$ are zero, or all of them are positive. Indeed, if a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,398 |
129. Given an arbitrary set of $N$ numbers (where $N$ is a power of two: $N=2^{k}$) $a_{1}, a_{2}, \ldots, a_{N}$, each of which is either +1 or -1. From these numbers, a new set of numbers, each also equal to +1 or -1, is formed as follows: $\quad a_{1}^{\prime}=a_{1} a_{2}, \quad a_{2}^{\prime}=$ $=a_{2} a_{3}, \ldot... | 129. Since the square of each of the numbers $a_{i}$ (where $i=1,2$, .., $N=2^{k}$) is equal to one, the first three sequences of numbers have the following form:
$$
\begin{aligned}
& \begin{array}{rrrrr}
a_{1}, & a_{2}, & \ldots & a_{N-1}, & a_{N} ; \\
a_{1} a_{2}, & a_{2} a_{3}, & \ldots & a_{N-1} a_{N}, & a_{N} a_{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,399 |
130*. Let $a_{1}, a_{2}, \ldots, a_{n}$, where $n>2$, be integers; from these numbers, a new system of numbers is formed: $a_{1}^{\prime}=\frac{a_{1}+a_{2}}{2}, \quad a_{2}^{\prime}=\frac{a_{2}+a_{3}}{2}, \ldots, \quad a_{n-1}^{\prime}=\frac{a_{n-1}+a_{n}}{2}, a_{n}^{\prime}=$ $=\frac{a_{n}+a_{1}}{2} ;$ from the number... | 130. Let us first show that when transitioning from the original set of numbers $a_{1}, a_{2}, \ldots, a_{n}$ to the "derived" set $a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{n}^{\prime}$, our numbers "level out", i.e., the difference between the largest and smallest numbers does not increase. Indeed, since the half-su... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,400 |
131. Let $x=1, y, z$ be any three numbers; $x_{1}$, $y_{1}, z_{1}$ be the absolute values $|x-y|,|y-z|,|z-x|$ of the pairwise differences of these numbers; $x_{2}, y_{2}, z_{2}$ be the absolute values of the pairwise differences of the numbers $x_{1}, y_{1}, z_{1}$ (i.e., the numbers $\left.\left|x_{1}-y_{1}\right|,\le... | 131. First of all, let's determine for which numbers \(x, y, z\) the triples \((x_n, y_n, z_n)\) and \((x, y, z)\) can coincide. Since all triples of numbers starting from \(x_1, y_1, z_1\) are non-negative, the numbers \(x\), \(y\), and \(z\) must also be non-negative. Let's agree to consider that \(x \geqslant y \geq... | (1,1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,401 |
133*. a) Arrange the numbers from 1 to 100 in such a way that no 11 (not necessarily consecutive!) of these numbers follow one another in ascending or descending order.
b) Prove that no matter how the numbers from 1 to 101 are arranged, it is always possible to select 11 (not necessarily consecutive) numbers that foll... | 133. a) It is not hard to see that the following arrangement of the first 100 integers satisfies the condition of the problem:
| 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 20 | 19 | 18 | 17 | 16 | 15 | 14 | 13 | 12 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,403 |
134. a) From the first 200 integers from 1 to 200, 101 numbers are selected. Prove that among the selected numbers, there will be a pair such that one divides the other.
b) Select 100 numbers from the first 200 integers such that no one of them divides another.
c) Prove that if at least one of 100 integers, not excee... | 134. a) First solution. Consider the greatest odd divisors of the 101 chosen numbers, i.e., the numbers obtained by dividing each number by the highest power of two that it contains as a factor. Since there are only 100 different odd numbers not exceeding 200, there will be two identical greatest odd divisors among the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,404 |
136. Given a non-increasing sequence $a_{1}$, $a_{2}, \ldots, a_{n}$ of positive numbers, the sum of which is 1, and the largest of which is $\frac{1}{2 k}$, where $k$ is an integer:
$\cdot \frac{1}{2 k}=a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant \ldots \geqslant a_{n}>0, a_{1}+a_{2}+\ldots+a_{n}=1$.
Prove that ... | 136. Since the sum of all our numbers is 1, and the largest of them is $\frac{1}{2 k}$, the total number of numbers is not less than $2 k$. We will prove our statement by contradiction, i.e., we will assume that in any group of $k$ numbers of our sequence, the smallest does not exceed half of the largest, and show that... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,406 |
137. Around a circle, $p$ crosses and $q$ zeros are written; the number of pairs of adjacent crosses is denoted by $a$, and the number of pairs of adjacent zeros - by $b$. Prove that $a-b=p-q$. | 137. We will traverse our circle, for example, clockwise, starting from zero, until we return to the same zero. During this traversal, the number of transitions from zero (to zero or to a cross) will be as many as there are zeros, i.e., q. Among these transitions, the transitions to zero will be as many as the pairs of... | p-q | Combinatorics | proof | Yes | Yes | olympiads | false | 29,407 |
138. Let $i_{1}, i_{2}, \ldots, i_{n}$ be the numbers $1,2, \ldots, n$, but arranged in some, generally new, order. Prove that when $n$ is even, the product $\left(1-i_{1}\right)\left(2-i_{2}\right)\left(3-i_{3}\right) \ldots\left(n-i_{n}\right)$ can be either even or odd; however, when $n$ is odd, it is necessarily ev... | 138. It is clear that if any number $i_{k}=k$, then the considered product is zero (even), and if we swap every two adjacent numbers in the sequence $1,2,3,4, \ldots, 2 m-1,2 m$, arriving at the sequence $i_{1}=2, i_{2}=1, i_{3}=4, i_{4}=3, \ldots, i_{2 m-1}=2 m$, $i_{2 m}=2 m-1$, then the product
$$
\begin{aligned}
&... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,408 |
139. Given $n$ numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$, each of which is equal to +1 or -1. Prove that if $x_{1} x_{2} + x_{2} x_{3} + \ldots + x_{n-1} x_{n} + x_{n} x_{1} = 0$, then $n$ is divisible by 4. | 139. It is clear that since each of the products $\boldsymbol{x}_{1} x_{2}, \boldsymbol{x}_{2} x_{2}, \ldots$, $x_{n} x_{1}$ equals +1 or -1, the sum of all these products can equal zero only if the number of terms in it $n=2 m$ is even and $m$ of these terms are +1, and $m$ others are -1. But if -1 equals exactly $m$ ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,409 |
