problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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461. The Game of 37. Here is a beautiful puzzle game that is simple yet incredibly engaging. Most of you might think that both players have equal chances of winning and that who wins is a matter of chance. However, there is a subtle trick in this game, knowing which can ensure a confident victory.
\) can always win, but to do so, he must start with 4. During the game, he must sequentially accumulate the following scores: \(4,11,17,24,30,37\). Below are three games. In the first, the second player ( \(B\) ) delays his defeat as long as possible. In the second game, he prevents \(A\) fro... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,376 | |
462. The Game of 22. Lay out 16 cards as shown in the figure. Two players take turns flipping one card, adding its value to the total score. The player who reaches 22 or forces the opponent to exceed this total wins.
. If you can score 17 and still have at least one 5-point pair of both types (4-1, 3-2), then you should win. If you can score 12 and still have two 5-point pairs of both types, then you should win. If you can score 7 ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,377 |
463. The Game of Nine Squares. Draw a simple diagram as shown in the figure, and take a box of matches. The side length of the large square is equal to three matches. The game consists of laying out one match at a time, in turns, to surround more small squares than your opponent. By completing a small square, you not o... | 463. I should go to \(M N\). My opponent can go to \(H L\), then I will respond with a move to \(C D\). (If he went to \(C D\), I would answer with \(H L\), and the positions would be the same.) The best he can do now is to go to \(D H\) (winning one point), but since he is forced to move again, I win the remaining eig... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,378 |
469. Groups of Dominoes. Is it known to any of my readers that if all 28 dominoes are laid out in a single line according to the usual rule (6 to 6, 2 to 2, blank to blank, etc.), the numbers at the ends will always match, so that the dominoes can actually be arranged in a circle? A very old trick involves hiding one o... | 469. The figure shows one of the solutions. The chain of dominoes is broken into 4 parts, each containing 7 pieces, and the sum of the points in each part is 22.
 | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,384 |
470. Dominoes. Here is an old French puzzle that I think will interest the readers. The task is to arrange a complete set of 28 domino tiles in the shape shown in the figure, such that all numbers form a series of squares. Thus, in the top two rows, we see a square of blanks, a square of threes, a square of fours, and ... | 470. The correct solution is shown in the figure: two squares, made up of dummies, are inside. If in the previous example not all numbers were on the boundary, it would have been enough to simply swap the missing number and the dummies. So, in this case, there would have been no puzzle. However, since all the numbers w... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,385 |
473. Sequences of Dominoes. A boy had a complete set of dominoes up to double 9, and he was struggling to arrange the tiles in a single line in the usual way - 6 to 6, 3 to 3, blank to blank, and so on. But his father told him:
- You are trying to do the impossible; however, if you allow me to remove four tiles, you wi... | 473. If we remove four tiles \(7-6, 5-4\), \(3-2, 1-0\) from the set, the remaining tiles can form a correct sequence. Any other combinations of these numbers would also work; for example, we could remove \(7-0, 6-1, 5-2\) and 4-3. The general rule is that from a domino set ending with a double of an odd number, we mus... | 7-6,5-4,3-2,1-0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,388 |
478. Setting up dominoes. One day someone reminded Professor Rackbrain about his promise to say how many ways there are to arrange 28 domino tiles in a single line according to the usual rules of the game, if arrangements from left to right and from right to left are considered different. After some time, he reported t... | 478. There are 126760 different ways to arrange 15 dominoes in a line, if the two directions are distinguished. | 126760 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,393 |
479. Matchstick Puzzle. Taking a box of matches, I found that I could form any pair of regular polygons shown in our diagram using all the matches each time. So, if I had 11 matches, I could use them to form, as shown, either a triangle and a pentagon, or a pentagon and a hexagon, or a square and a triangle (using only... | 479. The smallest possible number is 36 matches. We can form a triangle and a square from 12 and 24 matches, a triangle and a pentagon from 6 and 30 matches, a triangle and a hexagon from 6 and 30 matches, a square and a pentagon from 16 and 20 matches, a square and a hexagon from 12 and 24 matches, and a pentagon and ... | 36 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,394 |
481. Twenty Matches. The diagram shown here illustrates how 20 matches, divided into two groups (14 and 6 matches respectively), can be used to form enclosures for two plots, where the first plot has exactly three times the area of the second.
 arranged to form the number 57. The puzzle is to move two of the cigarettes without touching the others to get 0.
, and place them as shown in the figure. We have the square root of 1 minus 1 (i.e., \(1-1\)), which is obviously equal to 0. In the second case, we can move the same two cigarettes, placing one next to V and the other next to the second I, so that the word NIL (noth... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,399 |
485. Five Squares. Here is another not-so-difficult matchstick puzzle that will puzzle many readers, although they will laugh when they learn the answer.
You see in the picture how 12 matchsticks are arranged to form 4 squares. Can you arrange the same 12 matchsticks (all matchsticks must lie flat on the table) so tha... | 485. Arrange 12 matches as shown in the figure on the right; they will form 5 squares. Of course, one of them (indicated by the arrow) is very small, but there were no restrictions on the sizes of the squares in the conditions.
, and it falls into place. Of course, none of the originally shown m... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,401 |
487. Three times more. Lay out 20 matches on the table as shown in the figure. You can notice that 2 groups of 6 and 14 matches delimit 2 figures, the area of one of which is exactly 3 times the area of the other.
