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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
## 6. Coffee is served
Coffee is always served to me between 1 and 2 o'clock, at a moment when the bisector of the angle formed by the two hands of my clock points exactly at the "12" (hour) mark. At what exact time is my coffee served? | 6. Let's denote this unknown moment of time as 1 hour $x$ minutes (since it is between 1 hour and 2 hours!). The angle formed by the hour hand with the bisector of the angle between the two hands is (in degrees)
$$
\frac{360}{12}+\frac{x}{60} \cdot \frac{360}{12}=30+\frac{x}{2}
$$
(here we use the fact that the bisec... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,185 |
## 7. Bats, bears, poodles, and elephants
Attentive naturalists have established that the daily diet of 17 bears matches the diet of 170 poodles, the diet of 100000 bats matches the diet of
50 poodles, and 10 bears consume as much in a day as 4 elephants. How many bats can handle the diet of a dozen elephants? | 7. Let $y$ be the daily diet of an elephant, $x$ be that of a bat, $z$ be that of a Pekingese, and $t$ be the daily diet of a bear. We know that
$$
17 t=170 z, \quad 100000 x=50 z, \quad 10 t=4 y,
$$
or
$$
t=10 z, \quad z=2000 x, \quad y=(5 / 2) t ;
$$
what we want to express is $y$ in terms of $x$. This is not dif... | 600000x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,186 |
## 9. Cigarettes
Four married couples had dinner together. After dessert, Diana smoked three cigarettes, Elizabeth smoked two, Nicole smoked four, and Maud smoked one cigarette. Simon smoked as many as his wife, Pierre smoked twice as many as his wife, Louis smoked three times as many as his wife, and Christian smoked... | 9. If all the people present together smoked 32 cigarettes, then the share of the four men is 22 cigarettes. We need to represent the number 22 in the following form:
$$
22=1 \cdot x+2 \cdot y+3 \cdot z+4 \cdot t
$$
where $\{x, y, z, t\}=\{1,2,3,4\}$ (these are the numbers of cigarettes smoked by the four ladies). Bu... | Maud | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,188 |
## 10. Confession
The confessor has a habit of assigning penances "by the tariff" - in exact proportion to the gravity of the sins committed. Thus, before absolving the sin of pride, he requires the penitent to recite the "Te Deum" once and the "Pater Noster" twice. Slander is "priced" at two "Pater Noster" and seven ... | 10. Since I will have to repeat the "Te Deum" only nine times, it is clear that I am not guilty of adultery.
I confess to one instance of slander, for ten "Credo" would be too few for two instances of slander, and if I had not slandered at all, the "Credo" would correspond to such a number of sins of pride and slander... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,189 |
## 11. Typists
Two typists need to retype a certain text. First, one retyped half of it, and then the second retyped the second half; in total, they completed the work in 25 hours. How many hours would each typist need to complete the entire work on their own, if it is known that, working simultaneously, they would ty... | 11. Let $x$ be the number of hours required by the first typist, and $y$ by the second typist, to complete the entire work independently. To complete half of this work, the first typist would need $x / 2$ hours, and the second typist $y / 2$ hours; therefore,
$$
x / 2 + y / 2 = 25
$$
On the other hand, working togeth... | x=20,\quady=30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,190 |
## 12. Beautiful Lace
Two lacemakers need to weave a piece of lace. The first one would weave it alone in 8 days, and the second one in 13 days. How much time will they need for this work if they work together? | 12. The first lacemaker completes one eighth of the entire work in one day, while the second completes one thirteenth. Together, they will need a little less than 5 days:
$$
1:(1 / 8+1 / 13) \approx 4.95 \text { days. }
$$ | 4.95 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,191 |
## 13. Friendly Lunch
Ten married couples, who were friends with each other, met to have lunch together. First, they had an aperitif in the living room, and then all twenty people followed each other into the dining room. What is the minimum number of people that must go into the dining room so that among them there a... | 13. It is clear that as soon as thirteen people enter the dining room, there will inevitably be at least one married couple among them.
As soon as three people enter the dining room, there will certainly be at least two people of the same gender among them. | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,192 |
## 14. Dromedaries ${ }^{1}$
Some of the dromedaries are sleeping, while others are awake; the number of sleeping dromedaries is seven eighths of the number of awake dromedaries plus another seven eighths of a dromedary. If half of the sleeping dromedaries woke up, the number of awake dromedaries would be between 25 a... | 14. Let $d$ be the number of sleeping, and $r$ be the number of awake dromedaries. According to the problem, we have
$$
d=(7 / 8) r+7 / 8
$$
or
$$
8 d=7 r+7
$$
Thus, $d$ must be a multiple of 7.
On the other hand, this number must be such that it can be divided into two equal parts, i.e., it must be even; therefor... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,193 |
## 16. Grandmother, the Mark, and the Kittens
If you write down my mark, which I received in school for Latin ${ }^{1}$, twice in a row, you will get my grandmother's age. And what would you get if you divide this age by the number of my kittens? Imagine this - you would get my morning mark, increased by fourteen thir... | 16. Let $a$ be the age of my grandmother, $c$ be the number of my kittens, and $l$ be my grade in Latin. From the first condition of the problem, it follows that $l$ is an integer less than 10 (since it is clear that the age of my grandmother is less than 1010 years), and that $a=11 l$.
