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263. Find two triangular numbers, the ratio of which is $2: 1$.
Find two triangular numbers, the ratio of which is $2: 1$. | 263. Answer. ( 6,3 ), $(210,105),(7140,3570)$ and others. | (6,3),(210,105),(7140,3570) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,831 |
264. Solve the equation $x^{2}-2 y^{2}=2$ in integers. | 264. From the equation, it is clear that $x$ is an even number. Let $x=2 x_{1}$, we get $2 x_{1}^{2}-y^{2}=1, y^{2}-2 x_{1}^{2}=-1$, i.e., the Pell's equation.
Answer. $(2,1),(10,7),(58,41),(338,239)$ and others. | (2,1),(10,7),(58,41),(338,239) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,832 |
266*. Solve the equation $3 x^{2}-2 y^{2}=1$ in integers.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 266. It is easy to show that integer solutions of the equation $A x^{2}-B y^{2}=1$, where $A$ and $B$ are not perfect squares, can be found from the equation $x \sqrt{\bar{A}}+y \sqrt{B}=\left(x_{0} \sqrt{\bar{A}}+y_{0} \sqrt{\bar{B}}\right)^{k}$, where $k$ is an odd number. For the given equation, we have $x \sqrt{3}+... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,833 |
270. Find a three-digit number that is a multiple of 45, if the difference between this number and its reverse is 297. | 270. Instruction. The required number ends in zero or five.
Answer. $360 ; 855$. | 360;855 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,836 |
271. Find a four-digit number $\overline{a b c d}$, divisible by 11, under the condition: $b+c=a$ and $\overline{b c}$ is a perfect square. | 271. Since $\overline{b c}$ is a perfect square, then $c=0,1,4,5,6,9$.
Given that $\overline{a b c d}$ is divisible by 11, then $a+c-b-d$ can take the values 0 and $\pm 11$. Let's consider these cases:
1) $\left\{\begin{array}{l}a+c=b+d \\ b+c=a\end{array}\right.$, from this $2 c=d$, therefore, $c=0$, $d=0 ; c=1, d=2... | 9812,1012,4048,9361,9097 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,837 |
272. Find a three-digit number that is equal to the sum of the tens digit, the square of the hundreds digit, and the cube of the units digit.
$273 *$. Find the number $\overline{a b c d}$, which is a perfect square, if $\overline{a b}$ and $\bar{c} \bar{d}$ are consecutive numbers, with $\overline{a b}>\overline{c d}$... | 272. $100 x+10 y+z=y+x^{2}+z^{3}$, from here $9 y=(z-$ $-1)(z+1) z+x^{2}-x-99 x, \quad(z-1) z(z+1)+x(x-1)$ must be divisible by 9; since $z(z-1)(z+1)$ is divisible by $3, x(x-1)$ is also divisible by 3, therefore, $x=1$, $3,4,6,7,9$.
When $x=3 y=\frac{z^{3}-z-291}{9}, z^{3}>z+291$, hence $z=7$, 8,9, but only when $z=7... | 357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,838 |
276. Find the number $\overline{a b c d}$, which is a perfect square, under the condition: $\overline{a b}=\frac{1}{4} \overline{c d}$. | 276. From the condition $\overline{c d}=4 \overline{a b}$, we conclude that $\overline{a b}<25$. Letting $\overline{a b}=x$, we represent the desired number as $100 x + 4 x$, i.e., $8 \cdot 13 x$. Since $8 \cdot 13 x$ must be a perfect square, then $x=2 \cdot 13 t=26 t \geqslant 26$, which contradicts the condition $\b... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,839 |
277. Find a two-digit number that is equal to the sum of the cube of its tens digit and the square of its units digit. | 277. From the equality $10 x+y=x^{3}+y^{2}$ we find $x(10-$ $\left.-x^{2}\right)=y(y-1)$, hence $0<x<\sqrt{10}$; therefore, $x=1$, 2,3 ; but $y(y-1)$ is an even number, so $x$ is also even; $x=2$, then $y=4$.
Answer. 24. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,840 |
278. Find two numbers whose product is a three-digit number that is the cube of some number, and the quotient is the square of this number. | 278. From the system $\left\{\begin{array}{l}x y=t^{3} \\ \frac{x}{y}=t^{2}\end{array}\right.$ we find $x^{2}=t^{5}, y^{2}=t$;
thus, $t$ must be a perfect square; but since by the condition $x y$ is a three-digit number, then $100<t^{3}<1000, 5 \leqslant t \leqslant 9$. Therefore, $t=9, x=243, y=3$.
Answer. 243 and 3... | 2433 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,841 |
281. Find the six-digit number $\overline{x y 342 z}$, divisible by 396. | 281. Answer. 453420 and 413424. | 453420413424 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,844 |
283. Find a number of the form $7 x 36 y 5$, divisible by 1375. | 283. Answer. 713625. | 713625 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,846 |
285. The number is a perfect square and ends in 5. Prove that its third digit from the right is even. | 285. If a number is a perfect square and ends in an odd digit, then this digit is $1,5,9$; but
if the number ends in five and is a perfect square, then it must end in 25. It remains to consider the cases when the number ends in 1 and 9.
a) If a number ends in 1 and is a perfect square, then the square root of it ends ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,848 |
287. In a hundred-digit number, all digits except one are fives. Prove that it is not a perfect square. | 287. If the digit not equal to 5 is in the last place, then it can equal $1,4,6,9$. Further reasoning is as in № 286.
If the digit not equal to 5 is in the second-to-last place, then it must equal only 2; but then the given number can be represented as $(10 n+5)^{2}$; in it, the third digit from the right is even.
In... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,849 |
288. What whole number should the number 999999999 be multiplied by to get a number consisting of all ones. | 288. We have $999999999=10^{9}-1,\left(10^{9}-1\right) x=\frac{10^{k}-1}{9}$; $x=\frac{10^{k}-1}{9\left(10^{9}-1\right)}$.