\) with an area of 36 square inches. After using 6 matches, the remaining 30 formed a triangle \((13,12,5)\) with an area of 30 square inches, and using another 6 matches, he was able to form a triangle \((10,8,6)\) with an are... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,410 |
496. Card Puzzle. Take 13 cards of the diamond suit from the deck and, placing the five on top and the king at the bottom, stack them in the following order: five, jack, ten, ace, seven, eight, four, two, queen, six, nine, three, king. Now lay them out in a row according to the following rule. Name the cards in the cor... | 496. Arrange the cards face down in the following order (three of spades on top of the deck, ten of hearts at the bottom): three of spades, three of clubs, five of hearts, ace of diamonds, ten of diamonds, ten of spades, king of clubs, two of diamonds, king of spades, jack of hearts, five of clubs, three of diamonds, j... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,411 |
497. Shuffling Cards. An elementary method of shuffling cards consists of taking a deck face down in the left hand and transferring the cards one by one to the right hand; each successive card is placed on top of the previous one: the second on top of the first, the fourth on top of the third, and so on until all the c... | 497. To shuffle 14 cards in the manner described above and return them to their original order, it takes 14 shuffles, although in the case of 16 cards, only 5 are required. We cannot delve into the nature of this phenomenon here, but the reader may find it interesting to conduct an independent investigation of this que... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,412 |
500. Calendar Puzzle. According to our current calendar rules, the first day of a century can never fall on a Sunday, Wednesday, or Friday. Try to explain this mystery in the simplest way possible. | 500. Every year divisible by 4 is a leap year, except for those years that are divisible by 100; of these, only those divisible by 400 will be leap years, and the rest are not. This fact is often overlooked. Thus, the year 1800 was not a leap year, nor was 1900; however, the years \(2000, 2400, 2800\) and so on will be... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,415 |
505. The Engineer's Last Name. Three businessmen - Smith, Robinson, and Jones - live in the Leeds - Sheffield area. In the same area live three railway workers with the same names. Businessman Robinson and the conductor live in Sheffield, businessman Jones and the stoker live in Leeds, while businessman Smith and the r... | 505. It is clear that the conductor cannot be called Smith, since Mr. Smith is the businessman closest to the engineer, and his income, therefore, is definitely divisible by 3, while 10000 is not divisible by 3. Likewise, the stoker cannot be called Smith, since Smith beats him at billiards. Therefore, Smith must be th... | Smith | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,419 |
512. The Leap Year Puzzle. In February 1928, there were 5 Wednesdays. Of course, there is nothing particularly remarkable about this, but it would be interesting to find the nearest year preceding 1928 and the nearest year following 1928, in which February also had 5 Wednesdays. | 512. Since 1752, when the new style was adopted in England, the first year with five Wednesdays in February was 1764. Then such years were 1792 and 1804. By adding 28, we get 1832, 1860, 1888. Then we are forced to make a jump to 1928, 1956, 1984, and 2012. Therefore, the answer will be 1888 and 1956. Usually, this hap... | 18881956 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,426 |
1. The Absent-Minded Secretary. The typist typed ten letters and addresses on ten envelopes, but the absent-minded secretary placed these letters into the envelopes without any regard for the correspondence between the letter and the addressee. However, she did place only one letter in each envelope. What is the probab... | 1. If nine letters have gone into their intended envelopes, then the tenth letter will certainly do the same. Therefore, the probability that exactly nine letters have gone into their envelopes is zero.
$$
[M . M ., \mathbf{3 3}, 210(\text { March 1950).] }
$$ | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,434 |
2. Pythagorean Theorem. Prove that the square of the hypotenuse of a right triangle is equal to the sum of the squares of its legs. | 2. In the right triangle $A B C$, we drop a perpendicular from $C$ to the hypotenuse
to get the height $C H$ (see the figure). The triangles $A C B, A H C$, and $C H B$ are similar, hence ... | ^{2}+b^{2}=^{2} | Geometry | proof | Yes | Yes | olympiads | false | 41,435 |
3. Four equations with four unknowns. Find all solutions to the following system of equations:
$$
\left\{\begin{aligned}
x+y+z+w & =10 \\
x^{2}+y^{2}+z^{2}+w^{2} & =30 \\
x^{3}+y^{3}+z^{3}+w^{3} & =100 \\
x y z w & =24
\end{aligned}\right.
$$ | 3. By direct verification, we confirm that the set of numbers $(1,2,3,4)$ satisfies the first and fourth, as well as the second and third equations. Since all equations in the system are symmetric with respect to $x, y, z, w$, the other 23 permutations of the numbers $1,2,3,4$ are also solutions to the system. However,... | (1,2,3,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,436 |
4. Zero-sum test. A certain test consists of 26 questions. For each incorrect answer, five points are deducted from the test-taker, and for each correct answer, eight points are awarded.
The test-taker answered all the questions. How many questions did he answer correctly if the total number of points he received was ... | 4. The ratio of the number of answers of each type is equal to the inverse ratio of the corresponding points. Therefore, the number of correct answers is $\frac{5}{5+8} \cdot 26=10$.
[M. M., 31, 237 (March 1958).] | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,437 |
5. Ptolemy's Theorem. Prove that in any convex quadrilateral inscribed in a circle, the product of the diagonals is equal to the sum of the products of the opposite sides.
 | 5. In the inscribed quadrilateral $ABCD$, draw the segment $AE$ such that point $E$ lies on the diagonal $BD$ and the angle $BAE$ equals the angle $CAD$. Then, triangle $BEA$ will be similar to triangle $CDA$, and triangle $AED$ will be similar to triangle $ABC$. Consequently, $AC: AB = CD: BE$ and $AC: AD = BC: ED$. F... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,438 |
6. Simple factorization. Factor the expression without grouping terms:
$$
x^{8}-x^{7} y+x^{6} y^{2}-x^{5} y^{3}+x^{4} y^{4}-x^{3} y^{5}+x^{2} y^{6}-x y^{7}+y^{8}
$$ | 6. Let's write the following chain of equalities:
$$
\begin{aligned}
x^{9}+y^{9} & =(x+y)\left(x^{8}-x^{7} y+x^{6} y^{2}-\cdots+y^{8}\right)= \\
& =\left(x^{3}+y^{3}\right)\left(x^{6}-x^{3} y^{3}+y^{6}\right)= \\
& =(x+y)\left(x^{2}-x y+y^{2}\right)\left(x^{6}-x^{3} y^{3}+y^{6}\right)
\end{aligned}
$$
from which
$$
... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,439 | |
7. Mental arithmetic. Square 85 in your mind. | 7. Since $(10 a+5)^{2}=100 a^{2}+100 a+25=a(a+1) 100+25$, we get that $(85)^{2}=8 \cdot 9 \cdot 100+25=7225 \star$.
$$
[\text { M. M., 24, } 273 \text { (May 1951).] }
$$ | 7225 | Other | math-word-problem | Yes | Yes | olympiads | false | 41,440 |
8. Fourth-order equation. How many negative roots does the equation
$$
x^{4}-5 x^{3}-4 x^{2}-7 x+4=0 ?
$$
have? | 8. The equation $x^{4}-5 x^{3}-4 x^{2}-7 x+4=0$ can be rewritten as $\left(x^{2}-2\right)^{2}=5 x^{3}+7 x$. Since for any negative $x$ the left side of the equation is positive, while the right side is negative, the original equation cannot have negative roots.