The second condition of the pro... | 77 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,195 |
## 17. Mathematical Puzzle
This week's mathematical puzzle in the "Modern Values" magazine was particularly difficult. When I started thinking about it (it was between 9 and 10 o'clock), I noticed that the two hands of my clock were in a straight line, one continuing the other. And I found the solution only at the mom... | 17. Let 9 hours $x$ minutes be the time when I started solving the problem. The angle formed by the hour hand with the common direction of the clock hands at noon is (in degrees)
$$
2 \cdot 360 / 12 + [(60 - x) / 60] \cdot 30
$$
The angle formed by the minute hand with the same direction is
$$
(360 / 60) \cdot x
$$
... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,196 |
## 18. Seals
Once I entered a maritime museum and saw some seals there. There were not many of them - just seven eighths of all the seals plus another seven eighths of a seal that I counted in the pool. How many seals were there in the pool in total?
dili in two rounds). According to the French constitution, a second... | 18. Let $n$ be the total number of seals. In this case,
$$
n=(7 / 8) n+7 / 8
$$
from which it follows that $n=7$.
There were seven seals in the marine museum. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,197 |
## 19. Tram
On one tram line, trams depart regularly every 10 minutes throughout the day. A tram takes one hour to travel from one end of the line to the other. A passenger boards a tram at one terminal stop and rides to the last stop of the tram; out of boredom, he looks out the window and counts the oncoming trams o... | 19. The passenger will encounter all the tram cars that departed from the other end of the tram line less than an hour before his departure; generally speaking, there are six such trains. He will also encounter all the trains that will depart from the opposite end of the line within the next hour; generally speaking, t... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,198 |
## 20. Nicolas plays
## with tin soldiers
Nicolas has the same number of tin Indians, Arabs, cowboys, and Eskimos. After his cousin Sebastian's visit, he indignantly discovered the disappearance of a third of his soldiers. Suppose the number of Eskimos remaining is the same as the number of cowboys that disappeared, ... | 20. Let $x$ be the number of soldiers of each type that Nicolas had initially, and $y$ be the number of cowboys taken by Sebastian (or the number of remaining Eskimos); in this case, the number of Eskimos taken is $x-y$.
On the other hand, it is known that the number of Indians taken is $x / 3$. Let $z$ be the number ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,199 |
## 22. General Cleaning
A solid joint-stock company occupies three floors in a tower, which has the shape of a parallelepiped and is located in Paris in the La Défense district: the 13th and 14th floors, where the company's offices are located, and the 25th floor, where its board of directors is situated.
To clean th... | 22. Let $n$ be the unknown number of cleaners. The number of man-hours required to clean two floors of the bureau is
$$
4 \cdot(n+n / 2)
$$
The number of man-hours required to clean the 25th floor is
$$
4 \cdot(n / 2)+8
$$
But this last number is half of the previous one, since the area of one floor is half the are... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,201 |
## 23. Parisians
In Paris, there are 2,754,842 people. Each of them is assigned a specific number - from 1 to 2,754,842. How many digits are needed to write all these numbers, what is the sum of all these numbers, and what is the sum of all the digits that make up all these numbers? | 23. I. Let's determine the number $N_{1}$ of required digits:
1) for writing
2) " "
3) " "
4) " "
5) " "
6) " "
7) " "
 / 2 = 465$ coins, where one coin is taken from the offering of the first vassal, two coins from the offering of the second, three coins from the third, ..., and finally, 30 coins are taken from the offering of the thirtieth vassal (i.e., all the coins of the 30th vassal are ... | 4650-n=4650- | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,205 |
## 27. Pharmacies
New Town contains 13 groups of houses: $a, b, c, d, \ldots$, $l$ and $m$. It is known that $a$ neighbors $b$ and $d; b$ neighbors $a, c$ and $d$; $c$ neighbors $b; d$ neighbors $a, b, f$ and $e; e$ neighbors $d, f, j$ and $l; f$ neighbors $d, e, j$, $i$ and $g; g$ neighbors $f, i$ and $h; h$ neighbor... | 27. Pharmacies should be located in groups $b, i, l$ and $m$. This is easily discernible from the diagram by representing each group
of houses with a point and connecting adjacent groups w... | b,i,, | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,206 |
## 28. Fuel Tanks Full of Gasoline
Someone has two cars, a larger one and a small one. Their fuel tanks have a combined capacity of 70 liters. When both tanks are empty, the car owner must pay 45 francs to fill one tank and 68 francs to fill the other. Suppose this person uses regular gasoline for the small car and pr... | 28. Let $x$ be the capacity of the tank of a small car, and $X$ of a large car; let $p$ denote the price in francs of one liter of regular gasoline. We have the following system of equations:
$$
\begin{gathered}
x+X=70 \\
x p=45 \\
X(p+0.2)=68
\end{gathered}
$$
Multiplying the last equation by $x$ and eliminating $p$... | 30,X=40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,207 |
## 29. Apples
Amelie picked apples, ate a quarter of them herself, and gave the rest to her little sisters Bertha, Charlotte, and Dorothy. Bertha ate a quarter of her apples and gave the rest to her three sisters. Charlotte ate one apple and gave the rest to her three sisters. As for little Dorothy, she also ate a qua... | 29. Let $a$ be the number of apples Amelie ends up with, $c$ be the number of apples Charlotte has at this stage, and $2c - y$ be the number of apples Berta has (0 being the number of apples Dorothy has). In the following table, we will represent all events in reverse order of their actual occurrence:
| | Amelie | Be... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,208 |
## 30. Purchasing Power
While prices increased by $12 \%$, Mr. $X$'s salary increased by $22 \%$. By how much did his purchasing power increase? | 30. Let's consider the previous salary of Monsieur $X$. Without its increase, his purchasing power would have decreased in the ratio $1 /(1+0.12)$. After the salary increase, his purchasing power increased by
$$
(1+0.22) /(1+0.12)=1.22 / 1.12 \approx 1.089 \text { times. }
$$
Therefore, his purchasing power increased... | 8.9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,209 |
## 31. Child Psychology
The brightest hours of a little child are spent drawing. He depicts either Indians or Eskimos and draws a dwelling for each of them. Thus, next to the Eskimos, he usually draws an igloo, and next to the Indians - wigwams. But sometimes the child makes mistakes, and not infrequently, for example... | 31. Let $E W$ be the fraction of drawings depicting an Eskimo and a wigwam, $E I$ be the fraction of drawings depicting an Eskimo and an igloo, $I I$ be the fraction of drawings depicting Indians and an igloo, and finally, $I W$ be the fraction of drawings depicting Indians and a wigwam. Then the following equalities h... | \frac{7}{8} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,210 |