142. We have two sets of characters «十» and «-», each containing 1958 characters. Prove that after a certain number of "steps", where each step allows us to change any 11 characters of the 1st set, the 1st set can be transformed into the 2nd set. (Sets are considered the same if they have the same characters in the sam... | 142. It is clear that it is sufficient to ensure the possibility of changing any sign of the 1st set with the help of our "steps" without changing any of the other signs: for then, by sequentially changing all the signs of the 1st set that differ from the signs in the same positions of the 2nd set, we will transform th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,411 |
143*. A chess player, for training, plays no less than one game per day; at the same time, to avoid overexertion, he plays no more than 12 games in a week. Prove that it is possible to find several such consecutive days during which he will play exactly 20 games. | 143. Suppose that on some Monday a chess player played $a_{1}$ games, on Monday and Tuesday $-a_{2}$ games, over three days -- $a_{3}$ games, and so on, finally, over 77 days -- $a_{77}$ games. Now consider the following set of numbers: $a_{1}, a_{2}, a_{3}, \ldots, a_{77}; a_{1}+20$, $a_{2}+20, a_{3}+20, \ldots, a_{77... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,412 |
144. Let $N$ be an arbitrary positive integer. Prove that there exists an integer divisible by $N$ which, in the decimal system, is written using only the digits 0 and 1. Moreover, if $N$ is coprime with 10 (i.e., does not divide by 2 or 5), then there exists a number divisible by $N$ composed entirely of ones (if $N$ ... | 144. First solution. Consider the remainders of the division of the numbers
$$
1,11,111, \ldots, \underbrace{1111 \ldots 1}_{N \text { ones }}
$$
by $N$. Since there are $N$ such numbers, and there can be only $N-1$ different non-zero remainders when dividing by $N$, if none of these numbers is divisible by $N$ (whic... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,413 |
145. Arrange a system of non-overlapping (having no common internal points or common ends) segments of length 1 on a number line so that for any (infinite) arithmetic progression (with any first term and any difference l), at least one of its terms falls within one of the segments of our system. | 145. Let \(a_{N}\) and \(a_{N+1}=a_{N}+d\) be two consecutive terms of our progression; then the distance between the corresponding points \(A_{N}\) and \(A_{N+1}\) on the number line is equal to the difference \(d\) of the progression. Let \(d>0\); if the number \(d\) is not an integer, then denote by \(\alpha=\{d\}=d... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,414 |
146. Let \( m \) and \( n \) be two coprime positive integers. Prove that if the fractions:
\[
\begin{aligned}
& \frac{m+n}{m}, \frac{2(m+n)}{m}, \frac{3(m+n)}{m}, \ldots, \frac{(m-1)(m+n)}{m} \\
& \frac{m+n}{n}, \frac{2(m+n)}{n}, \frac{3(m+n)}{n}, \ldots, \frac{(n-1)(m+n)}{n}
\end{aligned}
\]
are represented by poin... | 146. First of all, it is obvious that none of the considered fractions is equal to an integer: indeed, if, for example, some fraction $\frac{k(m+n)}{m}$ (where $k$ is one of the numbers $1,2,3, \ldots, m-1$) were an integer, then $m+n$ would have to have common divisors with the number $m$ (since $k < m$ and cannot be ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,415 |
147*. Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be some positive integers such that each of these numbers is less than 1000, and the least common multiple of any two numbers is greater than 1000. Prove that the sum of the reciprocals of the numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ is less than two. | 147. First solution. If the number $a$ is in the interval $\frac{1000}{m} \geqslant a>\frac{1000}{m+1}, \quad$ then there are, obviously, $m$ integers not exceeding $10 \div 0$, multiples of $a$ (namely $a, 2 a, 3 a, \ldots$, ma). Therefore, if we denote by $k_{1}$ the number of our numbers that are between 1000 and $\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,416 |
148*. The fraction $\frac{q}{p}$ with an odd prime denominator $p \neq 5$ is expanded into an infinite periodic decimal fraction. Prove that if the number of digits in the period of the fraction is even, then the arithmetic mean of all the digits in the period is 4.5 (i.e., it coincides with the arithmetic mean of all ... | 148. Let's consider how a period of a simple fraction $\frac{q}{\rho}$ forms a periodic decimal fraction
$$
\frac{q}{p}=\overline{A, a_{1} a_{2} \ldots a_{k} a_{1} \alpha_{2} \ldots a_{k} a_{1} a_{2} \ldots}
$$
(here $A$ is an integer, and $a_{1}, a_{2}, \ldots, a_{k}$ are the digits of the fraction's period). Clearl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,417 |
149*. Prove that if the fractions
$$
\frac{a_{1}}{p}, \quad \frac{a_{2}}{p^{2}}, \quad \frac{a_{3}}{p^{3}}, \ldots, \frac{a_{n}}{p^{n}}, \ldots
$$
( $p$ - a prime number different from 2 and 5, $a_{1}, a_{2}, \ldots, a_{n} \ldots$ are any integers coprime with $p$) are expanded into infinite periodic decimal fraction... | 149. The number of digits in the periods of the fractions $\frac{a_{n}}{p^{n}}$ and $\frac{a_{n+1}}{p^{n+1}}$ is equal to the smallest positive integers $k$ and $l$, such that $10^{k}-1$ is divisible by $p^{n}$, and $10^{l}-1$ is divisible by $p^{n+1}$ (see the solution to the previous problem). Let us now form the dif... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,418 |
150. a) Find a four-digit number that is a perfect square and such that its first two digits are equal to each other and the last two digits are also equal to each other.