[P. E. Horton, M. M., 24, 114 (November 1950).] | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,441 |
9. A million on each side. Two million marked points are entirely located within a circle with a diameter of 1 cm. Is there a straight line such that exactly one million of these points are on each side of it? | 9. Consider all the lines determined by all possible pairs of our marked points. Take some point outside the given circle, not lying on any of these lines. Draw a line through this point so that all the marked points are to the right of it. Now let's rotate the obtained line around the given point from left to right. D... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,442 |
10. Infinity of the set of prime numbers. Show that there are infinitely many prime numbers. | 10. Suppose that there exists the largest prime number \( p \). Consider the number that exceeds the product of all prime numbers less than or equal to \( p \) by one, that is,
\[
Q=2 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot p+1
\]
Notice now that \( Q \) is not divisible by any prime number involved in the above p... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,443 |
12. Overlapping Circles. A circle of radius 15 intersects with a circle of radius 20 at a right angle. Consider the two regions that result after removing their common part from the respective circles. What is the difference in their areas?
common part is $x$, then the areas of the corresponding non-overlapp... | 175\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,445 |
13. Tennis Tournaments. In a tennis tournament, there are $n$ players. Each tennis player is eliminated from the tournament after their first loss. How many matches should be held to determine the winner? | 13. After each match, one of the players is eliminated. Since in the end $n-1$ tennis players are eliminated, a total of $n-1$ matches should be held.
[F. Mapep, M. M., 23, 287 (May 1950).] | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,446 |
14. Summation of Factorials. What is the sum
$$
1(1!)+2(2!)+3(3!)+\cdots+n(n!) ?
$$ | 14. We have
$$
\begin{aligned}
& 1(1!)+2(2!)+3(3!)+\ldots+(n-1)[(n-1)!]+n(n!)= \\
& \quad=2(1!)+3(2!)+4(3!)+\ldots+n[(n-1)!]+(n+1)(n!)- \\
& \quad-1!-2!-3!-\ldots-(n-1)!-n!=(n+1)!-1
\end{aligned}
$$ | (n+1)!-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,447 |
15. Intersecting Cylinders. The axes of symmetry of two straight circular cylinders, each with a diameter of 2 cm, intersect at a right angle. What is the volume of the common part of these cylinders? | 15. Consider the plane passing through the axes of symmetry of the given cylinders. The cross-section of the common part of the cylinders by this plane is a square. If we draw a plane parallel to this one, it is clear from the figure that the cross-section will also be a square, and the circle inscribed in this square ... | \frac{16}{3}\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,448 |
16. Representation of an integer as a sum. The number 3 can be represented in four ways as the sum of one or more positive numbers, namely: as $3, 1+2, 2+1$ and $1+1+1$. Show that any positive integer $n$ can similarly be expressed in $2^{n-1}$ ways. | 16. Let's write down a row of $n$ ones with spaces between them. It is clear that there is a one-to-one correspondence between the representations of $n$ as a sum and the ways of filling the $(n-1)$ spaces between the ones, where we either insert nothing or a plus sign +. Thus, for each of the $(n-1)$ spaces, we can ac... | 2^{n-1} | Combinatorics | proof | Yes | Yes | olympiads | false | 41,449 |
17. A fourth-degree polynomial with rational roots. Show that the fourth-degree equation
$$
() x^{4}+() x^{3}+() x^{2}+() x+()=0
$$
where the parentheses are filled in any order with the numbers $1, -2, 3, 4$, -6 (one number in each parenthesis), always has a rational root. | 17. For any polynomial $f(x)$, the number $f(1)$ is equal to the sum of the coefficients. If this sum is zero, then $f(x)$ is divisible by $x-1$. Since $1-2+3+4-6=0$, it follows that $x=1$ is a root of our equation regardless of how the parentheses are filled with these numbers. | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,450 |
18. Wire cube. Each edge of the wire cube has a resistance of 1 Ohm. What is the resistance between two opposite vertices of this cube?
 | 18. Suppose the edges of our cube are hinged at the vertices. If we lift the cube by one of its vertices, the edges will sag, forming a structure consisting of three sequentially connected groups of conductors, each of which in turn consists of three, six, and three parallel-connected conductors*. The ends of the resul... | \frac{5}{6} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,451 |
20. Flower Seller. A girl bought $x$ roses in the store, paying $y$ dollars for all ( $x$ and $y$ are integers). When she was about to leave, the seller told her: "If you bought 10 more roses, I would give you all the roses for 2 dollars, and you would save 80 cents on each dozen." Find $x$ and $y$. | 20. Since $y$ is an integer $<2, y=1^{\star}$. Then, expressing the cost of one rose in cents, we get
$$
\frac{100}{x}-\frac{200}{x+10}=\frac{80}{12}, \text{ or } x^{2}+25 x-150=0.
$$
The only positive root of this equation is $x=5$. This is the number of roses the girl initially bought. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,453 |
21. The ratio of areas of polygons. The perimeters of a certain equilateral triangle and a regular hexagon coincide. What is the ratio of their areas? | 21. The sides of a triangle and a hexagon are in the ratio 2:1. Therefore, the triangle can be cut into four, and
the hexagon can be divided into six congruent triangles. Therefore, their ar... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,454 |
22. Inverted Cups. It is required to turn over $n$ cups, following this rule: in one move, it is allowed to turn over exactly $n-1$ cups (any), and this procedure can be repeated several times. Show that the problem is solvable for even $n$ and unsolvable for odd $n$. | 22. If $n$ is even and we flip the cups $n$ times, leaving a new cup untouched at each subsequent manipulation, then in the end each cup will be flipped $n-1$ times and will end up upside down.
If $n$ is odd, we will write +1 next to each correctly standing cup and -1 next to each cup that is upside down. Then, at the... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 41,455 |
23. The End of the World. On April 1, 1946, the newspaper "Jericho Daily Lapse" reported: "A famous astrologer and numerologist from Guyasuela, Professor Euclid Paracelsus Bombast Umbugio, predicted that the end of the world will occur in 2141. His prediction is based on deep mathematical and historical research. Profe... | 23. All that is needed to solve the problem is that $x^{n}-y^{n}$ for $n=0,1,2, \ldots$ is divisible by $x-y$. Let the quantity the professor was calculating be $F(n)$. Then, since $2141-1863=$ $1770-1492=278, F(n)$ is divisible by 278 for any $n$. Similarly, $2141-1770=1863-1492=371$, a number coprime with 278. Thus, ... | 1946 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,456 |
24. Six integers. Find six different smallest integers in such a way that the product of any five arbitrarily chosen from this six equals one or several periods in the decimal representation of the number, which is the reciprocal of the remaining sixth number. For example, if we take the decimal representation of the r... | 24. If $b$ is the period in the decimal representation of the number reciprocal to $a$, then all digits of the product $a b$ are nines. Further, $99=$ $9 \cdot 11, 999=3 \cdot 9 \cdot 37, 9999=9 \cdot 11 \cdot 101, 99999=0 \cdot 41 \cdot 271$ and $999999=3 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 37$. The last six numbe... | 3,7,9,11,13,37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,457 |
25. Length of a spiral. A piece of wire is wound in a spiral around a cylindrical tube, forming 10 turns. The length of the tube is 9 cm, and the length of its outer circumference is 4 cm. The ends of the spiral lie on the same generatrix of the cylinder. Find the length of the wire.