To make $x$ an integer, it is sufficient to take $k=81$; indeed, $x=\frac{10^{81}-1}{9\left(10^{9}-1\right)}=\frac{\left(10^{9}\right)^{9}-1}{9\left(10^{9}-1\right)}=\frac{\left(10^{9}\right)^{8}+... | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,850 |
289*. Prove that there are no integers that double when the initial digit is moved to the last place on the right. | 289. Let $N=\overline{a b c d \ldots l}$ contain $n$ digits; therefore, it can be written as $a \cdot 10^{n-1}+k$, where $k=\overline{b c d \ldots l}$. According to the problem,
$$
10 k+a=2\left(a \cdot 10^{n-1}+k\right)
$$
from which $a=8 \frac{k}{2 \cdot 10^{n-1}-1}$. Since $a$ is an integer less than ten, and $2 \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,851 |
295. Find the number that is three times less than the square of the sum of its digits. | 295. $3 N=\underbrace{(a+b+c+\ldots+l)^{2}}_{n \text { digits }}$. Since a digit does not exceed 9, then $3 N \leqslant 81 n^{2}, \quad N \leqslant 27 n^{2} ;$ from this it is clear that the desired number $N$ should contain no more than three digits, and for $n=3 N \leqslant 243$.
Let $n=1$, then $3 a=a^{2}$, hence $... | 3,27,48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,856 |
296. The difference between a number and the product of its digits is equal to the sum of the digits of this number. Find this number.
## Chapter $I X$.
## VARIOUS PROBLEMS | 296. Let the desired number contain $n$ digits $a, b, c$, $\ldots, k, l$. According to the condition $a \cdot 10^{n-1}+b \cdot 10^{n-2}+\ldots+$ $+k \cdot 10+l-a b c d \ldots k l=a+b+c+\ldots+k+l, \quad$ from which $9 k=a\left(b c \ldots k l+1-10^{n-1}\right)+b\left(1-10^{n-2}\right)+c\left(1-10^{n-3}\right)+$ $+\ldots... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,857 |
298. Prove that for any integer $n$ there exists a number $x$ such that the number $n x+1$ is composite. | 298. If for some integer value $x=\alpha$ the number $n \alpha+1$ is a prime number $p$, then for $x=\alpha+p t$ $n(\alpha+p t)+1=(n \alpha+1)+p t n=p(n t+1)$, i.e., a composite number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,859 |
299. Prove that the sum of two numbers is divisible by 6 if and only if the sum of their cubes is divisible by 6. | 299. If $x+y$ is divisible by 6, then $x^{3}+y^{3}=(x+y)\left(x^{2}-\right.$ $-x y+y^{2}$ ) is also divisible by 6.
If $x^{3}+y^{3}$ is divisible by 6, then first of all, $x$ and $y$ are both even or both odd; therefore, $x+y$ is divisible by 2. Moreover, $x^{3}+y^{3}$ is divisible by 3; therefore, $x$ and $y$ are bot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,860 |
301. Show that the sum of the squares of ten consecutive numbers cannot be a perfect square. | 301. $N=(x-2)^{2}+(x-1)^{2}+x^{2}+\ldots+(x+7)^{2}=$ $=5\left(2 x^{2}+10 x+29\right)$. It is necessary for $2 x^{2}+10 x+29$ to be divisible by 5; however, $2 x^{2}+29$ is not divisible by 5, because otherwise the even number $2 x^{2}$ would end in either 1, which is impossible, or 6, but then $x^{2}$ would end in 3 or... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,861 |
307. If Fermat's theorem is valid for prime exponents, then it is also valid for composite ones. | 307. From the equation $(x-1)^{p}+x^{p}=(x+1)^{p}$, we observe that $x$ is an even number. By transforming the equation to $x^{p} - 2 p x^{p-1} - 2 C_{p}^{3} x^{p-3} - 2 C_{p}^{5} x^{p-5} - \ldots - 2 = 0$, we find $x^{p} = 2\left(p x^{p-1} + C_{p}^{3} x^{p-3} + \ldots + 1\right)$. Since $x$ is an even number, $p x^{p-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,867 |
311. Find the triangular numbers that are multiples of 39 (see No. 19). | 311. Answer. $u_{n}=\frac{n(n+1)}{2}, \quad$ where $n=39 t, \quad 39 t+12$, $39 t+26,39 t+38$, and $t$ is an arbitrary non-negative integer. | u_{n}=\frac{n(n+1)}{2},\quad | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,868 |
314. For what values of $x$ are the numbers $\frac{5 x^{3}-x+17}{15}$ and $\frac{2 x^{2}+x-3}{7}$ simultaneously integers? | 314. Answer. $x=105 t+22, x=105 t+37$. | 105+22,105+37 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,871 |
315. The numbers $2 \overline{a c}+1$ and $3 \overline{a c}+1$ are perfect squares. Find $\overline{a c}$. | 315. First method. By setting $\overline{a c}=x$, we obtain the system $\left\{\begin{array}{l}2 x+1=y^{2} \\ 3 x+1=z^{2}\end{array}\right.$, i.e., $\frac{y^{2}-1}{2}$ and $\frac{z^{2}-1}{3}$ represent equal integers.
It is easy to find $x=2 t^{2}+2 t$ and $x=3 t_{1}^{2} \pm 2 t_{1}$. According to the problem's condit... | 40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,872 |
316. For what natural value of $a$ will the number $a^{2}+$ $+a+1589$ be a perfect square? | 316. We find $a=\frac{-1+\sqrt{4 y^{2}-6355}}{2}$, from here $4 y^{2}-$ $-6355=z^{2} ;(2 y+z)(2 y-z)=5 \cdot 31 \cdot 41$.
Answer. $1588,28,316,43$. | 1588,28,316,43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,873 |
317. Prove that if an integer, which is no more than a four-digit number, is divisible by 99, then the sum of its digits is divisible by 18. | 317. Among two-digit numbers, only 99 is divisible by 99; $9+9=18$. A three-digit number divisible by 99 can be written as $99a=100(a-1)+90+(10-a)$, where $1<a \leqslant 10$, hence this number has digits $a-1, 9, 10-a$; since $(a-1)+9+(10-a)=18$, the theorem is true.
A four-digit number divisible by 99 can be written ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,874 |
318. When dividing a number by 72, the remainder is 68. How will the quotient change and what will we get as the remainder if we divide the same number by 24? | 318. $N=72 t+68, \frac{N}{24}=3 t+2+\frac{20}{24}$, therefore, the new remainder is 20, and the quotient $t_{1}=3 t+2$. | t_{1}=3+2,remainder20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,875 |
320. At the end of 1960, a student arrived at the school. When asked about his age, he replied: “My age is equal to the sum of the digits of the year of my birth.” How old is the student? | 320. Given the school age, we can write the year of birth as $1900+10 x+y$. According to the problem, $1960-(1900+10 x+y)=1+9+x+y$, or $11 x + 2 y=50$ (1), from which it is clear that $x=2 x_{1}$, then $y=25-11 x_{1}$.