## 32. What time is it?
To answer this question, it is enough to add two fifths of the time that has passed since midnight to the time remaining until noon. | 32. Let's denote the unknown time by $h$ (in hours). It is clear that $h$ is the time that has passed since midnight; therefore, the time remaining until noon is 12-h. The condition of the problem can be written as:
$$
h=(12-h)+(2 / 5) h,
$$
from which $h=60 / 8$ hours.
Thus, it is currently 7:30 AM.
218 | 7:30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,211 |
## 33. Reservoir
The reservoir has the shape of a rectangular parallelepiped, the width of which is half the length. It is filled to three eighths of its height. When 76 hl [1 hl (hectoliter) = 0.1 m³] is added, the water level rises by 0.38 m. After this, two sevenths of the reservoir remain to be filled.
What is th... | 33. As units of measurement, we will choose meters, square meters, and cubic meters.
Let $x$ be the desired height, and $y$ be the width of the reservoir. Then the length of the reservoir is $2 y$, and therefore, the area of the base is $2 y^{2}$; hence, the increase in the volume of water by 76 hl can be expressed as... | 1.12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,212 |
## 34. Family Meeting
Once, nine members of one family met. Each arrived on their own, but by chance, they all came at the same time. Without delving into the psychological state of the attendees or the complex family relationships, we can only note that each of the arrivals kissed five of their relatives and shook ha... | 34. The total number of handshakes is half the number of handshakes exchanged among all participants, i.e., it is
$$
(1 / 2)(9 \cdot 3)=27 / 2
$$
Thus, we obtain a non-integer number of handshakes, which is, of course, impossible. | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,213 |
## 35. Parent Meeting
At the end of the school year, the third-grade teachers met with some of their students' parents; exactly 31 people were present at this meeting. The Latin teacher was asked questions by 16 parents, the French teacher by 17, the English teacher by 18, and so on up to the math teacher, who was ask... | 35. Let $n$ be the number of parents present at the meeting, and $m$ be the number of teachers. The first teacher talked to $15+1$ parents, the second to $15+2$, and so on, ..., the $m$-th teacher talked to $15+m$ parents. But the last teacher was the math teacher, with whom all the parents talked. Therefore,
$$
15+m=... | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,214 |
## 36. Sacks of Flour
Two trucks are transporting identical sacks of flour from France to Spain. The first one carries 118 sacks, while the second one carries only 40. Since the drivers of these trucks do not have enough pesetas to pay the customs duty, the first one leaves 10 sacks for the customs officials, as a res... | 36. Let $x$ be the cost of one sack of flour, and $y$ be the duty charged per sack.
If the first driver left 10 sacks at the customs, he should pay duty only for 108 sacks. Similarly, the second driver should pay duty only for 36 sacks. As a result, we get the following system of equations:
$$
\begin{aligned}
10 x+80... | 1600 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,215 |
## 38. Simone and Her Complexes
Simone suffered from numerous complexes. Therefore, she decided to go to one psychoanalyst and after a course of treatment, she got rid of half of her complexes and half of one of the remaining complexes. Then she went to another psychoanalyst, thanks to which she got rid of half of the... | 38. Consider the following table:
| Stage of Treatment | Number of Complexes |
| :--- | :---: |
| | 1 |
| At the end | $(1+1 / 2) \cdot 2=3$ |
| Before the 3rd psychoanalyst | $(3+1 / 2) \cdot 2=7$ |
| Before the 2nd psychoanalyst | $(7+1 / 2) \cdot 2=15$ |
| Before the 1st psychoanalyst | $(7+1)$ |
Thus, at the beg... | 2758 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,217 |
## 39. Square Plot
Mathieu has a square plot of land. Surrounding this plot is an alley of constant width with an area of $464 \mathrm{~m}^{2}$. Walking around his plot, Mathieu noticed that the difference in length between the outer and inner edges of the alley is 32 m.
What is the total area of Mathieu's plot, incl... | 39. Let $x$ be the length of the outer edge of the alley, and $y$ be the length of the inner edge of the alley. We have the following system of equations:
$$
\begin{gathered}
x^{2}-y^{2}=464 \\
4 x-4 y=32
\end{gathered}
$$
From the second equation, it follows that $x=y+8$. Then the first equation gives $x=33$ and $y=... | 1089 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,218 |
## 40. Tennis Tournament
199 people have registered to participate in a tennis tournament. In the first round, pairs of opponents are selected by lottery. The same process is used to select pairs in the second, third, and all subsequent rounds. After each match, one of the two opponents is eliminated, and whenever the... | 40. After each match, one of the opponents is eliminated, and since 199 people are participating in the tournament, by the end, 198 athletes must be eliminated. Therefore, for the entire tournament, 198 boxes of balls will be needed. | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,219 |
## 41. Thermometer
A faulty thermometer shows a temperature of $+1^{\circ}$ in freezing water and $+105^{\circ}$ in the steam of boiling water. Currently, this thermometer shows $+17^{\circ}$; what is the actual temperature? | 41. The true temperature is
$$
\frac{17-1}{(105-1) / 100} \approx 15.38^{\circ} \mathrm{C}
$$ | 15.38\mathrm{C} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,220 |
## 42. Barrels
In how many ways can a 10-liter barrel be emptied using two containers with capacities of 1 liter and 2 liters? | 42. Let $u_{n}$ be the number of ways to empty a barrel of capacity $n$ liters using two vessels of capacity 1 and 2 liters.