, the tenfold repeated circumference ($10 \cdot 4$ cm), and the wire ($L$) now form a right triangle. Therefore, $L=(81+1600)^{\frac{1}{2}}=41$ cm.
\left(q^{2}+q+1\right)\left(r^{2}+r+1\right)\left(s^{2}+s+1\right)}{p q r s}
$$
is not less than 81. | 26. The given fraction can be rewritten as
$$
\left(p+1+\frac{1}{p}\right)\left(q+1+\frac{1}{q}\right)\left(r+1+\frac{1}{r}\right)\left(s+1+\frac{1}{s}\right)
$$
Further, the sum of two positive reciprocal numbers $\geqslant 2$. Therefore, each bracket $\geqslant 3$, and the entire product $\geqslant 81$.
[R. L. Moe... | 81 | Inequalities | proof | Yes | Yes | olympiads | false | 41,459 |
27. Digits of a perfect square. Show that any perfect square, which in the decimal system is written using two or more digits, contains at least two different digits. | 27. Studying the table of squares in the decimal system shows that
1) a square can only end in the digits $0,1,4,5,6$ or 9;
2) if the unit digit of our square is 6, then the tens digit is an odd number; otherwise, the corresponding digit is even.
| $N$ | $\ldots$ | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | $\l... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,460 |
29. Jigsaw Puzzle. Everyone is familiar with the puzzle that involves cutting a piece of plywood with a picture drawn on it into small pieces, and then proposing to reassemble the original picture from these mixed pieces. Let the picture be cut into $n$ small pieces. We will call a "move" the joining together of two pi... | 29. Since we start with $n$ pieces and end up with 1, and since each move reduces the number of pieces by 1, the puzzle can be assembled in $n-1$ moves, and this number clearly does not depend on the method of joining. Of course, this reasoning remains valid only under the condition that, once any parts are joined toge... | n-1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,462 |
30. Unknown remainder. Find the remainder of the division of $f\left(x^{5}\right)$ by $f(x)$, if
$$
f(x)=x^{4}+x^{3}+x^{2}+x+1
$$ | 30. Since $f(x)=x^{4}+x^{3}+x^{2}+x+1,(x-1) f(x)=x^{5}-1$. Further, $f\left(x^{5}\right)=\left(x^{20}-1\right)+\left(x^{15}-1\right)+\left(x^{10}-1\right)+\left(x^{5}-1\right)+4+1$. But $x^{5}-1$, and therefore $f(x)$, are divisors of each of the brackets. Hence,
$$
f\left(x^{5}\right)=[\text { multiple of } f(x)]+5
$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,463 |
33. Simple multiplication. Multiply 5746320819 by 125. | 33. Since $125=1000: 8, \quad 5746320819 \cdot 125=$ $5746320819000: 8=718290102375$
$[M$. M., 25, 289 (May 1952).] | 718290102375 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,466 |
34. Series with repetitions. What is the $n$-th term of the series
$$
-4+7-4+7-4+7-\ldots ?
$$ | 34. Subtracting $\frac{1}{2}(7-4)=1.5$ from each term of the series
$$
-4+7-4+7-4+7 \ldots
$$
we get
$$
-5.5+5.5-5.5+5.5-5.5+5.5-\ldots
$$
Therefore, the $n$-th term of the series is $1.5+5.5(-1)^{n}$. | 1.5+5.5(-1)^{n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,467 |
35. Simplifying Radicals. Simplify the expression
$$
\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}
$$ | 35. Let $\sqrt[3]{2+\sqrt{5}}=a, \quad \sqrt[3]{2-\sqrt{5}}=b, \quad a+b=x$. Then
$$
x^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=a^{3}+b^{3}+3 a b(a+b)=4+3(\sqrt[3]{-1}) x
$$
Thus, $x^{3}+3 x-4=0$, and the only real root of this equation is 1.
$[$ [K. Adler, A. M. M., 59, 328 (May 1952).] | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,468 |
36. Buried Treasures. A certain pirate decided to hide his treasures on the shore of an uninhabited island. Nearby were two boulders $A$ and $B$, and further from the shore grew three coconut palms $C_{1}$, $C_{2}, C_{3}$. Standing at $C_{1}$, the pirate laid out a segment $C_{1} A_{1}$, equal and perpendicular to the ... | 36. Since triangles $A B_{1} C_{1}$ and $A_{1} B C_{1}$ are equal to each other, angle $C_{1} A B_{1}$ is equal to angle $C_{1} A_{1} B$. Further, angle $A P_{1} A_{1}$ is equal to angle
$... | T | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,469 |
37. Overtime Work. A certain company offered 350 of its employees to work overtime, with each man being offered a reward of 10 dollars, and each woman - 8 dollars and 15 cents. All women agreed to this offer, while some men refused. Upon calculation, it was found that the total reward amount does not depend on the numb... | 37. Let $m$ be the total number of men, and $x$ be the fraction of men who refused to work overtime. Then the total amount of compensation paid is
$$
T=8.15(350-m)+10(1-x) m=2852.50+m(1.85-10 x)
$$
which is independent of $m$ only if $x=0.185$. It is also known that $m<350$ and that both $m$ and $0.185 m$ are integer... | 1222.50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,470 |
38. Product. Simplify the following product:
$$
\left(3^{2^{0}}+1\right)\left(3^{2^{1}}+1\right)\left(3^{2^{2}}+1\right) \ldots\left(3^{2^{n}}+1\right)
$$ | 38. Multiplying the product by $1=\frac{1}{2}\left(3^{2^{0}}-1\right)$, we get
$$
\begin{gathered}
\frac{1}{2}\left(3^{2^{n+1}}-1\right), \quad \text { since } \quad\left(3^{2^{0}}-1\right)\left(3^{2^{0}}+1\right)=3^{2^{1}}-1 \\
\left(3^{2^{1}}-1\right)\left(3^{2^{1}}+1\right)=3^{2^{2}}-1 \quad \text { and so on. }
\e... | \frac{1}{2}(3^{2^{n+1}}-1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,471 |
39. Untangled Ropes. Three ropes are tied to three nails hammered into board $A$, and are intertwined with each other as shown in the figure. To their free ends, it is required to tie three new ropes, the ends of which (which are also allowed to be intertwined with each other) should be attached to three nails on board... | 39. The required arrangement of the new ropes can be obtained by reflecting the given arrangement symmetrically with respect to the dashed line.