From the system of inequalities $\left\{\begin{array}{l}25-11 x_{1} \geqslant 0 \\ 25-11 x_{1}<10\en... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,877 |
321. Show that the equation $x^{2}+y^{2}=3 z^{2}$ has no solutions in integers. | 321. The numbers $x$ and $y$ must both be divisible by 3 (see No. 84).
Let $x=3 x_{1}$ and $y=3 y_{1}$, then $3 x_{1}^{2}+3 y_{1}^{2}=z^{2}, z$ is divisible by 3; assuming $z=3 z_{1}$, we get $x_{1}^{2}+y_{1}^{2}=3 z_{1}^{2}$, where $x_{1}<x, y_{1}<y, z_{1}<z$. Continuing this process further, we will arrive at an equ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,878 |
323. The quotient of dividing a certain three-digit number by 11 is equal to the sum of the squares of the digits of the dividend. Find this number. | 323. Given the condition $11\left(x^{2}+y^{2}+z^{2}\right)=100 x+10 y+z$, and either $x+z=y$ (1), or $x+z-y=11$ (2), or $y-x-z=11$ (3).
In case (1), we get $2 x^{2}+2 z^{2}+2 x z=10 x+z$, from which it is clear that $z$ is an even digit. Representing the last equation as $2 x^{2}+2(z-5) x+2 z^{2}-z=0$, we obtain $x=\f... | 550803 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,880 |
325. Show that $3^{x}-2$ is not divisible by 13. | 325. The number $x$ can be of the form $3 t, 3 t \pm 1$.
If $x=3 t$, then $3^{x}-2=\left(27^{t}-1\right)-1=26 n-1$, which is not divisible by 13. The other cases are considered similarly. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,882 |
326. Solve the equation $5^{x}-13 y=2$ in integers. | 326. The equation has no integer solutions.
Hint. Consider the cases $x=4 t, 4 t+1, 4 t+2$, $4 t+3$ | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,883 |
327. Solve the equation $5^{x}-13 y=8$ in integers. | 327. Answer. $x=4 t-1, y=\frac{5^{4 t-1}-8}{13}$, where $t-$ is an arbitrary positive integer. | 4-1,\frac{5^{4-1}-8}{13} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,884 |
328. Prove that the number $N=10 a+b$ is divisible by a number $d$ of the form:
a) $d=10 c+1$, if $a-b c$ is divisible by $d$;
b) $d=10 c+3$, if $a+b(3 c+1)$ is divisible by $d$;
c) $d=10 c+7$, if $a-b(3 c+2)$ is divisible by $d$;
d) $d=10 c+9$, if $a+b(c+1)$ is divisible by $d$. | 328. a) By the condition $a-b c=(10 c+1) k$, hence $10 a+b-b-10 b c=10 k(10 c+1), 10 a+b=b(10 c+1)+$ $+10 k(10 c+1), \frac{10 a+b}{10 c+1}=b+10 k$, i.e., an integer. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,885 |
329. The number $a^{100}$ when divided by 73 gives a remainder of 2, and the number $a^{101}$ when divided by the same number gives a remainder of 69. Find the remainder of the number $a$ when divided by 73. | 329. $a^{100}=73 m+2$, from which $a^{101}=73 n+2 a$ (1). In addition, $a^{101}=73 q+69(2)$. From (1) and (2) we find $73 k+2 a=$ $=69(3)$.
Noting that equation (3) is satisfied by the values $k=1$ and $a=-2$, we get $a=73 t-2$, which can be written as: $a=73 t+71$. The required remainder is 71. | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,886 |
330. Fermat's Little Theorem can be generalized as follows: if the number $a$ is coprime with $k$, then the difference $a^{\varphi(k)}-1$ is divisible by $k$, where $\varphi(k)$ is the number of numbers less than $k$ and coprime with it (Euler's theorem). Prove this theorem for the cases:
a) $k=p$ (prime number);
b) ... | 330. a) In this case $\varphi(k)=p-1, a^{p-1}-1$ is divisible by $p$ according to Fermat's little theorem.
b) $\varphi(k)=\varphi(p q)=(p-1)(q-1)$ (see № 170). Since $a$ is coprime with $k$, it is not divisible by either $p$ or $q$, so $a^{p-1}-1$ is divisible by $p$ and $a^{(p-1)(q-1)}-1$ is divisible by $p$; similar... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,887 |
331. Find the last two digits of the number $137^{42}$. | 331. The problem reduces to finding the remainder of the division of the number $137^{42}$ by 100. Since 137 and 100 are coprime, $137^{\varphi(100)}-1$ is divisible by $100, \quad$ but $\varphi(100)=$ $=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=40 ; \quad$ therefore, $\quad 137^{40}-1$ is divisible by 10... | 69 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,888 |
333. Find the remainder of the division of the number $\left(85^{70}+19^{32}\right)^{16}$ by 21. | 333. $85^{70}=(84+1)^{70}=21 n+1$; since $\varphi(21)=12$, then $19^{12}-1$ is divisible by 21; therefore, $19^{32}=$ $=19^{8}\left(19^{24}-1\right)+19^{8}=21 m+19^{8} . \quad$ Therefore $\quad\left(85^{70}+\right.$
$.\left.+19^{32}\right)^{16}=\left(21 n+1+21 m+19^{8}\right)^{16}=[21 k+1+(21-$
$.\left.-2)^{8}\right]^... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,890 |
335. Find the last two digits of the number $3^{5^{17}}$. | 335. Answer. 43.
Hint. The last digit of the number $a^{n}(n>4)$ is the same as the last digit of the number $a^{r}$, where $r>0$ is the remainder of the division of $n$ by 4; if $r=0$, then the last digit of the number $a^{n}$ coincides with the last digit of the number $a^{4}$ (see № 89). If $(a, 10)=1$, then the la... | 43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,892 |
336. Using Euler's theorem, find such values of $x$ for which $a x-b$ is divisible by $c$, if $(a, c)=1$. | 336. Since by condition $a$ and $c$ are coprime numbers, then $a^{\varphi(c)}-1$ is divisible by $c$, then $b a^{\varphi(c)}-b$ is also divisible by $c$, which can be written as: $a b a^{\varphi}(c)-1-b$, therefore, $x=b a^{\varphi}(c)-1$, hence $x=b a^{\varphi}(c)-1+c t$, where $t$ is any integer (see № 197). | ^{\varphi()-1}+ | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,893 |
337. For what values of $x$ is the number $7 x-3$ divisible by 12? | 337. $x=12 t+3 \cdot 7^{\varphi(12)-1}=12 t+1029 ; x=12 t_{1}+9$.