A barrel of $n$ liters will be emptied if we first use either the 1-liter vessel [and then we need to choose $(n-1)$ liters more from the barrel], or the 2-liter vessel [and then we need to choo... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,221 |
## 43. Calves, Cows, Pigs, Chickens
The dowry of a young peasant girl consists of a cow, a calf, a pig, and a chicken. Her fiancé knew that five cows, seven calves, nine pigs, and one chicken together cost 108210 francs, and also that a cow is 4000 francs more expensive than a calf, that three calves cost as much as t... | 43. Let $c$ be the price of a pig. Then the price of a calf is $(10 / 3) c$, and therefore, the price of a cow is $(10/3)c+4000$ and the price of a chicken is $(5 / 3000) \cdot(10 / 3) \cdot c$.
Considering the first piece of information the suitor had, we get
$$
(50 / 3) c+20000+(70 / 3) c+9 c+c / 180=108210
$$
fro... | 17810 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,222 |
## 45. Thirty-two Cards
Alfred, Bruno, Christoph, and Damien are playing cards using a deck of 32 cards. Damien dealt them all to the players, and when it was pointed out that the number of cards each player had was different, he said: "If you want us all to have the same number of cards, do as I say. You, Alfred, div... | 45. In the end, each player ends up with eight cards. Christoph had previously divided[^31]half of his cards equally between Alfred and Bruno; thus, before this, Christoph had $2 \cdot 8=16$ cards, and Alfred and Bruno each had 4 cards. But before this, Bruno had divided half of his cards between Christoph and Alfred, ... | 4,7,13,8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,224 |
## 2. Marel
Take a piece of cardboard and draw the following diagram on it:
Also, take three silver and three gold coins (if you are on the beach, draw the diagram directly in the sand and... | 2. Let's renumber our nine circles (nine positions of the scheme):
Let $C$ be the player who starts, and $F$ be the player who finishes placing their coins.
I. The player who starts the ga... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,226 |
## 3. The Game with a Pack of Cigarettes
Here we present you with a very ancient game - the game of "Nim" ${ }^{1}$.
Take a pack of cigarettes, preferably unopened. Divide it into two unequal piles (for example, seven and thirteen cigarettes). Choose a partner and start the game, taking turns to remove either a certa... | 3. Let's represent each situation encountered in our game by considering the number of cigarettes in the two corresponding piles as the coordinates $x$ and $y$ of a point on the plane. This will result in a diagram of the following type:
+(n+2)=3 n+3=3(n+1)
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,235 |
10. Prove that if the sum of two numbers is an odd number, then their product will be an even number. | Instruction. If the sum of two numbers is an odd number, then one of them is even and the other is odd, i.e., their product will be an even number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,237 |
11. Prove that if the product of three integers is an odd number, then their sum is also an odd number. | Instruction. All three numbers are odd, since if at least one of them were even, then the entire product would be even. Denoting the numbers as $2a+1, 2p+1, 2k+1$, find their sum. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,238 |
12. Prove that every natural number can be represented in one of three forms: $3 n, 3 n+1, 3 n+2$, where $n$ is a natural number or zero. | Instruction. Let's find the remainders of dividing any natural number by 3. The remainders must be less than the divisor, so they can be the numbers $0,1,2$. Then, based on the relationship between the dividend, divisor, quotient, and remainder, we have: $A=3 n$, or $A=3 n+1$, or $A=3 n+2$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,239 |
13. Prove that the product of two consecutive natural numbers is divisible by 2. | Instruction. Of two consecutive natural numbers, one is necessarily even.
Of two consecutive natural numbers, one is necessarily even. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,240 |
22. Prove that if two numbers give the same remainder when divided by a third number, then their difference is divisible by this third number ${ }^{1}$. | Let $a=b c+r, m=b k+r, a>m$, then
$$
a-m=b c+r-b k-r=b(c-k)
$$
is divisible by $b$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,242 |
23*. Prove that if the difference of two numbers is divisible by a third number, then the remainders of dividing each of these numbers by the third number are equal to each other. | Let $(a-b)$ be divisible by $k$, and $a$ and $b$ are not divisible by $k$. Then
$$
\begin{gathered}
a=k m+n, \quad b=k l+r \text{ or } a-b=k(m-l)+(n-r), \\
a>b, n<k, r<k .
\end{gathered}
$$
This means that the difference $n-r$ must be divisible by $k$, which is possible if $n=r$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,243 |
24*. Prove that if the dividend is the sum of several numbers, then the remainder of dividing this sum by a certain number will not change if one or several addends are decreased or increased by a number that is a multiple of the divisor. | Let $S=A B+r$, where $S$ is the sum of several terms, $A$ is the divisor, and $r$ is the remainder. Then:
$$
S \pm A K=A B+r \pm A K=A(B \pm K)+r
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,244 |
27*. Prove that if the dividend and divisor are multiplied and divided by the same number, the quotient does not change, while the remainder increases or decreases by a factor equal to this number. | Let $a=b q+r$. Multiply both sides of this equation by $k ; a k=b k q+r k$. But $r k<b k$, i.e., $r<b$. Therefore, $r k$ is the remainder of the division of $a k$ by $b k$. But the remainder $r k$ is $k$ times greater than the remainder $r$.