[N. Crossman, P. M. E. J., 2, 26 (November ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,472 |
40. The difference equals the quotient. Find two numbers whose difference and quotient are both equal to 5. | 40. Since the quotient of two numbers is 5, the difference between them is four times the smaller one. Therefore, the smaller number is $\frac{5}{4}$, and the larger one is $\frac{25}{4}$.
In the general case if $x-y=\frac{x}{y}=a$, then
$$
x=\frac{a^{2}}{a-1}, \quad \text { and } \quad y=\frac{a}{a-1}
$$ | \frac{25}{4},\quad\frac{5}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,473 |
41. The Dozing Schoolboy. A schoolboy, waking up at the end of an algebra lesson, heard only a fragment of the teacher's phrase: "… I will only say that all the roots are real and positive." Glancing at the board, he saw there a 20th-degree equation assigned as homework, and tried to quickly write it down. He managed t... | 41. The roots are positive; their arithmetic mean is $-\frac{(-20)}{20}$, and their geometric mean is $(+1)^{\frac{1}{20}}$. Since both these values coincide, it follows that all roots are equal to 1.
[D. S. Greenstein, A. M. M., 63, 493 (September 1956).] | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,474 |
42. Edges of a Polyhedron. Show that in three-dimensional space no polyhedron can have exactly seven edges, while there are polyhedra with the number of edges equal to any other integer greater than five. | 42. Let \( n > 3 \). A simple polyhedron with \( 2n \) edges is a pyramid with an \( n \)-sided polygon as its base. If the \( n \)-sided polygon is folded along a diagonal so that it lies in two different planes, and then its vertices are connected by straight lines to a point not in these planes, we obtain a polyhedr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,475 |
43. Simple comparison. Show that $63! \equiv 61! \pmod{71}$, that is, that $63! - 61!$ is divisible by 71. | 43. Since
$$
63!-61!=(63 \cdot 62-1)(61!)=5 \cdot 11 \cdot 71(61!)
$$
we obtain $63!\equiv 61!(\bmod 71)$.
$[$ M. M., 34, 358 (September 1961).] | 63!\equiv61!\pmod{71} | Number Theory | proof | Yes | Yes | olympiads | false | 41,476 |
44. The triangle turns out to be equilateral. Show that if \(a, b, c\) are the sides of some triangle and
$$
a^{2}+b^{2}+c^{2}=a b+b c+c a
$$
then this triangle is necessarily equilateral. | 44. The equation $a^{2}+b^{2}+c^{2}=a b+b c+c a$ is equivalent to the equation $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$. From this, we conclude that $a=b=c$, since each term must be zero.
$[$ [M. S. Klamkin, M. M., 27, 287 (May 1954).] | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,477 |
45. Simpler than it seems. Calculate the root
$$
\left(\frac{1 \cdot 2 \cdot 4+2 \cdot 4 \cdot 8+3 \cdot 6 \cdot 12+\ldots}{1 \cdot 3 \cdot 9+2 \cdot 6 \cdot 18+3 \cdot 9 \cdot 27+\ldots}\right)^{\frac{1}{3}}
$$ | 45. $\left(\frac{1 \cdot 2 \cdot 4+2 \cdot 4 \cdot 8+3 \cdot 6 \cdot 12+\ldots}{1 \cdot 3 \cdot 9+2 \cdot 6 \cdot 18+3 \cdot 9 \cdot 27+\ldots}\right)^{\frac{1}{3}}=$
$$
=\left(\frac{1 \cdot 2 \cdot 4(1^{3}+2^{3}+3^{3}+\ldots)}{1 \cdot 3 \cdot 9(1^{3}+2^{3}+3^{3}+\ldots)}\right)^{\frac{1}{3}}=\left(\frac{8}{27}\right)... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,478 |
47. Numerical equality. Show that
$$
1110 \cdot 1111 \cdot 1112 \cdot 1113=(1235431)^{2}-1
$$
in any number system with a base greater than 5. | 47. This is a special case of the algebraic identity
$$
a(a+b)(a+2 b)(a+3 b)=\left(a^{2}+3 a b+b^{2}\right)^{2}-b^{4}
$$
where $a=r^{3}+r^{2}+r$ ( $r$ is the base of the numeral system), and $b=1$, so that
$$
a^{2}+3 a b+b^{2}=r^{6}+2 r^{5}+3 r^{4}+5 r^{3}+4 r^{2}+3 r+1
$$
Since no coefficient exceeds 5, the equali... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,480 |
48. Book Series. A certain series of books was published at seven-year intervals. When the seventh book was released, the sum of all the years in which the books of this series were published was 13524. When was the first book of the series published? | 48. The arithmetic mean of the years of publication is $\frac{13524}{7}=$ 1932, or the middle term of our arithmetic progression. The first term of this progression differs from the middle term by three differences. Therefore, the first book was published in 1932 $-3 \cdot 7=1911$.
$[M$. M., 34, 372 (September 1961).] | 1911 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,481 |
49. Three Mean Values. Show geometrically that the geometric mean $G$ of two numbers $a$ and $b$ is equal to the mean proportional between the arithmetic mean $A$ and the harmonic mean $H$ of the same numbers. | 49. On the segment $B C$ (where $B E=a, E C=b$), as a diameter, construct a semicircle with center at $O$. From point $E$, draw a perpendicular to $B C$, intersecting the semicircle at point $D$. Draw the segment $O D$ and drop a perpendicular from point $E$ to it, denoted as $E F$. Then the radius $O D=\frac{(a+b)}{2}... | G^2=HA | Geometry | proof | Yes | Yes | olympiads | false | 41,482 |
## 50. Beauty Contest.
- Can you tell me in confidence, in what order did the five most beautiful girls take the first five places in the beauty contest organized by your magazine? - I asked the publisher. Of course, she refused to prematurely disclose the secret, but promised to tell me if I had guessed the places co... | 50. In the order $D-A-E-C-B$, the two participants who are in the correct positions must follow each other directly. Otherwise, since two participants correctly named their immediate predecessors (i.e., the girls who took the immediately preceding place), the correct positions would have been identified for three, not ... | E-D-A-C-B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,483 |
51. An equation containing sums. Find $n$, if
$$
\frac{1^{3}+3^{3}+5^{3}+\cdots+(2 n-1)^{3}}{2^{3}+4^{3}+6^{3}+\cdots+(2 n)^{3}}=\frac{199}{242}
$$ | 51. First, note that if \(a: b = c: d\), then obviously \((a+b): b = (c+d): d\). Applying such a transformation to the fractions in our equation, we get
\[
\frac{1^{3}+2^{3}+3^{3}+\cdots+(2 n)^{3}}{2^{3}\left(1^{3}+2^{3}+3^{3}+\cdots+n^{3}\right)}=\frac{441}{242}
\]
Then, applying the known formula for the sum of cub... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,484 |
53. Dice. A die, on the faces of which the numbers $0,1,2,3,4,5$ are depicted, is rolled until the total sum of the points rolled exceeds 12. What is the most likely value of this sum? | 53. Consider the penultimate throw. After this throw, the sum of points can be $12, 11, 10, 9$ or 8. If it is 12, then after the last throw, the final sum of points can, with equal probability, take the values $13, 14, 15, 16$ or 17. Similarly, if this sum is 11, then the final sum can, with equal probability, take the... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,486 |
## 54. System of Linear Equations.
- This system of $n$ linear equations with $n$ unknowns has an interesting property, - said the Great Mathematician.