337. $x=12 t+3 \cdot 7^{\varphi(12)-1}=12 t+1029 ; x=12 t_{1}+9$. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,894 | |
338. For what values of $x$ is the number $3 x+7$ divisible by 14?
$339^{*}$. Euler's Criterion. Given that $(a, p)=1$, where $p$ is a prime number. For there to exist values of $x$ such that the difference $x^{2}-a$ is divisible by $p$, it is necessary and sufficient that $a^{\frac{p-1}{2}}-1$ is divisible by $p$.
T... | 338. Answer. $x=14 t+7$.
Note. In the case of a large numerical value of $\varphi(c)$, the above method is cumbersome; in this case, it is advisable to proceed as in № 209. | 14t+7 | Number Theory | proof | Yes | Yes | olympiads | false | 40,895 |
342. Show that if each of the equations $x^{2} - a = p y$ and $x^{2} - b = p y$ has a solution in integers, then the equation $x^{2} - a b = p y$ also has a solution in integers ( $p$ is a prime, $a$ and $b$ are integers not divisible by $p$ ). | 342. The numbers $a^{\frac{p-1}{2}}-1$ and $b^{\frac{p-1}{2}}-1$ are divisible by $p$ (see No. 204), then $a^{\frac{p-1}{2}}=p t+1$ and $b^{\frac{p-1}{2}}=p t_{1}+1$, hence $(a b)^{\frac{p-1}{2}}=$ $=p^{2} t t_{1}+(t+t_{1}) p+1$. Therefore, $(a b)^{\frac{p-1}{2}}-1$ is divisible by $p$, so the equation $x^{2}-a b=p y$ ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,898 |
343. Prove that the number $6 \cdot 2^{2^{4 n}}+1$ is composite for $n$ natural.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 343. $6 \cdot 2^{2^{4 n}}+1=6 \cdot 2^{16^{n}}+1=9 \cdot 2^{2 \cdot 16^{n}}+2 \cdot 3 \cdot 2^{16^{n}}+1-9 \cdot 2^{2 \cdot 16^{n}}=(3 \cdot 2^{16^{n}}+1)^{2}-(3 \cdot 2^{16^{n}})^{2}$ and so on. | Number Theory | MCQ | Yes | Yes | olympiads | false | 40,899 | |
347. Find an arithmetic progression of five prime numbers: a) with a difference of 6, b) with a difference of 12. | 347. The first term of the progression cannot be 2 and 3, because then the other terms would not be prime. The first term can be of the form $6k+1$ or $6k+5$. Let's consider these cases.
1) $a=6k+1$, such numbers can end in $1,3,5,7,9$. According to the condition, the terms of the progression must be prime numbers, so ... | 5,17,29,41,53 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,902 |
348. If $8 n+1$ and $24 n+1$ are perfect squares, then $8 n+3$ is a composite number for $n>1$. | 348. $\left\{\begin{array}{r}8 n+1=x^{2} \\ 24 n+1=y^{2}\end{array}\right.$, from here $3 x^{2}-y^{2}=2$.
$$
\begin{gathered}
8 n+3=(8 n+1)+2=x^{2}+2=x^{2}+\left(3 x^{2}-y^{2}\right)= \\
=(2 x+y)(2 x-y)
\end{gathered}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,903 |
349. If the numbers $p$ and $8 p^{2}+1$ are prime, then $8 p^{2}-p+2$ is also prime. | 349. The number $p$ can be equal to 3 or $3 k \pm 1$, where $k>0$.
For $p=3$, the number $8 p^{2}+1$ is prime and $8 p^{2}-p+2=71$ is also prime; if $p=3 k \pm 1$, then $8 p^{2}+1=8(3 k \pm 1)^{2}+$ $+1=72 k^{2} \pm 48 k+9$ is a composite number.
Therefore, $p$ cannot be of the form $3 k \pm 1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,904 |
350. Show that if $a$ and $b$ are integers, then $\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$ cannot be an integer. | 350. Let $(a, b)=d$, then $a=d a_{1}, b=d b_{1}$, and $\left(a_{1}, b_{1}\right)=1 . \quad \frac{a^{2}+b^{2}}{a^{2}-b^{2}}=\frac{a_{1}^{2}+b_{1}^{2}}{a_{1}^{2}-b_{1}^{2}}, \quad a_{1}^{2}-b_{1}^{2} \neq 1 \quad$ (see № 11). Further, $\left(a_{1}^{2}+b_{1}^{2}, a_{1}^{2}-b_{1}^{2}\right)=\left(2 a_{1}^{2}, 2 b_{1}^{2}\r... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,905 |
351. If $2^{n}+1$ is a prime number, then $n=0$ or $n=2^{\alpha}$. | 351. For $n$ odd, $2^{n}+1$ is divisible by $2+1$, hence $2^{n}+1$ is a composite number. Furthermore, $n$ cannot have an odd divisor, because for $n=2^{\alpha} q$, where $q$ is an odd number, $2^{2 x} q+1=\left(2^{2 \alpha}\right)^{q}+1$ is divisible by $2^{2^{\alpha}}+1$, therefore $n=2^{\alpha}$. For $n=0$ we get th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,906 |
352. If $a+b+1$ is a prime number, then one of the numbers $a!b!+1$ and $a!b!-1$ is divisible by $a+b+1$. | 352. By Wilson's theorem (see No. 201) $(a+b)!+1$ is divisible by $a+b+1$, but $(a+b)!=b!a!C_{a+b}^{a}$, therefore, $b!a!C_{a+b}^{a}+1$ is divisible by $a+b+1$.