b) A two-digit number, when added to the number formed by the same digits in reverse order, results in a perfect square. Find all such numbers. | 150. a) Let $a$ be the first and $b$ the last digit of the number $N$ we are looking for. Then this number is equal to $1000 a + 100 a + 10 b + b = 1100 a + 11 b = 11(100 a + b)$. Since the number $N$ is a perfect square, it follows from its divisibility by 11 that it is also divisible by 121, i.e., $\frac{N}{11} = 100... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,419 |
151. Find a four-digit number that is equal to the square of the sum of two two-digit numbers formed by the first two and the last two digits of the number. | 151. Let the two-digit number formed by the first two digits of the desired number $N$ be denoted by $a$, and the number formed by its last two digits be denoted by $b$. In this case, $N=100a + b$, so the condition of the problem gives
$$
100a + b = (a + b)^2
$$
or
$$
99a = (a + b)^2 - (a + b) = (a + b)(a + b - 1)
$... | 9801,3025,2025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,420 |
152. Find all four-digit numbers that are perfect squares and can be written
a). with four even digits;
b) with four odd digits. | 152. a) A four-digit number, written with four even digits, can start with the digits 2, 4, 6, or 8; in other words, it is between 1999 and 3000, or between 3999 and 5000, or between 5999 and 7000, or between 7999 and 9000. Therefore, the square root of this number lies between 44 and 55, or between 63 and 71, or betwe... | 4624,6084,6400,8464 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,421 |
153. a) Find all three-digit numbers that are equal to the sum of the factorials of their digits.
b) Find all integers that are equal to the sum of the squares of their digits. | 153. a) Let the hundreds, tens, and units digits of the desired number \( N \) be denoted by \( x, y \), and \( z \), respectively, so that \( N = 100x + 10y + z \). In this case, the condition of the problem gives
\[
100x + 10y + z = x! + y! + z!
\]
Note that \( 7! = 5040 \) is a four-digit number; therefore, no dig... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,422 |
155. Solve the equations in integers:
a) $1!+2!+3!+\ldots+x!=y^{2}$;
b) $1!+2!+3!+\ldots+x!=\nu^{2}$. | 155. a) By direct verification, we find that for $x<5$, the only solutions to our equation are $x=1$, $y= \pm 1$ and $x=3, y= \pm 3$. Now let's prove that there are no solutions for $x \geqslant 5$. Notice that $1!+2!+3!+4!=33$ ends in the digit 3, and $5!, 6!, 7!, \ldots$ all end in zero. Thus, for $x \geqslant 5$, th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,424 |
156. In how many ways can $2^{n}$ be expressed as the sum of four squares of positive integers? | 156. Let
$$
a^{2}+b^{2}+c^{2}+d^{2}=2^{n}
$$
Denote the highest power of two by which all four numbers $a, b, c$ and $d$ are divisible as $p$. By dividing both sides of our equation by $2^{2 p}$, we get:
$$
a_{1}^{2}+b_{1}^{2}+c_{1}^{2}+d_{1}^{2}=2^{n-2 p}
$$
where at least one of the four numbers $a_{1}, b_{1}, c_... | 2^{2p}=(2^{p-1})^{2}+(2^{p-1})^{2}+(2^{p-1})^{2}+(2^{p-1})^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,425 |
157. a) Prove that the equality
$$
x^{2}+y^{2}+z^{2}=2 x y z
$$
is true for integers $x, y, z$ only when $x \Rightarrow y=z=0$.
b) Find integers $x, y, z, v$, such that
$$
x^{2}+y^{2}+z^{2}+v^{2}=2 x y z v
$$ | 157. a) First solution. The equation
$$
x^{2}+y^{2}+z^{2}=2 x y z
$$
is satisfied by the values $x=0, y=0, z=0$. In this case, if one of the numbers $x, y, z$ is equal to 0, then the others must also be equal to 0, because the sum of their squares in this case is 0.
Now let all the numbers $x, y, z$, satisfying our ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,426 |
158*. a) For which integer values of $k$ is the equality possible
$$
x^{2}+y^{2}+z^{2}=k x y z
$$
where $x, y, z$ are positive integers?
b) Find all possible triples of integers within the first thousand, the sum of the squares of which is divisible by their product. | 158. a) Let \(x, y, z\) be some three positive integers satisfying the equation
\[
x^{2} + y^{2} + z^{2} = k x y z
\]
First, we will show that it is always possible to assume that the inequalities
\[
x \leqslant \frac{k y z}{2}, \quad y \leqslant \frac{k x z}{2}, \quad z \leqslant \frac{k x y}{2}
\]
hold (i.e., tha... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,427 |
160. Solve the equation in integers
$$
x^{2}+x=y^{4}+y^{3}+y^{2}+y
$$ | 160. Let's complete the left side of the equation to a perfect square, for which it is sufficient to multiply both sides of the equation by 4 and add 1 to each:
$$
(2 x+1)^{2}=4 y^{4}+4 y^{3}+4 y^{2}+4 y+1
$$
But
$$
\begin{aligned}
4 y^{4}+4 y^{3}+4 y^{2} & +4 y+1=\left(4 y^{4}+4 y^{3}+y^{2}\right)+\left(3 y^{2}+4 y... | (0,-1),(-1,-1);(0,0),(-1,0);(5,2),(-6,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,428 |
161. Solve the equation in positive integers
$$
x^{2 y}+(x+1)^{2 y}=(x+2)^{2 y}
$$ | 161. For $y=1$ we arrive at the quadratic equation
$$
x^{2}+(x+1)^{2}=(x+2)^{2}, \text { or } x^{2}-2 x-3=0
$$
which has a single positive integer root $x=3$ (the second root of the equation $x=-1$ is negative). Let now $y>1$. Note that since $x^{2 y}$ and $(x+2)^{2 y}$ are numbers of the same parity, the number $(x+... | 1,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,429 |
163*. Prove that the equation $x^{2}+x+1=p y$ has solutions in integers $x, y$ for infinitely many prime values of the coefficient $p$. | 163. Let's use the method of proof by contradiction: suppose that the equation given in the problem has solutions in integers $x$ and $y$ only for a finite number of prime values $p$, and that the largest of these is the $n$-th prime number $p_{n}$. Now, let's form the number $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,431 |
164*. Find four whole (positive) numbers such that the square of each of them, added to the sum of the other three, is also a perfect square.