- Oh my God! - exclaimed the Poor Thing. - What is this property?
- Pay attention, - said the Great Mathematician, - the coefficients form an arithmetic progression.
... | 54. If $n \leqslant 3$, then the equations of our system are dependent and, consequently, its solution is not unique*. Therefore, $n=1$. Consequently, $n=2$.
Since our system has the form
$$
\begin{gathered}
a x+(a+d) y=a+2 d \\
(a+3 d) x+(a+4 d) y=a+5 d
\end{gathered}
$$
then
$$
x+y=1 \quad \text { and } \quad x=-... | -1,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,487 |
55. Property of the bisector. Prove that the bisector of any angle of a triangle divides the opposite side into parts proportional to the adjacent sides. | 55. The ratio of the areas of those parts into which the bisector of angle \( C=2 \varphi \) divides the given triangle:
\[
S_{B C D}: S_{D C A}=\left(\frac{1}{2} a t \sin \varphi\right):\left(\frac{1}{2} b t \sin \varphi\right)=a: b
\]
But the areas of two triangles with equal height are proportional to their bases,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,488 |
56. Probability of divisibility. Find the probability that if the digits $0,1,2, \ldots, 9$ are placed in random order in the empty spaces in the sequence of digits
$$
5-383-8-2-936-5-8-203-9-3-76
$$
then the resulting number will be divisible by 396. | 56. The number 76, formed by the last two digits, is divisible by 4. The difference between 73 (the sum of all digits in even positions) and $17+45$ (the sum of all digits in odd positions) is divisible by 11 regardless of the order in which the empty places are filled*. The sum of all digits, $90+45$, is divisible by ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,489 |
57. Equation without integer solutions. Prove that the equation $x^{2}-3 y^{2}=17$ has no solutions in integers. | 57. Any integer $x$ can be written in the form $3 n$ or $3 n \pm 1$. If we substitute numbers of this form into the equation $x^{2}-3 y^{2}=17$, we get respectively
$$
3\left(3 n^{2}-y^{2}\right)=17 \quad \text { and } \quad 3\left(3 n^{2} \pm 2 n-y^{2}\right)=16 .
$$
Since neither 17 nor 16 is divisible by 3, our eq... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,490 |
58. The son of a mathematics professor. The mathematics professor wrote a polynomial $f(x)$ with integer coefficients on the board and said:
- Today is my son's birthday. If his age $A$ is substituted into this polynomial instead of $x$, then the equation $f(A)=A$ holds. Note also that $f(0)=P$, where $P$ is a prime n... | 58. Since $f(0)=P$,
$$
f(x)=x \cdot q(x)+P \quad \text { and } \quad f(A)=A \cdot q(A)+P=A .
$$
Therefore, $P$ is divisible by $A$. Since $P>A$ and $P$ is prime, $A=1$. Thus, the professor's son is 1 year old. The professor could have written down any polynomial from an infinite class of such polynomials, for example... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,491 |
59. Position in space. Let us be given a plane $E$ and three non-collinear points $A, B, C$, located on one side of $E$ and lying in a plane intersecting with $E$. Take three arbitrary points $A^{\prime}, B^{\prime}$ and $C^{\prime}$ on the plane $E$. Denote by $L, M, N$ the midpoints of the segments $A A^{\prime}, B B... | 59. Place a particle of unit mass at each of the points $A, B, C, A^{\prime}, B^{\prime}, C^{\prime}$. Let $R$ be the center of mass of the particles located at points $A, B, C$, and $T$ be the analogous center of mass for the particles at points $A^{\prime}, B^{\prime}, C^{\prime}$. We can then consider that $S$
=2$ completely clear days, so the period under consideration covered $9+2=11$ days.
$[$ [M. M., 34, 244 (March 1961).] | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,493 |
61. Steiner-Lehmus Theorem. Prove that if the angle bisectors of two internal angles of a certain triangle are equal, then the triangle is isosceles. | 61. To prove this, we will use the well-known formulas expressing the lengths of the angle bisectors in terms of the sides of the triangle*:
$$
\frac{b c(a+b+c)(b+c-a)}{(b+c)^{2}}=\beta_{a}^{2}=\beta_{b}^{2}=\frac{a c(a+b+c)(c+a-b)}{(a+c)^{2}}
$$
After simplifications, we arrive at
$$
c(a+b+c)(a-b)\left[(a+b)\left(c... | b | Geometry | proof | Yes | Yes | olympiads | false | 41,494 |
62. Hula Hoop. Imagine a slender girl who has momentarily stopped while a hula hoop is rotating (without slipping) around her waist. The girl's waist is in the shape of a circle, with a diameter that is half the diameter of the hula hoop. Show that after one complete revolution of the hoop, a point on it that was initi... | 62. Since motion is relative, we can consider the hoop as stationary and the poor girl as moving inside it.
The initial point of contact on the waist will travel along the diameter of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,495 |
63. Division of a Circle into Eight Parts. Show that if $k-$ is an arbitrary real number, then the curve
$$
x^{4}+k x^{3} y-6 x^{2} y^{2}-k x y^{3}+y^{4}=0
$$
divides the circle $x^{2}+y^{2}=1$ into eight equal parts. | 63. Applying the standard rotation formulas for $45^{\circ}$:
$$
\begin{aligned}
& x=X \cos 45^{\circ}-Y \sin 45^{\circ}=\frac{X-Y}{\sqrt{2}} \\
& y=X \sin 45^{\circ}+Y \cos 45^{\circ}=\frac{X+Y}{\sqrt{2}}
\end{aligned}
$$
Substituting these expressions for $x$ and $y$ into our equation, we get
$$
X^{4}+k X^{3} Y-6 ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,496 |
64. Arithmetic progression deprived of powers. Find an arithmetic progression consisting of integers, with an arbitrarily large number of terms, none of which is an exact $r$-th power for $r=2,3, \ldots, n$. | 64. The progression $2,6,10, \ldots,(4 k+2), \ldots$ does not contain any powers, since the power of any odd number is odd, and the power of any even number is divisible by 4.