If we take $a+b+1=p$, then $C_{a+b}^{a} \pm 1=C_{p-1}^{a} \pm 1=$ $=\frac{(p-1)(p-2)(p-3) \ldots(p-a) \pm 1 \cdot 2 \cdot 3 \cdots \cdot a}{1 \cdot 2 \cdot 3 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,907 |
353. If $n$ is odd, then $n^{3}+1$ cannot be a perfect square. | 353. Let $n^{3}+1=x^{2}$, hence $n^{3}=(x+1)(x-1)$. From the condition of the problem, it follows that $x+1$ and $x-1$ are odd numbers; their common divisor must be a divisor of their difference, which is 2; hence we conclude that the numbers $x+1$ and $x-1$ are coprime, therefore, $x+1=a^{3}$ and $x-1=b^{3}$, from whi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,908 |
355. The number $a^{2 q}-b^{2 q}$ either is not divisible by 2, or is divisible by 8, where $q>1$. | 355. For even $a$ and $b$, the given difference is divisible by 8; for odd, denoting $a=2 a_{1}+1$ and $b=2 b_{1}+1$, we get $a^{2 q}-b^{2 q}=\left[\left(2 a_{1}+1\right)^{2}\right]^{q}-\left[\left(2 b_{1}+1\right)^{2}\right]^{q}=(8 m+1)^{q}-(8 n+$ $+1)^{q}=8 r$ (by the binomial theorem), i.e., divisible by 8; for $a$ ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,910 |
1. Bank Check. A person came to the bank to get money by check. The cashier, paying the check, made a mistake and instead of the dollars owed, gave the same number of cents and accordingly, instead of cents - dollars. The person, without counting the money, put it in their pocket, and even dropped a 5 cent coin, and up... | 1. The check was issued for the amount of 31 dollars and 63 cents. The person received 63 dollars and 31 cents. After losing a five-cent coin, 63 dollars and 26 cents remained, which is twice the amount specified on the check. | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,912 |
3. Разменная монета. На какую наибольшую сумму можно взять мелкой монеты, чтобы не быть в состоянии разменять доллар, полдоллара, четверть доллара, 10 центов и 5 центов?* | 3. Наибольшая сумма равна 1 доллару 19 центам и составлена из одной монеты в полдоллара, одной монеты в четверть доллара, четырех монет по 10 центов и четырех монет по 1 центу. | 1доллар19центов | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,918 |
4. Charity. A generous person distributed the same amount of money equally among those who asked him for assistance every week. One day he noticed:
- If the number of petitioners is five less next week, each will receive two dollars more.
But, alas, after a week, the number of petitioners did not decrease but increas... | 4. At first, there were 20 people and each received 6 dollars. Fifteen people (five fewer) would have received 8 dollars each. But their number increased to 24 (by four people), and each received only 5 dollars. Thus, the weekly donation amount is 120 dollars. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,920 |
5. At the bakery. The bakery has three types of buns. For 1 cent, you can buy either one bun of the first type, or two buns of the second type, or, finally, three buns of the third type.[^1]
Children (among whom there were an equal number of boys and girls) received 7 cents to buy buns, with each child receiving the s... | 5. The group of children consisted of three boys and three girls. Each child received two rolls of the third grade and one roll of the second grade, the total cost of all the rolls is 7 cents. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,921 |
7. An Unusual Client. A certain man brought 1000 dollars in one-dollar bills and 10 empty bags to the bank and, addressing the clerk, said:
- I would appreciate it if you could distribute this money into the bags in such a way that any amount of money I need can always be given out in one or several bags without openi... | 7. Ten bags should contain respectively \(1,2,4,8\), \(16,32,64,128,256\) and 489 one-dollar bills. The first nine numbers form a geometric progression. If the sum of this progression is subtracted from 1000, the result is the contents of the tenth bag. | 1,2,4,8,16,32,64,128,256,489 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,923 |
8. The "Reverse" Game. Seven friends decided to play cards by somewhat unusual rules. The one who won had to pay each of the other players as much money as they had in their pocket. The players played seven rounds and, strangely enough, they won in alphabetical order of their names, starting with $A, B, C, D, E, F$ and... | 8. Players \(A, B, C, D, E, F\) and \(G\) had, before the game began, 4 dollars 49 cents, 2 dollars 25 cents, 1 dollar 13 cents, 57 cents, 29 cents, 15 cents, and 8 cents, respectively. The answer can be obtained by working from the end of the problem to the beginning, but a simpler method is as follows: \(7+1=8 ; 2 \t... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,924 |
9. Two Diggers. This curious problem is actually more difficult than it might seem at first glance. Abraham, a frail old man, contracted to dig a ditch for 2 dollars. He hired Benjamin, a strapping young man, to help him. The money was to be divided in proportion to each man's digging abilities. Abraham digs as fast as... | 9. Abraham \((A)\) should receive one third of the total sum, while Benjamin (B) should receive two thirds. For instance, let \(B\) be able to dig a ditch in 2 hours and throw out all the soil in 4 hours. Then \(A\) can dig a ditch in 4 hours and throw out all the soil in 8 hours. Therefore, when digging the ditch, the... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,925 |
10. Name their wives. A man left three of his relatives and their wives 1000 dollars in his will. The wives together received 396 dollars. Jane received 10 dollars more than Catherine, and Mary received 10 dollars more than Jane. John Smith received as much as his wife, Henry Snooks received one and a half times more t... | 10. Katherine, Jane, and Mary received 122, 132, and 142 dollars respectively, which together make up the total amount of their share of the inheritance, 396 dollars. According to the problem, John Smith receives the same amount as his wife Katherine (122 dollars). Henry Snooks receives one and a half times more than h... | Katherine,Jane,Mary | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,926 |
11. Market transactions. A farmer bought 100 heads of cattle at a total cost of 1000 dollars on the market. One cow cost 50 dollars, one sheep - 10 dollars, and one rabbit - 50 cents. How much money did the farmer spend on buying cows, sheep, and rabbits separately?
This problem can be solved through more or less meti... | 11. A farmer bought 19 cows for 950 dollars, 1 sheep for 10 dollars, and 80 rabbits for 40 dollars, which totals 100 heads with a combined value of 1000 dollars. The problem can be easily solved arithmetically using the method of averages: the average cost per head of livestock is the same as the cost of one sheep. Alg... | 19,1,80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,927 |
12. Seven apple vendors. Seven vendors had 20, 40, 60, 80, 100, 120, and 140 apples respectively. They went to the market and sold all their apples at the same price, earning the same revenue. At what price did the vendors sell the apples? | 12. All seven vendors sold apples at 1 cent for 7 pieces; in cases where fewer than seven apples remained, they were sold at 3 cents per piece. Thus, each vendor earned 20 cents. Without in any way disputing the ingenuity of this problem, I have always found its solution unsatisfactory due to its ambiguity, even if we ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,928 |
15. Division of inheritance. One person left an inheritance of 100 dollars, which had to be divided between his sons Alfred and Benjamin. If one third of Alfred's share is subtracted from one quarter of Benjamin's share, 11 dollars will remain. What is the share of each son? | 15. Alfred's and Benjamin's shares are equal to 24 and 76 dollars, respectively. Indeed, if 8 (one third of 24) is subtracted from 19 (one quarter of 76), then 11 remains. | 2476 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,931 |
16. A New Partner. Two partners, Smag and Williamson, decided to take on a third partner, Mr. Rogers. Smag invested $1 \frac{1}{2}$ times more capital than Williamson. Rogers must contribute 2500 dollars, which should be divided between the two other partners so that the shares of all three partners become equal after ... | 16. The $2500 that Rogers contributed to the business clearly constitutes a third of the total capital, which, therefore, before his entry, amounted to $7500. Consequently, Smaga's share was $4500 (1.5 times more than Williamson's share), and Williamson's share was $3000. Since their shares should become equal, Smaga w... | Smaga\,will\,receive\,2000,\,\,Williamson\,will\,receive\,500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,932 |
18. Distribution of Money. Nine friends $A, B, C, D, E, F, G$, $H, K$, gathering one evening to spend time together, did the following. First, $A$ gave each of the other eight people as much money as they already had. Then $B, C$, and so on up to $K$ did the same. After this, it turned out that all nine friends had the... | 18. The smallest sum (in cents) that one of the participants could have must exceed the number of participants by one. The sums belonging to the other participants can be found by sequential doubling and subtracting 1. Thus, we get \(10,19,37,73,145,289,577,1153\) and 2305 cents. Let the one with the most money start f... | 512 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,934 |