Note. The solution to most problems 28-37 is based on the possibility of writ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,247 |
28. Given a three-digit number, the digits of which are consecutive natural numbers. Let's form a new three-digit number by reversing the digits. Prove that the difference between the larger and smaller number will be 198. | Instruction. Let the three-digit number be: $100 n+10(n+1)+$ $+(n+2)$. The second number will then be: $100(n+2)+10(n+1)+n$. Find their difference. | 198 | Number Theory | proof | Yes | Yes | olympiads | false | 42,248 |
29*. Prove that the difference between a three-digit number and the number formed by the same digits but in reverse order is expressed by a number whose middle digit is 9, and the sum of the other digits is 9. | Let the given number be: $A \cdot 100 + B \cdot 10 + C$, where $A > C$. Then,
$(100 A + 10 B + C) - (100 C + 10 B + A) = (A - C) 100 + (C - A)$.
Breaking down one hundred into tens, and one ten into units, we get:
$$
\begin{aligned}
&(A - C - 1) 100 + 9 \cdot 10 + (10 + C - A), \text{ but } (A - C - 1) + \\
&+(10 + ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,249 |
30. Prove that every three-digit number written with the same digit is divisible by 37. | Instruction. A three-digit number written with the same digit has the form: $100A + 10A + A = 111A$, but 111 is divisible by 37, so the given number is divisible by 37. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,250 |
31*. Prove that if a three-digit number is divisible by 37, then there are other three-digit numbers, composed of the digits of the given number, that are also divisible by 37. | Instruction. If a three-digit number is divisible by 37, then
$$
100 A+10 B+C=37 p
$$
where $p$ is an integer. Let's form another three-digit number from these same digits: $100 B+10 C+A$. From the condition that this number is divisible by 37, we have:
$$
10 B=37 p-100 A-C
$$
Thus,
$100 B+10 C+A=370 p-1000 A-10 C... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,251 |
32. Prove that if there are two three-digit numbers, each of which is not divisible by 37, but their sum is divisible by 37, then the six-digit number formed from these two numbers is divisible by 37. | Let $A$ and $B$ be three-digit numbers, neither of which is divisible by 37, but their sum $A+B$ is divisible by 37. We form a six-digit number from these two:
$$
10^{3} A+B=1000 A+B=999 A+(A+B)
$$
Each term is divisible by 37, so the sum is also divisible by 37, i.e., $1000 A+B$ is divisible by 37.
Note. Problems 3... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,252 |
34. To prove: for a two-digit number, the sum of whose digits is greater than or equal to 10, to multiply it by 11, one needs to write the excess of the sum of the digits over 10 between the tens digit, increased by 1, and the units digit. | If $10 A+B$ is a two-digit number where $A+B=10+k$, then by multiplying it by 11 and substituting $A+B$ with $10+k$, we get:
$(10 A+B) \cdot 11=100 A+10(A+B)+B=100(A+1)+10 k+B$. | 100(A+1)+10k+B | Number Theory | proof | Yes | Yes | olympiads | false | 42,254 |
35. To multiply a two-digit number by 9, you need to subtract from the given number the number of tens increased by one, and append to the result the complement of the number of units to 10. | Instruction. Multiplying a two-digit number $10 A+B$ by $9=10-1$, we will have
$$
100 A+10 B-10 A-B
$$
or
$$
[(10 A+B)-(A+1)] \cdot 10+(10-B)
$$ | [(10A+B)-(A+1)]\cdot10+(10-B) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,255 |
36. To multiply a two-digit number by 99, you need to decrease this number by one and append the complement of the number to 100. | Instruction. Take a two-digit number $10 A+B$, multiply it by $99=100-1$ and, by adding and subtracting 100, transform the resulting expression into the form:
$$
100(10 A+B-1)+[100-(10 A+B)]
$$ | 100(10A+B-1)+[100-(10A+B)] | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,256 |
38. Prove that the sum of a two-digit number and the number written with the same digits but in reverse order is divisible by 11. | Instruction. Find the sum of the numbers: $10 A+B$ and $10 B+A$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,258 |
39. Prove that the product of two consecutive even numbers is always divisible by 8. | Instruction. Consecutive even numbers have the form: $2 n$ and $2 n+2$. Find their product. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,259 |
40. Prove that if two natural numbers, when divided by 3, both give a remainder of 1, then their product, when divided by 3, also gives the same remainder. | Instruction. If numbers give a remainder of 1 when divided by 3, then they are of the form $3 n+1$ and $3 p+1$. Find their product and determine the remainder it gives when divided by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,260 |
41. Prove that if one of two natural numbers gives a remainder of 1 when divided by three, and the other gives a remainder of 2, then their product gives a remainder of two when divided by three. | Instruction. The given numbers have the form $3 n+1$ and $3 p+2$.