[A. Rosenfeld, A. M. M., 62, 185 (March 1955).]
There is also another trivial solution, where the first term is not a power, and the differenc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,497 |
66. Equation in factorials. Find all solutions of the equation
$$
n!(n-1)!=m!
$$ | 66. We have $n!(n-1)!=[(n-1)!]^2=m$! Obviously, $1!0!=1$! and $2!1!=2!$ are solutions to this equation. In all other cases, if $n \neq m$, then there are no solutions*. Further, for $m>10$, there always exist two prime numbers $p$ and $q$ such that $>\frac{m}{2}$ and $\leqslant m$. In this case,
$$
p q \geqslant\left(... | 7!6!=10! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,499 |
67. Two triangles. Show that if segments $a, b, c$ form a triangle, then segments $\sqrt{a}, \sqrt{b}, \sqrt{c}$ also form some triangle. | 67. Since $|\sqrt{b}-\sqrt{c}|(\sqrt{b}+\sqrt{c})=|b-c|<(\sqrt{a})^{2}<(b+c)<$ $(\sqrt{b}+\sqrt{c})^{2}$, it follows that $|\sqrt{b}-\sqrt{c}|<\sqrt{a}<(\sqrt{b}+\sqrt{c})$.
[Wang Ji-yi, A. M. M., 67, 82 (January 1960).] | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,500 |
68. The Greatest Angle in a Circle. Let two points $A$ and $B$ be given inside a certain circle. For which point $C$, located on the circumference, does the angle $A B C$ take the greatest value? | 68. To obtain the desired angle, draw two circles through points $A$ and $B$ that are tangent internally to the given circle. The point of tangency $C$ of the smaller of these circles with the given circle is the point for which the angle $A C B$ is maximized. Indeed, if we take any other point on the given circle, loc... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,501 |
69. Determinant of a Magic Square. Let $S$ be the sum of all elements of a magic square of the third order, composed of integers, and $D$ be the determinant of this square, if the latter is considered as a matrix. Show that $\frac{D}{S}$ is an integer.[^2] | 69. Let $N=S: 3$ - the sum of the numbers in each column, row, and on each diagonal of the magic square
$$
\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}
$$
Then
$$
\begin{gathered}
N=(a+e+i)+(d+e+f)+(g+e+c)- \\
-(a+d+g)-(c+f+i)=3 e
\end{gathered}
$$
and $S=9 e$. From this, by adding the rows an... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,502 |
70. Non-square five-digit numbers. Prove that no perfect square can be written in the decimal system using exactly five different digits, all of which are either even or all odd. | 70. There are only two sets, each consisting of five different digits, comparable to each other modulo 2, namely: $0,2,4,6,8$ and $1,3,5,7,9$. The sum of the digits of each square, when divided by 9, gives a remainder of $0,1,4$ or 7. However, the sum of the digits of the first set, when divided by 9, gives a remainder... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,503 |
71. Cutting a Sphere. Into what maximum number of congruent pieces can a sphere be cut so that each side of each piece represents an arc of a great circle that is less than a quarter of such a circle? | 71. Inscribed in this sphere is a regular dodecahedron or icosahedron, and perpendiculars are dropped from the center of the sphere to each face. Construct 60 isosceles triangles whose vertices are at the bases of these perpendiculars and whose bases are the sides of the corresponding faces. Now project these triangles... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,504 |
72. Trisection of a side of a triangle. Prove that if a line drawn from vertex $C$ of some triangle $A B C$ bisects the median dropped from vertex $A$, then it divides the side $A B$ in the ratio $1: 2$.
$M$ be the midpoint of the median $A D$ and extend $C M$ to inte... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,505 |
73. Factorization. Factorize $a^{15}+1$. | 73. Let's factor the given expression in different ways:
\[
\begin{aligned}
a^{15}+1 & =\left(a^{3}+1\right)\left(a^{12}-a^{9}+a^{6}-a^{3}+1\right)= \\
& =(a+1)\left(a^{2}-a+1\right)\left(a^{12}-a^{9}+a^{6}-a^{3}+1\right)= \\
& =\left(a^{5}+1\right)\left(a^{10}-a^{5}+1\right)= \\
& =(a+1)\left(a^{4}-a^{3}+a^{2}-a+1\ri... | (+1)(^{4}-^{3}+^{2}-+1)(^{2}-+1)(^{8}+^{7}-^{5}-^{4}-^{3}++1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,506 |
74. A curious number. Find such a positive number that $\frac{1}{5}$ of it, multiplied by its $\frac{1}{7}$, equals this number. | 74. If $(N: 5)(N: 7)=N$, then $N(N-35)=0$, so $N=35$. In other words, if the given number is multiplied by itself, the result will be 35 times greater than if we multiply $\frac{1}{5}$ of this number by $\frac{1}{7}$ of it. Therefore, the number we are looking for is 35.
In the general case, if a number equals $\prod_... | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,507 |
75. Two regular hexagons. Find, without using radicals, the ratio of the areas of two regular hexagons, one of which is inscribed in a given circle, and the other is circumscribed around it. | 75. Consider a regular inscribed hexagon, whose vertices divide the corresponding sides of the circumscribed hexagon in half. Connect the center of the circle with the vertices of the inscribed hexagon, and then connect the centers of the resulting equilateral triangles with the vertices of these triangles. As a result... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,508 |
76. Common terms of sequences. Find expressions for the common terms of each of the following sequences:
(a) $0,3,26,255,3124, \ldots$;
(b) $1,2, \quad 12,288,34560, \ldots$ | 76. In both cases, the sixth term can be equal to any number. Therefore, if we manage to describe the first five terms of the given sequence with a single formula, we can use the same formula to define the rest of the terms of this sequence. For example, we can set
(a) $0,3,26,255,3124, \ldots,\left(\left(n^{n}-1\righ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,509 |
77. Heat Flow. The temperature of three sides of a square metal sheet is maintained at $0^{\circ} \mathrm{C}$, while the temperature of the fourth side is maintained at $100^{\circ} \mathrm{C}$. Neglecting heat loss due to radiation, find the temperature at the center of the sheet. | 77. If four such sheets are superimposed so that each side has one edge at a temperature of $100^{\circ} \mathrm{C}$, then the average temperature of each side will be $25^{\circ} \mathrm{C}^{\star}$. Therefore, the temperature at the center of the sheet is $25^{\circ} \mathrm{C}$.