Find four whole (positive) numbers such that the square of each of them, added to the sum of the other three, is also a perfect square. | 164. The task is to solve the following system of equations in positive integers ($x, y, z$ and $u$ are the unknowns):
$$
\begin{aligned}
& x^{2}+y+z+u=(x+v)^{2} \\
& y^{2}+x+z+u=(y+w)^{2} \\
& z^{2}+x+y+u=(z+t)^{2} \\
& u^{2}+x+y+z=(u+s)^{2}
\end{aligned}
$$
or
$$
\left.\begin{array}{l}
y+z+u=2 v x+v^{2} \\
x+z+u=2... | 96,57,u=40;11,u=6;k(3k\2),u=1;u=1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,432 |
165. Find all pairs of integers, the sum of which is equal to their product. | 165. Denoting the required numbers by $x$ and $y$, we have:
$$
x+y=x y
$$
or
$$
\begin{aligned}
& x y-x-y+1=1 \\
& (x-1)(y-1)=1
\end{aligned}
$$
Since 1 can only be factored into integer factors in two ways, we immediately obtain:
$$
x-1=1, y-1=1 ; x=2, y=2
$$
or
$$
x-1=-1, y-1=-1 ; x=0, y=0
$$ | 2,2or0,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,433 |
166. The sum of the reciprocals of three positive integers is equal to one. What are these numbers? | 166. First, we establish that at least one of the three numbers $x, y$, and $z$, such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
must be less than 4: if all these numbers were $\geqslant 4$, then the sum $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ would be $\leqslant \frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$. ... | \frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1;\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1;\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,434 |
167. a) Prove that the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$, where $n>1$ is a natural number, has exactly three solutions in natural numbers $x, y$ (where solutions $x=a, y=b$ and $x=b, y=a$ for $a \neq b$ are considered different) when $n$ is prime, and more than three solutions when $n$ is composite.
b) Fi... | 167. a) From the equation, it is clear that $x, y > n$; let us set $x = n + x_1, y = n + y_1$. In this case, our equation transforms as follows:
$$
\begin{aligned}
\frac{1}{n + x_1} + \frac{1}{n + y_1} = \frac{1}{n}, & \text{ or } (n^2 + n x_1) + (n^2 + n y_1) = \\
& n^2 + n x_1 + n y_1 + x_1 y_1, \text{ i.e., } x_1 y... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,435 |
168. a) Find all positive, distinct integers \(x, y\) that satisfy the equation
\[
x^{y}=y^{x}
\]
b) Find all positive, distinct rational numbers \(x, y\) that satisfy the equation
\[
x^{y} \doteq y^{x}
\]
'(provide a formula that gives all solutions). | 168. a) From the equality $x^{y}=y^{x}$, it is clear that the prime divisors of the numbers $x$ and $y$ are the same:
$$
x=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{n}^{\alpha_{n}}, y=p_{1}^{\beta_{1}} p_{2}^{R_{2}} \ldots p_{n}^{\beta_{n}}
$$
where $p_{1}, p_{2}, \ldots, p_{n}$ are prime numbers. Then, due to ... | 2,4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,436 |
169. In a chess tournament, two 7th-grade students and a certain number of 8th-grade students participated. Each participant played one game with every other participant. The two 7th-graders together scored 8 points, and all the 8th-graders scored the same number of points (in the tournament, each participant earns 1 p... | 169. Let $n$ be the number of eighth-graders, and $m$ be the number of points earned by each of them. In this case, the number of points scored by all participants in the tournament is $m n + 8$. This number is equal to the number of games played. Since the number of participants in the tournament is $n + 2$ and each p... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,437 |
170. In a chess tournament, students from the 9th and 10th grades participated. Each participant played against every other participant once. There were 10 times more 10th graders than 9th graders, and they scored 4.5 times more points in total than all the 9th graders. How many 9th grade students participated in the t... | 170. Let the number of ninth graders be $n$ and they scored $m$ points. Then the number of tenth graders was $10 n$ and they scored $4.5 m$ points. The total number of participants in the tournament was $11 n$ and they scored 5.5 m points.
The total number of points scored by everyone is equal to the number of games p... | n=1,=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,438 |
173. A square table of $n^{2}$ cells is filled with integers from 1 to $n$ such that each row and each column of the table contains all numbers from 1 to $n$. Prove that if the table is symmetric with respect to the diagonal leading from its upper left corner to the lower right corner, and $n$ is odd, then all numbers ... | 173. Let $n$ be odd. If our table is symmetric with respect to the diagonal specified in the condition (we will call this diagonal "main" and denote it by the letter $d$), then each number located above $d$ will correspond to an equal number located below $d$: this number occupies a position symmetric to the position o... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,441 |
174. $n^{2}$ integers from 1 to $n^{2}$ are written in a square table of size $n \times n$: in this case, the number 1 is placed at any position in the table; 2 belongs to the row whose ordinal number is equal to the column number containing 1; the number 3 belongs to the row whose number matches the column number cont... | 174. Let us agree to denote the number at the intersection of the $i$-th row and $j$-th column of the table by $a_{ij}$; in this case $a_{ij} = 1, \ldots, n^2$, where $i, j = 1, \ldots, n$. Let now $1 = a_{ji}$; then according to the problem's condition, $2 = a_{j_2 j_2}, \quad 3 = a_{j_3 j_4}$, and so on, up to $n^2 =... | n^2-n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,442 |
181*. In all cells of an $n \times n$ chessboard, except one, there are signs «+», and in one cell there is a sign «-»; moreover,
a) $n=4$ and the sign «-» is on the edge of the board, but not in its corner;
b) $n=8$ and the sign «-» is not in the corner of the board.