---
Note: I've maintained the original formatting and line breaks as requested. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,261 |
42. Prove that half the product of two consecutive integers when divided by 3 can leave a remainder of 0 or 1. | Let the given numbers be $A$ and $A+1$. When $A$ is an even number, i.e., $A=2k$, then $\frac{A(A+1)}{2}=k(2k+1)$. Consider the cases when $k=3p, k=3p+1, k=3p+2$. Similar reasoning should be applied when $A$ is an odd number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,262 |
43. Prove that for any numbers $A$ and $B$, where $A>B$, one of the numbers $A, B, A+B, A-B$ is divisible by 3. | Instruction. The number $A$ can be of one of the forms: $3 n, 3 n+1$, $3 n+2$. The number $B$ can be of one of the forms: $3 p, 3 p+1, 3 p+2$. Consider all possible cases depending on the conditions of the problem. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,263 |
46.* Let $A, B, C, D$ denote the units, tens, hundreds, and thousands of a certain number, respectively. Prove that this number is divisible by 4 if $A+2B$ is divisible by 4; it is divisible by 8 if $A+2B+4C$ is divisible by 8, and it is divisible by 16 if $A+2B+4C+8D$ is divisible by 16, and $B$ is an even number. | Instruction. The given number has the form: $1000 D+100 C+10 B+A$. The first two terms are divisible by 4, the second term gives a remainder of $2 B$ when divided by 4, thus the remainder will be $A+2 B$. If this number is divisible by 4, then the remainder will be 0 and the number will be divisible by 4. Conduct simil... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,264 |
47. Prove that if a number is divisible by 8, then the sum of the units digit, twice the tens digit, and four times the hundreds digit is divisible by 8, and vice versa. | Instruction. Any number can be represented as $1000a +$ $+100b+10c+d$. The first term is divisible by 8, the second term gives a remainder of $4b$ when divided by 8, the third - gives a remainder of $2c$. Therefore, if the sum of all remainders and the units digit is divisible by 8, then the entire number is divisible ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,265 |
49.* A two-digit number, when added to the number written with the same digits but in reverse order, gives a perfect square. Find all such numbers. | Let $N=10 a+b$. Then, according to the problem:
$$
10 a+b+10 b+a=11(a+b)=k^{2}
$$
Thus, $k^{2}$ is divisible by $11$, $k$ is divisible by $11$, and then $k^{2}$ is divisible by $121$. Therefore, $a+b=11$. So, when $a=2,3,4,5,6,7,8,9$, respectively, $b=9,8,7,6,5,4,3,2$, i.e., the sought numbers are: $29,38,47,56,65,74... | 29,38,47,56,65,74,83,92 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,267 |
54. Find the difference between the squares of two consecutive even numbers and two consecutive odd numbers. | Instruction. Consecutive even numbers have the form: $2 n$ and $2 n+2$, consecutive odd numbers: $2 n+1, 2 n+3$. In the first case, the answer is: the quadrupled odd number enclosed between the two given even numbers; in the second case - the quadrupled even number enclosed between the given odd numbers. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,268 | |
60. Find the general form of numbers whose squares end in the digit 9. | Instruction. Numbers whose squares end in the digit 9 have the digits 3 or 7 in the units place, i.e., they have the form:
$10a+7, 10a+3; 10a+7=5(2a+1)+2, 10a+3=5(2a+1)-2$, so these numbers have the form: $5(2a+1) \pm 2$ or $10b \pm 3$. | 10b\3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,271 |
61. Prove: 1) that every number not divisible by 2 and 3 can be represented in the form $6 n+1$ or $6 n+5$, where $n$ is a natural number or zero; 2) that the product of two numbers of the form $6 n+1$ or of the form $6 n+5$ is a number of the form $6 n+1$;[^1]
3) that the product of a number of the form $6 n+1$ and $6... | Instruction. A natural number that is not divisible by 2 and 3 is also not divisible by 6, i.e., it can have one of the forms:
$$
6 n+1, 6 n+2, 6 n+3, 6 n+4, 6 n+5
$$
according to the condition of the problem, only two forms can be possible: $6 n+1$ and $6 n+5$, since the others are divisible by either 2 or 3. Consid... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,272 |
62. Prove that the square of an integer, when divided by 4, leaves a remainder of 0 or 1. | Instruction. Consider the cases when the integer is even and when it is odd: $N=2 n ; N=2 n+1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,273 |
63. Prove that a number representing a perfect square either is divisible by 3 or gives a remainder of 1 when divided by 3. | Instruction. Any integer can be represented in one of the forms: $3 n, 3 n+1, 3 n+2$, where $n$ is any natural number or zero. Consider all cases. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,274 |
64*. Prove that if a number is not divisible by 5, then its square, increased or decreased by 1, is divisible by 5. | Instruction. A number not divisible by 5 has one of the forms: $5 n+1, 5 n+2, 5 n+3, 5 n+4$, where $n$ is any natural number or zero. The square of the first number has the form:
$$
(5 n+1)^{2}=25 n^{2}+10 n+1=5 k+1
$$
Similarly, the squares of the other numbers can be represented in the form: $5 a+4, 5 b+9, 5 c+6$, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,275 |
65*. Prove that if a number is not divisible by 7, then its cube, increased by 1 or decreased by 1, is divisible by 7. Determine in which case the cube of the number should be increased, and in which case it should be decreased, to make the division possible. | Instruction. Numbers that are not divisible by 7 have one of the forms:
$$
7 n+1,7 n+2,7 n+3,7 n+4,7 n+5,7 n+6
$$
or
$$
7 k \pm 1,7 k \pm 2,7 k \pm 3
$$
since
$$
7 n+4=7 n+7-7+4=7 k-3
$$
similarly
$$
\begin{aligned}
& 7 n+5=7 k-2 \\
& 7 n+6=7 k-1
\end{aligned}
$$
The cube of such a number equals the sum of numb... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,276 |
66. Prove that if $a$ is an even number, then $a\left(a^{2}+20\right)$, $a\left(a^{2}-20\right)$ and $a\left(a^{2}-4\right)$ are divisible by 8. | Instruction. Transform the given expressions, taking into account that an even number has the form $2 n$.
