p1. It is known that $m$ and $n$ are two positive integer numbers consisting of four digits and three digits respectively. Both numbers contain the number $4$ and the number $5$. The number $59$ is a prime factor of $m$. The remainder of the division of $n$ by $38$ is $ 1$. If the difference between $m$ and $n$ is not more than $2015$. determine all possible pairs of numbers $(m,n)$. p2. It is known that the equation $ax^2 + bx + c = $0 with $a> 0$ has two different real roots and the equation $ac^2x^4 + 2acdx^3 + (bc + ad^2) x^2 + bdx + c = 0$ has no real roots. Is it true that $ad^2 + 2ad^2 <4bc + 16c^3$ ? p3. A basketball competition consists of $6$ teams. Each team carries a team flag that is mounted on a pole located on the edge of the match field. There are four locations and each location has five poles in a row. Pairs of flags at each location starting from the far right pole in sequence. If not all poles in each location must be flagged, determine as many possible flag arrangements. p4. It is known that two intersecting circles $L_1$ and $L_2$ have centers at $M$ and $N$ respectively. The radii of the circles $L_1$ and $L_2$ are $5$ units and $6$ units respectively. The circle $L_1$ passes through the point $N$ and intersects the circle $L_2$ at point $P$ and at point $Q$. The point $U$ lies on the circle $L_2$ so that the line segment $PU$ is a diameter of the circle $L_2$. The point $T$ lies at the extension of the line segment $PQ$ such that the area of the quadrilateral $QTUN$ is $792/25$ units of area. Determine the length of the $QT$. p5. An ice ball has an initial volume $V_0$. After $n$ seconds ($n$ is natural number), the volume of the ice ball becomes $V_n$ and its surface area is $L_n$. The ice ball melts with a change in volume per second proportional to its surface area, i.e. $V_n - V_{n+1} = a L_n$, for every n, where a is a positive constant. It is also known that the ratio between the volume changes and the change of the radius per second is proportional to the area of the property, that is $\frac{V_n - V_{n+1}}{R_n - R_{n+1}}= k L_n$ , where $k$ is a positive constant. If $V_1=\frac{27}{64} V_0$ and the ice ball melts totally at exactly $h$ seconds, determine the value of $h$.
Problem
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Tags: algebra, geometry, combinatorics, number theory, indonesia juniors
BackToSchool
10.11.2021 04:58
First Step, find possible value of $n$.
$$n\equiv 1 \pmod {38} \implies n \equiv 1 \pmod 2$$$$\implies n=450+k \text { or } 540+k, k\in \{1, 3, 5, 7. 9\}$$$$\boxed {n=457} = 38 \cdot 12+1$$
Second Step, find possible value of $m$.
$$1000\le m \le 2472$$We can do casework for various places of $4$ and $5$.
Casework #$1$
$m=1450+k, k\in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, 1450 = 1416+34=24\cdot59+34$. There is no solution in this case.
Casework #$2$
$m=1540+k, k\in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, 1540 = 1534+6=26\cdot59+6$. There is no solution in this case.
Casework #$3$
$m=2450+k, k\in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, 2450 = 2419+31=41\cdot59+31$. There is no solution in this case.
Casework #$4$
$m=2540+k, k\in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, 2450 = 2537+3=43\cdot59+3$. There is no solution in this case.
Casework #$5$
$m=1405+k, k\in \{0, 10, 20, 30, 40, 50, 60, 70, 80, 90\}, 1405 = 1357+48=23\cdot59+48$. Thus, $1405+70=1475$ is a solution.
Casework #$6$
$m=1504+k, k\in \{0, 10, 20, 30, 40, 50, 60, 70, 80, 90\}, 1504 = 1475+29=25\cdot59+29$. Thus, $1504+30=1534$ is a solution.
Casework #$7$
$m=2405+k, k\in \{0, 10, 20, 30, 40, 50, 60, 70, 80, 90\}, 2405 = 2360+45=49\cdot59+45$. There is no solution in this case.