## 19. Price Reduction.
- It often puzzles me, - said Colonel Crackham, - the curious system of price reduction one sometimes encounters, and I have been trying to understand its pattern. For example, two years ago a man offered me a motorcycle for $1024. A year later he reduced the price to $640, a little later he as... | 19. The seller reduced the price by \(\frac{3}{8}\) of the motorcycle's cost each time. Consequently, with the next reduction, he will offer a price of 156 dollars and 25 cents. | 156.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,935 |
20. Horses and Oxen. A livestock trader bought a certain number of horses at $344 each and a certain number of oxen at $265 each. He found that the total cost of the horses was $33 more than the total cost of the oxen. What is the smallest number of horses and oxen he could have bought under these conditions? | 20. The task reduces to solving the indeterminate equation \(344 x=265 y+33\). The methods for solving such equations are well known, so we will not dwell on them. Solving the equation, we find that \(x=252\) and \(y=327\). Thus, if the trader buys 252 horses for 344 dollars each and 327 oxen for 265 dollars each, the ... | 252 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,936 |
21. Turkeys. A farmer bought a batch of turkeys, which cost 60 dollars. Leaving 15 birds for himself, the farmer sold the rest of the turkeys for 54 dollars. In doing so, he made a profit of 10 cents per bird. How many turkeys did he buy? | 21. Seventy-five turkeys were bought at 80 cents each, for a total of 60 dollars. Leaving 15 birds for himself, the farmer sold the remaining 60 turkeys at 90 cents each, for a total of 54 dollars, as required. Thus, he made a profit of 10 cents on each of the 60 resold birds. | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,937 |
22. The Unfortunate Grocer. A grocer, the owner of a small shop, decided to set aside a small amount of money - all in dollar bills and coins of fifty cents and twenty-five cents. He distributed the entire amount into 8 bags, so that in each bag there was the same number of dollar bills and coins of each denomination. ... | 22. The grocer set aside 168 paper dollars, 168 half-dollar coins, and 168 quarter-dollar coins for a rainy day, totaling 294 dollars. In each of the six bags, there should be 28 monetary units of each type; in each of the seven bags, 24; and in each of the eight bags, 21 monetary units of each type. | 294 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,938 |
23. The Lost Cent. This is an old problem that can still stump many people today. Two traders were selling apples, one at three for a cent, and the other at two for a cent. They had to leave for a while. Each still had 30 unsold apples, which they entrusted to a friend to sell at 2 cents for five apples. If the traders... | 23. The explanation is simple. Each method of selling will lead to the same results only if the number of apples sold at three for a cent is to the number of apples sold at two for a cent as 3 to 2.
For example, if the first vendor had 36 apples left, and the second had 24, the revenue would be 24 cents regardless of ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,939 |
24. The League of Red Death. During a raid on the headquarters of a secret organization, the police found a scrap of paper, depicted in the image.
- "I've been working on this piece of pap... | 24. The total amount of contributions, expressed in cents, is 300737. This number can be represented as the product of two prime factors: 311 and 967. Since we know that the League of Red Death has no more than 500 members, the number of members is 311, and the contribution is 967 cents, or 9 dollars and 67 cents.
The... | 311 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,940 |
26. Boys and Girls. Nine boys and three girls decided to share their pocket money equally. Each boy gave the same amount to each girl, and each girl also gave the same amount (but a different one) to each boy. After this, all the children had the same amount of money. What is the smallest amount of money that each of t... | 26. Each boy initially had 12 cents, and he gave 1 cent to each girl. Each girl had 36 cents, from which she gave 3 cents to each boy. After this, each child had 18 cents. | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,942 |
30. A Thriving Business. A businessman initially invested $2000 in his business. Every 3 years, he increased his capital by $50\%$. What was the amount of his capital after 18 years? | 30. After 18 years, the capital amounted to 22,781 dollars and 25 cents. | 22781.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,946 |
31. The Banker and the Counterfeit Bill. A banker was walking down the street of a small provincial town when he suddenly saw a $5 bill on the sidewalk. He picked it up, noted the number, and went home for breakfast. Over breakfast, his wife informed him that the butcher had sent a bill for $5. Since the banker had no ... | 31. Since the same counterfeit banknote was involved in all transactions, they all turned out to be invalid. Consequently, each person remained in the same position relative to their debtor as they were before the banker found the banknote. Moreover, the butcher still owes the farmer 5 dollars for the calf*. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,947 |
33. The Wilson Family. Mrs. Wilson had three children: Edgar, James, and John. Half of her age was equal to the sum of the ages of all the children. Five years later, when another daughter, Ethel, was born, Mrs. Wilson's age became equal to the sum of the ages of all her children. Another ten years passed, and a daught... | 33. Mrs. Wilson is 39 years old, Edgar is 21, James is 18, John is 18, Ethel is 12, Daisy is 9 years old. It is clear that James and John are the same age. | Mrs.Wilsonis39old,Edgaris21,Jamesis18,Johnis18,Ethelis12,Daisyis9old | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,949 |
34. De Morgan and Others. Mathematician Augustus De Morgan, who died in 1871, used to say that he was $x$ years old in the year $x^{2}$. Jasper Jenkins, wishing to outdo him, told me in 1925 that he was $\mathrm{a}^{2}+b^{2}$ years old in the year $\mathrm{a}^{4}+b^{4}$, that his age was $2 m$ in the year $2 m^{2}$, an... | 34. De Morgan was born in 1806. When he was 43 years old, the current year equaled the square of his age - 1849. Jenkins was born in 1860. He was \(5^{2}+6^{2}(61)\) years old in the \(5^{4}+6^{4}(1921)\) year. In the \(2 \times 31^{2}\) (1922) year, he turned \(2 \times 31\) (62) years old. Finally, his age was \(3 \t... | 18061860 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,950 |
38. Mike's Age. "To Pat O'Connor," said Colonel Crackham, "it is now $1 \frac{1}{3}$ times as many years as it was when he built the pigsty under the window of his drawing-room. Little Mike, who was 3 years and 4 months old when Pat built the pigsty, is now 2 years older than half the age Biddee, Pat's wife, was when P... | 38. Mike is currently \(10 \frac{16}{21}\), Pete \(29 \frac{16}{21}\), and Biddy \(24 \frac{20}{21}\) years old. When Pete built a pigsty under the window of his living room ( \(7 \frac{9}{21}\) years ago), Mike was \(3 \frac{7}{21}\), Pete \(22 \frac{6}{21}\), and Biddy \(17 \frac{11}{21}\) years old. In \(11 \frac{11... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,954 | |