Translate the text above into English, keep the line breaks and format of the source text, and output the translation result directly. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,277 |
67. Prove that if $a$ is an odd number, then the expression $a^{4}+9\left(9-2 a^{2}\right)$ is divisible by 16. | Instruction. Transform the given expression by representing $a$ as $2 n+1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,278 |
68. Prove that the expression $2 a^{2}+2 b^{2}$ can be represented as the sum of two squares. | Instruction. Represent each of the expressions $2 a^{2}$ and $2 b^{2}$ as the sum of two terms, then add and subtract $2 a b$:
$$
2 a^{2}+2 b^{2}=a^{2}+a^{2}+b^{2}+b^{2}+2 a b-2 a b=(a+b)^{2}+(a-b)^{2} .
$$ | (+b)^{2}+(-b)^{2} | Algebra | proof | Yes | Yes | olympiads | false | 42,279 |
72. Prove that the semi-sum of the squares of two even or two odd numbers is equal to the sum of the squares of two integers. | If the given numbers are even, they have the form: $2 a$ and $2 b$, the half-sum of their squares is
$$
\frac{(2 a)^{2}+(2 b)^{2}}{2}=2 a^{2}+2 b^{2}=(a+b)^{2}+(a-b)^{2}
$$
(see problem 68). If the given numbers are odd, they have the form: $2 a+1$ and $2 b+1$, the half-sum of their squares is
$$
\begin{gathered}
\f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,282 |
73. Prove that if each of two numbers is the sum of two squares, then their product will also be the sum of two squares. | Let the given numbers be $K=a^{2}+b^{2}, P=n^{2}+k^{2}$. Form their product, add and subtract $2 a b n k$, we get $(a n+b k)^{2}+(a k-b n)^{2}$. | (+)^{2}+(-)^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 42,283 |
74. Represent the product of numbers $a$ and $b$ as a difference of squares.
$$
\begin{gathered}
\text { Hint. } a b=\frac{a b}{2}+\frac{a b}{2}=\frac{a^{2}}{4}+\frac{a b}{2}+\frac{b^{2}}{4}-\frac{a^{2}}{4}+ \\
+\frac{a b}{2}-\frac{b^{2}}{4}=\left(\frac{a+b}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}
\end{gathered}
... | Instruction. Square the given expression, represent $2 a^{2} b^{2}$ as a sum of two terms. Then:
$$
\left(a^{2}+a b+b^{2}\right)^{2}=\left(a^{2}+a b\right)^{2}+\left(b^{2}+a b\right)^{2}+a^{2} b^{2}
$$ | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,284 |
76*. Prove that $x^{4}-2 x^{8}+6 x^{2}-2 x+1$ is a sum of three squares. | Instruction. Represent $6 x^{2}$ as the sum of three terms: $x^{2}, x^{2}, 4 x^{2}$, then
$x^{4}-2 x^{3}+6 x^{2}-2 x+1=\left(x^{4}-2 x^{3}+x^{2}\right)+\left(x^{2}-2 x+1\right)+4 x^{2}=$ $=\left(x^{2}-x\right)^{2}+(x-1)^{2}+(2 x)^{2}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,285 |
77*. Prove that the binomial $3 a^{4}+1$ is the sum of three squares. | Instruction. Add and subtract $2 a^{8}+2 a^{2} ; 3 a^{4}$ and represent it as a sum of three terms:
$$
\begin{gathered}
3 a^{4}+1=\left(a^{4}+2 a^{3}+a^{2}\right)+\left(a^{4}-2 a^{3}+a^{2}\right)+\left(a^{4}-2 a^{2}+1\right)= \\
=\left(a^{2}+a\right)^{2}+\left(a^{2}-a\right)^{2}+\left(a^{2}-1\right)^{2}
\end{gathered}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,286 |
78*. Prove that the sum of the product of four consecutive natural numbers and one is a perfect square. | Instruction. $n(n+1)(n+2)(n+3)+1=$
$$
=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1=\left(n^{2}+3 n+1\right)^{2}
$$ | (n^{2}+3n+1)^{2} | Algebra | proof | Yes | Yes | olympiads | false | 42,287 |
80*. Find the condition under which the polynomial $x^{3}+p x^{2}+q x+n$ is a perfect cube. | Instruction. Compare the coefficients of the given polynomial with the coefficients of a complete cube and establish the relationship between $n, p, q ; q=\frac{p^{2}}{3} ; n=\frac{p^{3}}{27}$. | q=\frac{p^{2}}{3};n=\frac{p^{3}}{27} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,289 |
82*. Find the value of $x$ for which the expression
$$
(a+b-x)^{2}+(b+c-x)^{2}+(c+a-x)^{2}
$$
becomes a perfect square. | Given. Denoting $(a+b-x)^{2}=A, \quad(b+c-x)^{2}=B$, $(c+a-x)^{2}=C$, and applying the condition obtained in the previous problem, we get: $A C=\left(\frac{B}{2}\right)^{2}$. Substituting the values of $A, B$ and $C$ and solving the resulting equation with respect to $x$:
$$
x=2 a \pm \sqrt{(a-b)^{2}+(a-c)^{2}}
$$ | 2\\sqrt{(-b)^{2}+(-)^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,291 |
83*. Prove that $(x+a)(x+2 a)(x+3 a)(x+4 a)+a^{4}$ is a perfect square. | Instruction. See the solution of problem 78.
$$
\begin{gathered}
(x+a)(x+2 a)(x+3 a)(x+4 a)+a^{4}= \\
=\left(x^{2}+5 a x+4 a^{2}\right)\left(x^{2}+5 a x+6 a^{2}\right)+a^{4}=
\end{gathered}
$$
(multiplied the factors: the first and fourth, the second and third)
$$
\begin{gathered}
=\left(x^{2}+5 a x\right)^{2}+10 a^... | (x^{2}+5+5^{2})^{2} | Algebra | proof | Yes | Yes | olympiads | false | 42,292 |
84*. Prove that the sum of the squares of two odd numbers cannot be a square of an integer. | Let $2 n+1$ and $2 m+1$ be two odd numbers. The sum of their squares is an even number. If this sum is a square of an integer, then this integer can only be even, i.e., the sum itself must be divisible by 4, which is impossible, since
$$
\begin{gathered}
(2 n+1)^{2}+(2 m+1)^{2}=4 n^{2}+4 n+1+4 m^{2}+4 m+1= \\
=4\left(... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,293 |
85*. Prove that for any integer $n>4$ the expression $n^{2}-3 n$ cannot be a square of a natural number. | Instruction. Transform the given expression: $n^{2}-3 n=$ $=n(n-3)$. The obtained factors $n$ and $n-3$ can either be coprime or have a common factor of 3, since the common factor of the numbers $n$ and $n-3$ must also be a common factor of their difference: $n-(n-3)=3$. Let's consider both cases.