Casework #$8$
$m=2504+k, k\in \{0, 10, 20, 30, 40, 50, 60, 70, 80, 90\}, 2504 = 2478+26=51\cdot59+26$. There is no solution in this case.
Casework #$9$
$m=1045+100k, 0 \le k \le 24$. The units digit of the multiple should be $5$.
$15\cdot59=885 $.This is not a solution.
$25\cdot59=1475$, same as above.
$35\cdot59=2065$.This is not a solution.
$45\cdot59=2655$.This is not a solution.
Casework #$10$
$m=1054+100k, 0 \le k \le 24$. The units digit of the multiple should be $6$.
$16\cdot59=944 $.This is not a solution.
$26\cdot59=1534$, same as above.
$36\cdot59=2124$.This is not a solution.
$46\cdot59=2714$.This is not a solution.
To sum up, There are two pairs of $\boxed{(m, n) \in \{(1475, 457), (1534, 457)\}}$.
Anybody has a smarter solution?
amogususususus
19.07.2022 12:01
BackToSchool wrote:
First Step, find possible value of $n$.
$$n\equiv 1 \pmod {38} \implies n \equiv 1 \pmod 2$$$$\implies n=450+k \text { or } 540+k, k\in \{1, 3, 5, 7. 9\}$$$$\boxed {n=457} = 38 \cdot 12+1$$
Second Step, find possible value of $m$.
$$1000\le m \le 2472$$We can do casework for various places of $4$ and $5$.
$n=495$ ?
Casework #$1$
$m=1450+k, k\in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, 1450 = 1416+34=24\cdot59+34$. There is no solution in this case.
Casework #$2$
$m=1540+k, k\in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, 1540 = 1534+6=26\cdot59+6$. There is no solution in this case.
Casework #$3$
$m=2450+k, k\in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, 2450 = 2419+31=41\cdot59+31$. There is no solution in this case.
Casework #$4$
$m=2540+k, k\in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, 2450 = 2537+3=43\cdot59+3$. There is no solution in this case.
Casework #$5$
$m=1405+k, k\in \{0, 10, 20, 30, 40, 50, 60, 70, 80, 90\}, 1405 = 1357+48=23\cdot59+48$. Thus, $1405+70=1475$ is a solution.
Casework #$6$
$m=1504+k, k\in \{0, 10, 20, 30, 40, 50, 60, 70, 80, 90\}, 1504 = 1475+29=25\cdot59+29$. Thus, $1504+30=1534$ is a solution.
Casework #$7$
$m=2405+k, k\in \{0, 10, 20, 30, 40, 50, 60, 70, 80, 90\}, 2405 = 2360+45=49\cdot59+45$. There is no solution in this case.
Casework #$8$
$m=2504+k, k\in \{0, 10, 20, 30, 40, 50, 60, 70, 80, 90\}, 2504 = 2478+26=51\cdot59+26$. There is no solution in this case.
Casework #$9$
$m=1045+100k, 0 \le k \le 24$. The units digit of the multiple should be $5$.
$15\cdot59=885 $.This is not a solution.
$25\cdot59=1475$, same as above.
$35\cdot59=2065$.This is not a solution.
$45\cdot59=2655$.This is not a solution.
Casework #$10$
$m=1054+100k, 0 \le k \le 24$. The units digit of the multiple should be $6$.
$16\cdot59=944 $.This is not a solution.
$26\cdot59=1534$, same as above.
$36\cdot59=2124$.This is not a solution.
$46\cdot59=2714$.This is not a solution.
To sum up, There are two pairs of $\boxed{(m, n) \in \{(1475, 457), (1534, 457)\}}$.
Anybody has a smarter solution? $n=495$ ?
muhallif
18.07.2023 09:18
Notice that $ac^2x^4+2acdx^3+(bc+ad^2)x^2+bdx+c=a(c^2x^4+2cdx+d^2x^2)+b(cx^2+dx)+c=a(cx^2+dx)^2+b(cx^2+dx)+c=0$.