It is allowed to simultaneously change all signs ... | 181. a) On a board of size $4 \times 4$, there are 4 columns, 4 rows, and $2 \cdot 7=14$ parallel diagonal rows (of which 4 "rows" contain only one corner cell); it is easy to verify that each of these $4+4+14=22$ vertical, horizontal, or slanted rows contains an even number (either 0 or 2) of the shaded cells shown in... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 29,448 |
182. a) In all cells of an $8 \times 8$ chessboard, there are signs «+» or «-». It is allowed to select any $3 \times 3$ or $4 \times 4$ square on the board and change the signs in all cells of this square. We want to achieve, using such operations, that all signs on the board become «+»; is this always possible?
b) I... | 182. a) First, let's calculate the number of all possible squares of size $3 \times 3$ and size $4 \times 4$ that can be placed on
a)
 There are three piles of stones; a boy playing with them is allowed to take one stone simultaneously from all three piles or to double the number of stones in one (any) pile. Will the boy be able to remove all the stones from the piles after several such "moves"?
b) In all cells of a rectangular table with 8 r... | 183. a) He can. It is clear that by choosing one stone from each pile, the boy can achieve that at least one pile will have a single stone left. Then, by doubling the number of stones in that pile (or those piles) where there is 1 stone, and again subtracting one stone from each pile, the boy will reduce by one the num... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 29,450 |
184. A table with two columns of positive integers and some number of rows is written according to the following rule. In the top row, we write arbitrary numbers $a$ and $b$; then under $a$ we write (a positive integer) number $a_{1}$, equal to $\frac{a}{2}$ if $a$ is even, and equal to $\frac{a-1}{2}$ if $a$ is odd, a... | 184. Let us write the number \( a \) in "binary notation", i.e., in the form
$$
a=\alpha_{n} \cdot 2^{n}+\alpha_{n-1} \cdot 2^{n-1}+\alpha_{n-2} \cdot 2^{n-2}+\ldots+\alpha_{1} \cdot 2+\alpha_{0} \cdot 1
$$
where all "digits" \(\alpha_{0}, \alpha_{1}, \ldots, \alpha_{n-2}, \alpha_{n-1}, \alpha_{n}\) are equal to 0 or... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,451 |
185. Prove that every natural number is either a Fibonacci number (a member of the Fibonacci sequence ( ${ }^{*}$ ), see p. 37), or can be represented as the sum of several (distinct) Fibonacci numbers. | 185. The shortest solution to this problem is provided by the method of mathematical induction. Let us agree to denote the $k$-th Fibonacci number, where $k=1,2,3, \ldots$, by $u_{k}$. Suppose that our statement has already been proven for all natural numbers $n$ less than the $k$-th Fibonacci number $u_{k}$; the valid... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,452 |
187. Prove that if a natural number $n$ is divisible by 5, then the $n$-th Fibonacci number $u_{n}$ (see problem 185) is also divisible by 5. | 187. Let us agree to denote the remainders of the division of Fibonacci numbers $u_{1}, u_{2}, u_{3}, \ldots$ by 5 through $\alpha_{1}, \alpha_{2}, \alpha_{3}, \ldots$ It is clear that from the equality $u_{k}=$ $=u_{k-1}+u_{k-2}$, where $k=3,4,5, \ldots$, it follows that
$$
\alpha_{k}= \begin{cases}\alpha_{k-1}+\alph... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,454 |
188. Among the first 100000001 Fibonacci numbers (see problem 185), is there a number that ends with four zeros? | 188. Let's leave in each member of the Fibonacci series, written with five or more digits, only the last four digits. We will get a sequence of numbers, each of which is less than $10^{4}$. Let's denote by $a_{k}$ the member of this sequence standing at the $k$-th place. Note that if we know $a_{k+1}$ and $a_{k}$, then... | 7501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,455 |
190. The sequence of numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ is such that
$a_{1}=0,\left|a_{2}\right|=\left|a_{1}+1\right|,\left|a_{3}\right|=\left|a_{2}+1\right|, \ldots,\left|a_{n}\right|=\left|a_{n-1}+1\right|$. Prove that the arithmetic mean $\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}$ of all the numbers is not less ... | 190. First solution. Let's add one more number $a_{n+1}$ to our sequence such that $\left|a_{n+1}\right|=\left|a_{n}+1\right|$, and square all the given equalities in the problem (including the "additional" equality connecting $a_{n}$ and $a_{n+1}$):
$$
\begin{aligned}
& a_{1}^{2}=0, \quad a_{2}^{2}=\left(a_{1}+1\righ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,457 |
191*. The sequence of natural numbers $a_{0}, a_{1}$, $a_{2}, a_{3}, \ldots$ is formed according to the following rule:
$$
a_{0}, a_{1}, a_{2}=\left|a_{0}-a_{1}\right|, a_{3}=\left|a_{1}-a_{2}\right|, \ldots
$$
(and generally $a_{n}=\left|a_{n-2}-a_{n-1}\right|$ for all $n \geqslant 2$); the sequence continues until ... | 191. It is not difficult to understand that the largest number in our sequence can only be one of the first two numbers $a_{0}$ and $a_{1}$ (since each $a_{k}$ for $k \geqslant 2$ is necessarily $a_{2}$, the sequence can be "extended backwards" without changing it, by agreeing to start as follows: $a_{0}=a_{1}-a_{2}, \... | 1967 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,458 |
192. An infinite sequence of digits $\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}, \ldots$ is written consecutively, where $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \ldots$ denotes any digit except nine. Prove that among the numbers $\alpha_{1}, \overline{\alpha_{1} \alpha_{2}}, \overline{\alpha_{1} \alpha_{2} \a... | 192. We will prove our statement by contradiction. Suppose that in the sequence of numbers $\alpha_{1}, \overline{\alpha_{1} \alpha_{2}}, \overline{\alpha_{1} \alpha_{2} \alpha_{3}}, \ldots$ there are only a finite number of composite numbers. Clearly, in this case, the sequence of digits $\alpha_{1}, \alpha_{2}, \alph... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,459 |
193. In the sequence $1975 \ldots$ each digit, starting from the fifth, is equal to the last digit of the sum of the preceding four digits. Will the sequence contain
a) the digits 1234 in a row; b) the four digits 1975 a second time? | 193. a) The evenness or oddness of the last digit of the sum of four digits (in other words, of the sum itself) depends only on the evenness or oddness of the considered digits. If we denote an odd number by the letter n and an even number by the letter c, it is easy to see that our sequence starts as follows:
## nnnn... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,460 |
194. All numbers divisible by nine are written in a row:
$\left.9,18,27,36,45,54,63,72,81,90,99,108,117, \ldots \quad(*)^{*}\right)$ and for each of these numbers, the sum of its digits is found:
$$
9,9,9,9,9,9,9,9,9,9,18,9,9, \ldots
$$
At which position in the sequence (**) will the number 81 first appear, and what... | 194. The answer to the first question of the problem is quite simple: it is clear that among 8-digit and even smaller numbers, we cannot find one whose sum of digits equals $9 \cdot 9=81$, while among 9-digit numbers, such a number is 999999999; accordingly, in the sequence (**) the number 81 will first appear at the $... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,461 |
195. Let's consider the following sequence of sets of (natural) numbers. The first set $I_{0}$ consists of
two ones 1,1. Then we insert their sum $1+1=2$ between these numbers; we get the set $I_{1}: 1,2,1$. Next, between every two numbers in the set $I_{1}$, we insert the sum of these numbers; we get the set $I_{2}: 1... | 195. First of all, note that each of the sets $I_{0}, I_{1}$, $I_{2}, \ldots$ is obtained from the preceding one by adding a certain number of new numbers; at the same time, all the numbers that were previously present are retained in the new set. It is easy to see, furthermore, that all the new numbers appearing in th... | 1972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,462 |
196. There is a (finite) sequence of zeros and ones such that all quintets of consecutive digits that can be extracted from it are distinct (and these quintets can, of course, overlap like the quintets 01011 and 01101, arranged as follows: ...0101101...). Prove that if the sequence cannot be continued while preserving ... | 196. Let $\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}$ (where all $\alpha_{i}, i=1,2,3,4$ are digits 0 or 1) be the last four digits of the sequence. If the sequence did not contain the digits $\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4} 0$ anywhere, it could be continued by appending 0 at the end; similarly, if the se... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,463 |
197. In one line, all divisors of the number $N=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37$ are written. Under the number 1 and those numbers that represent the product of an even number of prime factors, the number +1 is written, and under the numbers that represe... | 197. The number $N$ represents the product of all prime numbers from 2 to 37 inclusive; each divisor of the number $N$ represents the product of some of these numbers. Let's show that the statement of the problem is true for any number $N_{k}=$ $=2 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot p_{k}$ - the product of the ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,464 |
198. Let $p$ and $q$ be two coprime natural numbers. A natural number $n$ will be called "good" if it can be represented as $p x + q y$, where $x$ and $y$ are non-negative integers, and "bad" otherwise.
a) Prove that there exists a number $A$ such that if the sum of two integers is equal to $A$, then one of them is ne... | 198. From the fact that $p$ and $q$ are coprime, using the Euclidean algorithm, it is established that any integer $n$ can be represented in the form:
$$
n=p x+q y, \text { where } x \text { and } y \text { are integers. }
$$
Indeed, let $p>q$; then $p=q d+r$, where $00$ (since the number $n$ is positive). On the pla... | Number Theory | proof | Yes | Yes | olympiads | false | 29,465 | |
199. Prove that if a number $n$ is non-negative, then it can (and in a unique way!) be represented in the form $n=\frac{(x+y)^{2}+3 x+y}{2}$, where $x$ and $y$ are non-negative integers. | 199. First solution. We need to prove that each non-negative integer $n$ can be uniquely represented in the form:
$$
\begin{aligned}
n=\frac{(x+y)^{2}+3 x+y}{2}= & \frac{(x+y)^{2}+(x+y)+2 x}{2}= \\
& =\frac{(x+y)^{2}+(x+y)}{2}+x=\frac{X(X+1)}{2}+x
\end{aligned}
$$
where $X=x+y$ and, therefore, $X \geqslant 0$ and $0 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,466 |
200. Let $t$ be an arbitrary positive number; the number of irreducible fractions $\frac{p}{q}$, where the numerator $p$ and the denominator $q$ do not exceed $t$, is denoted by $d(t)$,
## What is the sum
$-S=d\left(\frac{100}{1}\right)+d\left(\frac{100}{2}\right)+d\left(\frac{100}{3}\right)+\ldots+d\left(\frac{100}{... | 200. To each irreducible fraction $\frac{p}{q}$, where
$$
0<p \leqslant 100, \quad 0<q \leqslant 100
$$
we associate a point $M$ with coordinates $p$, $q$; the inequalities ( ${ }^{*}$ ) indicate that the point $M$ is located inside the shaded square $\mathscr{K}=O A C B$ bounded by the coordinate axes and the lines ... | 10000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,467 |
201. Prove the following properties of the integer part of a number (see p. 37):
1) $[x+y] \geqslant[x]+[y]$
2) $\left[\frac{[x]}{n}\right]=\left[\frac{x}{n}\right], \quad n$-integer
3) $\left[x+\frac{1}{2}\right]=[2 x]-[x]$
4) $[x]+\left[x+\frac{1}{n}\right]+\left[x+\frac{2}{n}\right]+\ldots+\left[x+\frac{n-1}{n}\righ... | 201. 202) Let $x$ be any real number. Then it can be represented as the sum $x=[x]+\alpha$, where $\alpha$ is a non-negative number less than one. Now let's represent $y$ as $y=[y]+\beta(0 \leqslant \beta<1)$. Then $x+y=[x]+[y]+\alpha+\beta$. Since $\alpha+\beta \geqslant 0$, it is clear from this equality that $[x]+[y... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,468 |
202. Simplify the sum
$$
\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\left[\frac{n+4}{8}\right]+\ldots+\left[\frac{n+2^{k}}{2^{k+1}}\right]+\ldots
$$
where $n$ is a positive integer. | 202. First solution. In the result of problem 201 3), the sum of interest to us (which, obviously, contains only a finite number of non-zero terms) is equal to
$$
\begin{aligned}
& {\left[\frac{n}{2}+\frac{1}{2}\right]+\left[\frac{n}{4}+\frac{1}{2}\right]+\left[\frac{n}{8}+\frac{1}{2}\right]+\ldots=} \\
& =\left(\lflo... | n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,469 |
203*. Prove that if $p$ and $q$ are coprime integers, then
\[
\begin{aligned}
& {\left[\frac{p}{q}\right]+\left[\frac{2 p}{q}\right]+\left[\frac{3 p}{q}\right]+\ldots+\left[\frac{(q-1) p}{q}\right]=} \\
& =\left[\frac{q}{p}\right]+\left[\frac{2 q}{p}\right]+\left[\frac{3 q}{p}\right]+\ldots+\left[\frac{(p-1) q}{p}\rig... | 203. Let us mark all points on the plane with both integer coordinates ("integer" points), for which $1 \leqslant x \leqslant q-1, 1 \leqslant y \leqslant p-1$ ($x$ and $y$-coordinates of the points). These points lie inside the rectangle $O A B C$, the lengths of whose sides are $O A = q$ and $O C = p$.