41. "Square" family. One person had 9 children, all of whom were born at equal intervals, and the sum of the squares of their ages equaled the square of his own age. How many full years old was each child? | 41. The children were respectively \(2, 5, 8, 11, 14, 17, 20, 23, 26\) years old, and the father was 48 years old. | 2,5,8,11,14,17,20,23,26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,957 |
43. Find out the birthday. One reader informed us that by noon on November 11, 1928, he had lived in the 19th century exactly as long as in the 20th. Of course, we wanted to find out his date of birth. Can you do it too? We will assume that he was born at noon. | 43. The reader was born at noon on February 19, 1873, and by noon on November 11, 1928, had lived \(10176 \frac{1}{2}\) days in each century. Of course, the 19th century ended at midnight on December 31, 1900, which was not a leap year, and on November 11, 1928, the reader was 55 years and (approximately) 9 months old. | February19,1873 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,959 |
44. The Birth of Boudicca. Boudicca ${ }^{*}$ died 129 years after the birth of Cleopatra. The sum of their ages (that is, the sum of the duration of each of their lives) was one hundred years. Cleopatra died in 30 BC. When was Boudicca born? | 44. Between the birth of Cleopatra and the death of Boudicca, 129 years passed, but since their combined age was only 100 years, there was a period of 29 years when neither of them was alive (i.e., the period between Cleopatra's death and Boudicca's birth). Therefore, Boudicca was born 29 years after Cleopatra's death,... | 1AD | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,960 |
46. The Clocks from the Land of Dreams. In a dream, I traveled to a country where wondrous things happened. One incident was so memorable that I didn't forget it even after I woke up. In my dream, I saw a clock and spoke aloud the time it showed, but my guide corrected me. He said:
- Clearly, you don't know that our m... | 46. If these were ordinary clocks, they would show 4 hours \(23 \frac{1}{13}\) minutes. But since the minute hand moved in the direction opposite to the hour hand, the true time was 4 hours \(36 \frac{12}{13}\) minutes. To get the true time, you need to subtract from 60 the number of minutes that the clock shows. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,962 |
47. When does this happen? When the clock hands are arranged in such a way that if the distance is taken as the number of minute divisions after XII, then the path traveled by one of the hands is equal to the square of the path traveled by the other? | 47. This happens at 9:06 \(\frac{3}{4}\) when the hour hand has traveled a distance of \(45 \frac{9}{16}\) (which is \(6 \frac{3}{4}\) squared) minute divisions (after XII). If we allowed fractions less than one minute, there would be another solution, namely: 12:05 ( \(\frac{1}{12}\) min). | 9:06\frac{3}{4} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,963 |
48. Indistinguishable Clock Hands. A person had a clock on which it was impossible to distinguish the hour hand from the minute hand. If this clock was started at noon, when would it first be impossible to tell the exact time?
The reader should remember that in such clock puzzles, there is a convention that we are abl... | 48. This will first occur at 12 hours \(5 \frac{5}{143}\) minutes, which could be incorrectly interpreted (due to the identical positions of the hands) as 1 hour \(\frac{60}{143}\) minutes. | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,964 |
## 50. When did the ball start?
- At the last ball, - Dora said during breakfast, - the guests thought the clocks had stopped: the hands were in the same position as at the beginning of the evening. It turned out that the hour and minute hands had simply switched places. As you remember, the ball started between ten a... | 50. The evening began at 10:59 \( \frac{83}{143} \) min, and when the guests looked at the hands, which had switched places, they showed 11:54 \( \frac{138}{143} \) min. | 10:59\frac{83}{143} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,966 |
## 51. Mixed-Up Hands.
- Yesterday, between two and three o'clock, - said Colonel Crackham, - I looked at my watch and, mistaking the hour hand for the minute hand, I was wrong in my estimate of the time. The erroneous time was 55 minutes less than the actual time. What was the actual time? | 51. The true time was 2 hours \(5 \frac{5}{11}\) minutes. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,967 |
52. Equal Distances. A few days ago, Professor Rackbrain stunned his students with the following puzzle:
- When between three and four o'clock is the minute hand at the same distance from VIII as the hour hand is from XII? | 52. At 3 hours \(23 \frac{1}{13}\) minutes. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,968 |
53. To the right and to the left. At what time between three and four o'clock is the minute hand the same distance to the left of XII as the hour hand is to the right of XII? | 53. At 3 hours \(41 \frac{7}{13}\) minutes. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,969 |
54. At a Right Angle. Once during breakfast, Professor Rackbrain asked his young friends a simple question:
- When between five and six o'clock will the hour and minute hands be exactly at a right angle? | 54. For the angle between the hands to be a right angle, the minute hand must be exactly 15 minutes ahead of or behind the hour hand. Each of these positions will occur 11 times in 12 hours, that is, every 1 hour \(5 \frac{5}{11}\) minutes. If eight such intervals pass after 9 o'clock, the clock will show 5 hours \(43 ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,970 |
55. Westminster Clocks. One morning, a man was walking across Westminster Bridge, judging by the clock tower (which is often mistakenly called Big Ben, although this name refers only to the large bell; but this is by the way) between eight and nine o'clock. Returning between four and five o'clock, he noticed that the h... | 55. At 8:23 \(\frac{71}{143}\) min and at 4:41 \(\frac{137}{143}\) min. In clock puzzles, we assume that fractional parts of a minute can be determined on the clock. | 8:23\frac{71}{143} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,971 |
56. Up the Hill. Willie-Lazybones climbed up the hill at a speed of $1 \frac{1}{2}$ km/h, and descended at a speed of $4 \frac{1}{2}$ km/h, so the entire journey took him exactly 6 hours. How many kilometers is it from the base to the top of the hill? | 56. To the top of the hill is \(6 \frac{3}{4}\) km. Upwards, Willy-Lazybones climbed for \(4 \frac{1}{2}\) hours, and downwards he descended in \(1 \frac{1}{2}\) hours. | 6\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,972 |
## 57. The Speed of a Car.
- I was walking down the road at a speed of $3 \frac{1}{2}$ km/h,- said Mr. Pipkins,- when suddenly a car sped past me, nearly knocking me off my feet ${ }^{2}$.