$$
\text { [L. Moser, M. M., 24, } 2... | 25\mathrm{C} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 41,510 |
80. Cryptarithm with "ham". In the following cryptarithm, each letter stands for a decimal digit (its own for each letter):
$$
7(F R Y H A M)=6(H A M F R Y)^{*}
$$
Determine which digit each letter represents. | 80. Let $F R Y=x$, and $H A M=y$; then
$$
\begin{aligned}
7(1000 x+y) & =6(1000 y+x) \\
6994 x & =5993 y \\
538 x & =461 y
\end{aligned}
$$
Since in the last equation the numerical coefficients are coprime, we get $x=F R Y=461$ and $y=H A M=538$. | FRY=461,HAM=538 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,513 |
81. Sheep Buyers. A farmer died, leaving a herd of cows to his two sons. The sons sold the herd, receiving as many dollars for each head as there were heads in the herd. With the money, the brothers bought sheep at 10 dollars each and one lamb, which cost less than 10 dollars. Then they divided the sheep and the lamb b... | 81. Let $x$ be the number of cows in the herd, $y$ the number of sheep, and $z$ the cost of a lamb. Then $x^{2}=10 y+z$, where $y$ is an odd number, and $z<10$. But the second-to-last digit of a square is odd if and only if the last digit is $6^{*}$. Thus, $z=6$, and the luckier son should pay his brother 2 dollars.
[... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,514 |
82. Constant Volume. Consider a tetrahedron defined by two segments that belong to skew lines. Prove that the volume of the tetrahedron will not change if we slide our segments (without changing their lengths) along the corresponding lines. | 82. We will shift the segments sequentially. Let, for example, segment $C D$ be fixed, while segment $A B$ moves to a new position $A^{\prime} B^{\prime}$. The area of triangle $A B C$ is equal to the area of triangle $A^{\prime} B^{\prime} C$, since they have equal bases and the same height. Moreover, the distance fro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,515 |
83. What is in the period? Determine what stands in the period of the decimal expansion of the fraction $\frac{1}{49}$.[^3] | 83. If we limit ourselves to the first period in the decimal expansion of the number $\frac{1}{7}$, we get 0.142857. Repeating this period 7 times and dividing the resulting number by 7, there will be no remainder*. The result is the first period of the decimal expansion of the number $\left(\frac{1}{7}\right)^{2}$. Th... | 0.020408163265306122448979591836734693877551 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,516 |
84. Equation containing radicals. Solve the equation
$$
(6 x+28)^{\frac{1}{3}}-(6 x-28)^{\frac{1}{3}}=2
$$ | 84. If $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c^{\star}$; therefore
$$
\begin{aligned}
(6 x+28)-(6 x-28)-8 & =3[(6 x+28)(6 x-28)(8)]^{\frac{1}{3}} \\
48 & =6\left(36 x^{2}-784\right)^{\frac{1}{3}} \\
512 & =36 x^{2}-784 \\
x^{2} & =36 \\
x & = \pm 6
\end{aligned}
$$
[F. E. Németh, S. S. M., 41, 291 (March 1941).] | \6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,517 |
85. Contest. Euclid Paracelsus Bombast Umbugio tries to supplement his meager professor's salary by participating in company-sponsored soap production contests. In one such competition, it was required to determine the number of paths along which the word MATHEMATICIAN* could be read on a given diagram. Umbugio counted... | 85. A path can be traced by moving "backwards" from $N$. If we consider the left half of the diagram, including the central column, then at each step backward we have a choice between two possible directions, giving us $2^{12}$ paths. Doubling this number and subtracting 1 (to avoid counting the central column twice), ... | 8191 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,518 |
86. Constant Sum. Let $A B C$ be an equilateral triangle, and $P$ be an arbitrary point on the inscribed circle. Prove that the value $(P A)^{2}+(P B)^{2}+(P C)^{2}$ is constant.
[^4] | 86. Introduce a Cartesian coordinate system in space, and let $A(1,0,0), B(0,1,0), C(0,0,1)$ and $P(x, y, z)$ be the vertices of our triangle and a point on the circle, respectively. The circle inscribed in this triangle is the intersection of the sphere $x^{2}+y^{2}+z^{2}=c_{1}$ and the plane $x+y+z=c_{2}$.
Therefore... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,519 |
87. Two numbers. Show that if $x$ and $y$ are positive integers and $y>2$, then $2^{x}+1$ is not divisible by $2^{y}-1$. | 87. Let's write our numbers in binary:
$$
2^{x}+1=100 \ldots 001, \quad 2^{y}-1=111 \ldots 111
$$
Now let's try to divide the first one by the second one using long division.
```
\(100 \ldots . . . . . .001 \quad 111 \ldots 111\)
\(111 \ldots 111 \quad 1 \ldots\)
\(100 \ldots 001\)
```
The numbers will divide evenl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,520 |
88. Choosing a pair. Among the pairs of numbers listed below, one and only one does not satisfy the equation $187 x-104 y=41$. Which one exactly
1) $x=3, \quad y=5$
2) $x=107, \quad y=192$
3) $x=211, \quad y=379$
4) $x=314, \quad y=565$
5) $x=419, \quad y=753$. | 88. Since the difference between the two terms on the left side of the equation is the odd number 41, one of these terms must be odd and the other even. Since $104 y$ is even, $187 x$ is odd, and therefore $x$ is odd. Thus, the pair $x=314$, $y=565$ does not satisfy our equation.
[D. Woods, S. S. M., 64, 242 (March 19... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false | 41,521 |
89. Intersection of Medians. Prove that the medians $A A^{\prime}, B B^{\prime}$, $C C^{\prime}$ of an arbitrary triangle $A B C$ intersect at one point. | 89. The converse of the well-known Ceva's theorem states: if three points taken on the sides of a triangle divide these sides into six segments such that the product of three of them, which do not have
<B^{4}+B^{3}+B^{2}+B+1<\left(B^{2}+\frac{B}{2}+1\right)^{2}
$$
and if the middle term is a perfect square, then the equality must hold
$$
\left(B^{2}+\fra... | 11111=(102)^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,523 |
91. A polygon is inscribed in an ellipse. Show that in an ellipse with unequal axes, it is impossible to inscribe a regular polygon with more than four sides. | 91. If such a polygon existed, the circle circumscribed around it would intersect the ellipse in more than four points - the vertices of the polygon, which is impossible.
$[M . S$. Klamkin, M. M., 34, 58 (September 1960).] | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,524 |
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