(q-1)}{2} | Number Theory | proof | Yes | Yes | olympiads | false | 29,470 |
205. Does there exist a positive integer $n$ such that the "fractional part" of the number $(2+\sqrt{2})^{n}$, i.e., the difference
$$
\left\{(2+\sqrt{2})^{n}\right\}=(2+\sqrt{2})^{n}-\left[(2+\sqrt{2})^{n}\right]
$$
(see p. 37) is greater than 0.999999?
206*, a) Prove that for any positive integer $n$, the number $... | 205. It is obvious that $(2+V \sqrt{2})^{n}+(2-V \sqrt{2})^{n}$ is an integer; indeed, if $(2+\sqrt{2})^{n}=a_{n}+b_{n} \sqrt{2}$, where $a_{n}$ and $b_{n}$ are integers, then $(2-\sqrt{2})^{n}=a_{n}-b_{n} \sqrt{2}$ (this follows from the binomial formula, and can also be proven by mathematical induction). Since $(2-\s... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,472 |
207. Prove that if $p$ is a prime number greater than 2, then the difference
$$
\left[(2+\sqrt{5})^{p}\right]-2^{p+1}
$$
is divisible by $p$. | 207. From the fact that $(2+\sqrt{5})^{p}+(2-\sqrt{5})^{p}-$ is an integer and $-1<$ $(2-\sqrt{5})^{p}<0$ (since $p$ is odd), it follows that
$$
\left[(2+\sqrt{5})^{p} \right] = (2+\sqrt{5})^{p}+(2-\sqrt{5})^{p}
$$
(compare with the solutions to problems 205 and 206). By the binomial formula, we have:
$$
\begin{alig... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,473 |
208*. Prove that if $p$ is a prime number, then the difference
$$
C_{n}^{p}-\left[\frac{n}{p}\right]
$$
is divisible by $p$ ( $C_{n}^{p}$ is the number of combinations of $n$ elements taken $p$ at a time; $n$ is an arbitrary positive integer not less than $p$ ).
For example, $C_{11}^{5}=\frac{11 \cdot 10 \cdot 9 \cd... | 208. $C_{n}^{p}=\frac{n(n-1)(n-2) \cdots(n-p+1)}{. p!}$. Among $p$ consecutive integers $n, n-1, n-2, \ldots, n-p+1$ one and only one number is divisible by $p$; let us denote this number by $N$. In this case,
$\left[\frac{n}{p}\right]=\frac{N}{p}$ and the difference in the problem's condition takes the form
$$
\frac{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,474 |
209. Indicate all such numbers $\alpha$ that the numbers $[\alpha],[2 \alpha]$, $[3 \alpha], \ldots,[N \alpha]$, where $N$ is a fixed natural number, are all distinct, and the numbers $\left[\frac{1}{\alpha}\right],\left[\frac{2}{\alpha}\right],\left[\frac{3}{\alpha}\right], \ldots,\left[\frac{N}{\alpha}\right]$ are al... | 209. Let $\alpha>0$. It is clear that the value $\alpha=1$ satisfies the condition of the problem; if, however, $\alpha>1$ and, accordingly, $\frac{1}{\alpha}<1$, then $\frac{N-1}{N} \leqslant \beta=\frac{1}{\alpha}<1$; if $\alpha<1$ and, accordingly, $\frac{1}{\alpha}>1$, then $\frac{N-1}{N} \leqslant \beta=\frac{1}{\... | \frac{N-1}{N}\leqslant\alpha\leqslant\frac{N}{N-1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,475 |
210*. Prove that in the equality
$$
N=\frac{N}{2}+\frac{N}{4}+\frac{N}{8}+\ldots+\frac{N}{2^{n}}+\ldots
$$
${ }^{\prime}(N$ - any positive integer) all fractions can be replaced by the nearest integers to them:
$$
N=\left(\frac{N}{2}\right)+\left(\frac{N}{4}\right)+\left(\frac{N}{8}\right)+\ldots+\left(\frac{N}{2^{n... | 210. First solution. It is obvious that ( $a$ ) $=\left[a+\frac{1}{2}\right]$; thus, the equality we need to prove takes the form
$$
N=\left[\frac{N}{2}+\frac{1}{2}\right]+\left[\frac{N}{4}+\frac{1}{2}\right]+\left[\frac{N}{8}+\frac{1}{2}\right]+\ldots
$$
Let now
$$
N=a_{n} \cdot 2^{n}+a_{n-1} \cdot 2^{n-1}+\ldots+a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,476 |
211. How many digits does the number $2^{100}$ have? | 211. It is easy to check that $2^{10}=1024$; thus, $2^{100}=$ $=1024^{10}$. Since $1000^{10}=10^{30}$ represents a number consisting of a one followed by 30 zeros, and $1024^{10}>1000^{10}$, the number $2^{100}=$ $=1024^{10}$ cannot have fewer than 31 digits. On the other hand,
$$
\begin{aligned}
\frac{1024^{10}}{1000... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,477 |
212. a) Prove that
$$
\frac{1}{15}<\frac{1}{10 \sqrt{2}}<\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{3}{6} \ldots \frac{99}{100}<\frac{1}{10}
$$
b) * Prove that
$$
\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{99}{100}<\frac{1}{12}
$$
Note The result of part b) clearly strengthens the resul... | 212. a) First solution. Let the product $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \ldots \frac{99}{100}$ be denoted by $A$ and consider another product $B=\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdots \frac{98}{99}$. Since