- How fast was it going? - asked his friend.
- I'll tell you. From the moment it sped past me until it disappeared around the bend... | 57. Since a person takes 27 steps in the time it takes a car to travel a distance of 162 steps, it is clear that the car is moving 6 times faster than the person. The person moves at a speed of \(3 \frac{1}{2}\) km \(/ h\); therefore, the speed of the car is 21 km \(/ h\). | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,973 |
59. A Walk. One man set out for a walk from Appleminster to Boniham at noon, and his friend left Boniham for Appleminster at two o'clock of the same day. On the way, they met. The meeting took place at five minutes to five, after which the friends arrived simultaneously at their final destinations. When did they finish... | 59. It should be noted (and this is the key to the solution) that the person from B. covers 7 km in the same time that the person from E. covers 5 km. Let's assume, for example, that the distance between the cities is 24 km, then they met 14 km from E. The person from E. was moving at a speed of \(3 \frac{3}{7}\) km/h,... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,975 |
60. Riding in windy weather. A cyclist takes 3 minutes to travel 1 km if the wind is at his back, and 4 minutes if he has to ride against a headwind. How long will it take him to travel 1 km if the wind dies down? Someone might say that since the arithmetic mean of 3 and 4 is $3 \frac{1}{2}$, the cyclist will need $3 \... | 60. The cyclist will cover one kilometer in \(3 \frac{3}{7}\) minutes, or at a speed of \(\frac{7}{24}\) km/min. The wind changes his speed by \(\frac{1}{24}\) km/min. Therefore, with the wind, he moves at a speed of \(\frac{8}{24}\) km/min, and against the wind - at a speed of \(\frac{6}{24}\) km/min, so he covers 1 k... | 3\frac{3}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,976 |
61. The rowers' puzzle. A team of rowers can cover a given distance against the current in $8 \frac{4}{7}$ minutes. In the absence of the current, the same distance is covered in 7 minutes less than it would take to cover it with the current. How many minutes does it take the team to cover the given distance with the c... | 61. For \(3 \frac{9}{17}\) min. The team in still water covers \(\frac{1}{5}\) of the total distance per minute, while the current contributes \(-\frac{1}{12}\) of the total distance per minute. The difference and sum of these fractions are \(\frac{7}{60}\) and \(\frac{17}{60}\), respectively. Therefore, the journey ag... | 3\frac{9}{17} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,977 |
62. Escalator. While on one of the escalators in the London Underground, I found that by walking 26 steps, I would reach the platform in $30 \mathrm{s}$. But if I walked 34 steps, the entire descent would take 18 s. How many steps are there on the escalator? Time is measured from the moment the top step begins to desce... | 62. If I walk down 26 steps, it will take me 30 seconds to descend, and if 34, it will take -18 seconds. Multiplying 30 by 34 and 26 by 18, we get 1020 and 468, the difference between these numbers is 552. Dividing this by the difference between 30 and 18 (that is, by 12), we get 46, the number of steps on the escalato... | 46 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,978 |
63. One bicycle for two. Two brothers had to set off on a journey and arrive at their destination simultaneously. They had only one bicycle, on which they took turns riding, and when the rider's time was up, he would dismount and leave the bicycle by the fence, then walk forward without waiting for his brother, while t... | 63. Let Anderson travel \(11 \frac{1}{9}\) km, leave the bicycle, and walk the remaining part of the journey. Brown will walk until he picks up the bicycle and then ride the remaining part of the way. In this case, he will arrive at the destination simultaneously with Anderson, and the entire journey will take them 3 h... | 11\frac{1}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,979 |
64. Back to the bicycle. Let's add a third participant who uses the same bicycle. Suppose Anderson and Brown have brought a person named Carter. They walk at 4.5, 3, and 3 km/h respectively, and ride the bicycle at 10.8, 12, and 12 km/h. How should they use the bicycle so that they all cover the distance of 20 km in th... | 64. Anderson travels \(7 \frac{11}{27}\), Brown \(1 \frac{13}{27}\), and Carter \(11 \frac{3}{27}\) km, which in total is 20 km. They can travel in any order, but each must use the bicycle only once, and the second rider must walk both before and after the ride. The journey will take each \(3 \frac{8}{9}\) hours, and t... | 3\frac{8}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,980 |
65. Motorcycle with a sidecar. Atkins, Baldwin, and Clark decided to go on a trip. Their journey will be 52 km. Atkins has a motorcycle with a one-person sidecar. He needs to take one of his companions for some distance, drop him off to walk the rest of the way, return, pick up the other companion who started walking a... | 65. Atkins drives Clark 40 km and drops him off to walk the remaining 12 km. Then he returns and picks up Baldwin 16 km from the start and drives him to the end. The three of them spend 5 hours on the road. Another solution is for Atkins to first drive Baldwin 36 km and return for Clark, who by this time has walked 12 ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,981 |
66. Communication. An army column 40 km long travels 40 km. How many kilometers will the communication officer travel, sent with a package from the rearguard to the vanguard and returning back? | 66. The distance traveled by the messenger is equal to the square root of twice the square of 40, added to 40, which amounts to 96.568 km, or approximately \(96 \frac{1}{2}\) km. | 96.568 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,982 |
67. Two Trains. Two railway trains, one 400 feet long and the other 200 feet long, are moving on parallel tracks. When they move in opposite directions, each passes the other in 5 seconds, and when they move in the same direction, the faster one passes the other in 15 seconds. A curious passenger, using this data, mana... | 67. The relative speed of two oncoming trains is 600 feet in 5 s, or \(81 \frac{9}{11}\) miles/h. When the trains are moving in the same direction, their relative speed is 600 feet in \(15 \mathrm{s}\), or \(27 \frac{3}{11}\) miles/h. From this, we find that the speed of the faster train is \(54 \frac{6}{11}\) miles/h,... | 54\frac{6}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,983 |
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