1) If the numbers $n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,294 |
86*. Prove that if all sides of a right-angled triangle are expressed in integers, then one of them or all three numbers are even, and there cannot be more than two odd numbers. | Let $a, b$ and $c$ be numbers representing the lengths of the legs and the hypotenuse, then $a^{2}+b^{2}=c^{2}$. It is known that the square of an even number is even, and the square of an odd number is odd. If $a^{2}$ and $b^{2}$ are both even numbers, then $c^{2}$ is also an even number, i.e., all three numbers are e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,295 |
88. Prove that if in a right-angled triangle all sides are expressed in integers, then its area is expressed in an integer.
untranslated part:
(There is no additional text to translate, as the original text only contains one sentence.) | Indication. From the previous problem, it follows that one of the legs is an even number. The area of a right triangle is half the product of its legs.
## § 3. Factorization of Polynomials | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,296 |
89. Prove that the difference between the square of a natural number and the number itself is divisible by 2. | Instruction. $a^{2}-a=a(a-1)$ - the product of two consecutive integers, one of which is necessarily even. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,297 |
91. Prove that the difference between the cube of an odd number and the number itself is divisible by 24. | Instruction. $(2 n+1)^{3}-(2 n+1)=(2 n+1) 2 n(2 n+2)-$ the product of three consecutive integers is divisible by 6. Consider the cases when $n$ is even and $n$ is odd. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,299 |
92. Prove that the difference between the square of a number not divisible by 3 and one is divisible by 3. | Instruction. Numbers that are not divisible by 3 have the form: $3 n+1$, $3 n+2$. Consider both cases. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,300 |
93. Prove that the square of any odd number, decreased by one, is divisible by 8. | Instruction. $(2 n+1)^{2}-1=4 n(n+1)$ is divisible by 8, since $n(n+1)$ is divisible by 2. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,301 |
94. Prove that the difference of the squares of two consecutive odd numbers is divisible by 8. | Instruction. $(2 n+3)^{2}-(2 n+1)^{2}=8(n+1)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,302 |
95. Prove that the difference of the squares of two odd numbers is divisible by 8. | Instruction. $(2 n+1)^{2}-(2 p+1)^{2}=4(n-p)(n+p+1)$. Consider the cases when $n$ and $p$ are both even numbers, both odd, one is an even number and the other is odd. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,303 |
96. Prove that the sum of the cubes of three consecutive integers is divisible by 3. | Instruction. $n^{3}+(n+1)^{3}+(n+2)^{3}=3\left(n^{3}+3 n^{2}+5 n+3\right)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,304 |
97. Prove that the sum of three consecutive powers of the number 2 is divisible by 7. | Instruction. $2^{n}+2^{n+1}+2^{n+2}=2^{n}\left(1+2+2^{2}\right)=7 \cdot 2^{n}$. | 7\cdot2^{n} | Number Theory | proof | Yes | Yes | olympiads | false | 42,305 |
98. Prove that the sum of two consecutive powers of the number 2 is divisible by 6. | Instruction. $2^{n}+2^{n+1}=2^{n} \cdot 3, n \neq 0$.
Translate the text above into English, keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in a form that can be considered as both the instruction and the content to be translated. Here ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,306 |
99. Prove that the sum of two consecutive powers of some number $a$ is divisible by the product $a(a+1)$. | Instruction. $a^{n}+a^{n+1}=a^{n}(1+a)=a^{n-1} \cdot a(a+1)$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,307 |
100. Prove that the product of the square of an integer and the integer preceding this square is always divisible by 12. | Instruction. $n^{2}\left(n^{2}-1\right)=(n-1) n(n+1) n$. Determine why this product will be divisible by 12. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,308 |
101*. Prove that for any integer $a$ the expression $\left(a^{2}+3 a+\right.$ $+1)^{2}-1$ is divisible by 24. | Instruction. $\left(a^{2}+3 a+1\right)^{2}-1=a(a+1)(a+2)(a+3)$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,309 |
102*. Prove that if $n$ is a prime number, different from 2 and 3, then $n^{2}-1$ is divisible by 24. | Instruction. $n^{2}-1=(n-1)(n+1)$, among three consecutive integers, one is divisible by $3$, $n$ is a prime number, therefore, $(n-1)$ or $(n+1)$ is divisible by 3. If $n$ is a prime number, then $n-1$ and $n+1$ are consecutive even numbers, thus one is divisible by 2, the other by 4. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,310 |
103*. Prove that for any odd $x$ the expression $x^{3}+3 x^{2}-x-3$ is divisible by 48. | Instruction. Factorize the given polynomial:
$$
A=x^{3}+3 x^{2}-x-3=(x+3)(x+1)(x-1)
$$
If $x$ is odd, then
$$
x=2 n+1 \text { and } A=8(n+2)(n+1) n
$$
but the product of three consecutive natural numbers is divisible by 6. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,311 |
104*. Prove that for any integer value of $n$ the numerical value of the expression $n^{4}+6 n^{3}+11 n^{2}+6 n$ is divisible by 24. | Instruction. $\quad n^{4}+6 n^{3}+11 n^{2}+6 n=n\left(n^{3}+6 n^{2}+11 n+6\right)=$ $=n\left(n^{3}+n^{2}+5 n^{2}+5 n+6 n+6\right)=n(n+1)(n+2)(n+3)$ is divisible by 24. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,312 |
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