$$
\frac{2}{3}>\frac{1}{2}, \frac{4}{5}>\frac{3}{4}, \frac{6}{7}>\frac{5}{6}, \l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,478 |
215. Prove that in the decimal number system, the records of the numbers $1974^{n}$ and $1974^{n}+2^{n}$ contain the same number of digits for any natural number $n$. | 215. Let the number of digits in the number $1974^{n}$ be $k$; this means that $10^{h-1} \leqslant 1974^{n}1000^{n}=10^{3 n}$, so it is clear that $k \geqslant 3$ p.) If the number of digits in the number $1974^{n}+2^{n}$ is more than $k$, then $1974^{n}+2^{n} \geqslant 10^{k}$. Since $1974^{n}=2^{n} \cdot 987^{n}$, an... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,480 |
218. What is greater, $99^{n}+100^{n}$ or $101^{n}$ ( $n-$ a positive integer) ? | 218. It is required to determine which is greater: $101^{n}-99^{n}$ or $100^{n}$. Let's form the ratio
$$
\begin{aligned}
\frac{101^{n}-99^{n}}{100^{n}} & =\frac{(100+1)^{n}-(100-1)^{n}}{100^{n}}=\frac{2\left(C_{n}^{1} \cdot 100^{n-1}+C_{n}^{3} \cdot 100^{n-3}+\ldots\right)}{100^{n}}= \\
& =2\left(\frac{n}{100}+\frac{... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 29,482 |
219. What is greater, $100^{300}$ or 300!? | 219. Let's prove preliminarily that the product of \( n \) consecutive integers is greater than the square root of the product of the extreme numbers raised to the power of \( n \). Let's denote these numbers as \( a, a+1, \ldots, a+n-1 \). Then the \( k \)-th number from the beginning is \( a+k-1 \), and the \( k \)-t... | 300!>100^{300} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,483 |
221. What is greater, $(1.000001)^{1000000}$ or 2? | 221. By the result of the previous task
$$
(1,000001)^{10000000}=\left(1+\frac{1}{1000000}\right)^{1000000}>2
$$ | (1.000001)^{1000000}>2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,485 |
223. Prove that for any integer $n>6$
$$
\left(\frac{n}{2}\right)^{n}>n!>\left(\frac{n}{3}\right)^{n}
$$ | 223. Let the inequalities of the problem be valid for some $n$. To show their validity for $n+1$, it is sufficient to check the validity of the following inequalities:
$$
\left(\frac{n+1}{2}\right)^{n+1}:\left(\frac{n}{2}\right)^{n} \geqslant n+1 \geqslant\left(\frac{n+1}{3}\right)^{n+1}:\left(\frac{n}{3}\right)^{n}
$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,487 |
224*. Prove that for $m > n$ ($m, n$ - positive integers)
a) $\left(1+\frac{1}{m}\right)^{m}>\left(1+\frac{1}{n}\right)^{n}$
For example, $\left(1+\frac{1}{2}\right)^{2}=\frac{9}{4}=2 \frac{1}{4}$, and $\left(1+\frac{1}{3}\right)^{3}=$ $=\frac{64}{27}=2 \frac{10}{27}>2 \frac{1}{4}$.
b) $\left(1+\frac{1}{m}\right)^{m... | 224. a) By the binomial formula of Newton, we have:
$$
\begin{gathered}
\left(1+\frac{1}{n}\right)^{n}=1+C_{n}^{1} \frac{1}{n}+C_{n}^{2} \frac{1}{n^{2}}+\ldots+C_{n}^{n-1} \frac{1}{n^{n-1}}+\frac{1}{n^{n}}= \\
=1+n \cdot \frac{1}{n}+\frac{n(n-1)}{2!} \frac{1}{n^{2}}+\frac{n(n-1)(n-2)}{3!} \frac{1}{n^{3}}+\cdots \\
\ld... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,488 |
229*. Find the integer part of the number
$$
\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\frac{1}{\sqrt[3]{6}}+\ldots+\frac{1}{\sqrt[3]{1000000}}
$$ | 229. First, note that from the comparison of two expressions
$$
\left(1+\frac{1}{n}\right)^{2}=1+2 \frac{1}{n}+\frac{1}{n^{2}}
$$
and
$$
\left(1+\frac{2}{3} \frac{1}{n}\right)^{3}=1+2 \frac{1}{n}+\frac{4}{3} \frac{1}{n^{2}}+\frac{8}{27} \frac{1}{n^{3}}
$$
it follows that for each positive integer $n$
$$
\left(1+\f... | 14996 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,491 |
230. a) Determine the sum
$$
\frac{1}{10^{2}}+\frac{1}{11^{2}}+\frac{1}{12^{2}}+\ldots+\frac{1}{1000^{2}}
$$
with an accuracy of 0.006.
b) Determine the sum
$$
\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\ldots+\frac{1}{1000!}
$$
with an accuracy of 0.000000015. | 230. a) Obviously, we have:
$$
\begin{aligned}
& \frac{1}{10^{2}}+\frac{1}{11^{2}}+\ldots+\frac{1}{1000^{2}}>\frac{1}{10 \cdot 11}+\frac{1}{11 \cdot 12}+\ldots+\frac{1}{1000 \cdot 1001}= \\
& =\left(\frac{1}{10}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{12}\right)+\ldots+\left(\frac{1}{1000}-\frac{1}{1001}\right... | 0.00000029 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,492 |
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