Beginner's Marathon
Re: Beginner's Marathon
Solution to Problem $18$:
The $21$ answer the host received were different. Then all the handshake numbers occur from $0$ to $20$. Let's name the people according to their handshake number  $H_0, H_1,....,H_{20}$.
Here, $H_0$ shakes hands with no one. And $H_{20}$ shakes hands with everyone except its spouse. So, $H_0$ and $H_{20}$ must be each other's spouses.
Eliminating $H_0$ and $H_{20}$ from the party, we now have a party of $9$ couples. $H_1$ is now least social person, his only handshake being with $H_{20}$. And $H_{19}$ shook hands with everyone except his spouse and $H_0$. So, $H_1$ and $H_{19}$ must be spouses.
Going on and on, only $H_{11}$ remains single. So, $H_{11}$ must be the hostess.
The $21$ answer the host received were different. Then all the handshake numbers occur from $0$ to $20$. Let's name the people according to their handshake number  $H_0, H_1,....,H_{20}$.
Here, $H_0$ shakes hands with no one. And $H_{20}$ shakes hands with everyone except its spouse. So, $H_0$ and $H_{20}$ must be each other's spouses.
Eliminating $H_0$ and $H_{20}$ from the party, we now have a party of $9$ couples. $H_1$ is now least social person, his only handshake being with $H_{20}$. And $H_{19}$ shook hands with everyone except his spouse and $H_0$. So, $H_1$ and $H_{19}$ must be spouses.
Going on and on, only $H_{11}$ remains single. So, $H_{11}$ must be the hostess.
Re: Beginner's Marathon
Problem $19$:
If $127$ people play in a singles tennis tournament, prove that at the end of the tournament, the number of people who have played an odd number of games is even.
If $127$ people play in a singles tennis tournament, prove that at the end of the tournament, the number of people who have played an odd number of games is even.

 Posts: 57
 Joined: Sun Dec 11, 2016 2:01 pm
Re: Beginner's Marathon
$\text{Problem 19}$
Let,the number of games played by a person $i$ be $g_i$.Therefore,the sum $g_1$+$g_2$+......+$g_{127}$ is even as each game is counted twice.We,assume that the number of people who played an odd numbered games is odd.In that case,if we eliminate their played games from the sum,we get that the left over teams played an odd numbered games as there are different parities.But that contradicts the fact that there are in that case even numbered leftover people who played even number of games.So,our assumption was wrong and there is an even number of people who played an odd numbered games.
Let,the number of games played by a person $i$ be $g_i$.Therefore,the sum $g_1$+$g_2$+......+$g_{127}$ is even as each game is counted twice.We,assume that the number of people who played an odd numbered games is odd.In that case,if we eliminate their played games from the sum,we get that the left over teams played an odd numbered games as there are different parities.But that contradicts the fact that there are in that case even numbered leftover people who played even number of games.So,our assumption was wrong and there is an even number of people who played an odd numbered games.

 Posts: 57
 Joined: Sun Dec 11, 2016 2:01 pm
Re: Beginner's Marathon
An easy problem:$\text{Problem 20}$
The fractions $\frac{7x+1}{2}$, $\frac{7x+2}{3}$,......., $\frac{7x+2016}{2017}$ are irreducible.Find all possible values of $x$ such that $x$ is less than or equal to $300$.This Problem was recommended by Zawad bhai in a mock test at CMC.
The fractions $\frac{7x+1}{2}$, $\frac{7x+2}{3}$,......., $\frac{7x+2016}{2017}$ are irreducible.Find all possible values of $x$ such that $x$ is less than or equal to $300$.This Problem was recommended by Zawad bhai in a mock test at CMC.
 Atonu Roy Chowdhury
 Posts: 64
 Joined: Fri Aug 05, 2016 7:57 pm
 Location: Chittagong, Bangladesh
Re: Beginner's Marathon
Here's a hint for you guys.aritra barua wrote:An easy problem:$\text{Problem 20}$
The fractions $\frac{7x+1}{2}$, $\frac{7x+2}{3}$,......., $\frac{7x+2016}{2017}$ are irreducible.Find all possible values of $x$ such that $x$ is less than or equal to $300$.This Problem was recommended by Zawad bhai in a mock test at CMC.
This was freedom. Losing all hope was freedom.
Re: Beginner's Marathon
And since $x \le 300$, the answer are all the primes of the form $7x1$ between $2018$ and $2099$.Atonu Roy Chowdhury wrote:Here's a hint for you guys.aritra barua wrote:An easy problem:$\text{Problem 20}$
The fractions $\frac{7x+1}{2}$, $\frac{7x+2}{3}$,......., $\frac{7x+2016}{2017}$ are irreducible.Find all possible values of $x$ such that $x$ is less than or equal to $300$.This Problem was recommended by Zawad bhai in a mock test at CMC.
$\boxed{\text{Problem }21}$
Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ for which $a_i +a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.
Re: Beginner's Marathon
$Solution to Problem 21$
Let $a_1 < a_2 < a_3 < a_4$. Then, $(a_3+a_4)$ can't divide $(a_1+a_2)$ and $(a_2+a_4)$ can't divide $(a_1+a_3)$.
So, $n_A$ < 4. So, to get the max value 4, $(a_1+a_3)(a_2+a_4)$, $(a_1+a_2)(a_3+a_4)$, $(a_1+a_4)(a_2+a_3)$ and $(a_2+a_3)(a_1+a_4)$. So, $a_1+a_4 = a_2+a_3$.
Since, $(a_1+a_3)(a_2+a_4)$, we can write, $w(a_1+a_3) = a_2+a_4 = a_2+(a_2+a_3a_1) = 2a_2+a_3a_1$.
=> $a_1(w+1) + a_3(w1) = 2a_2$. w > 1, else the numbers aren't distinct. If w > 2, $w(a_1+a_3)$ is larger than $2a_2+a_3a_1$. So, w = 2. And $a_3=2a_23a_1$.
Moving on, as $(a_1+a_2)(a_3+a_4)$, $y(a_1+a_2)=a_3+a_4=2a_23a_1+a_2+a_3a_1=3a_24a_1+2a_23a_1=5a_27a_1$. => $a_1(y+7) = a_2(5y)$.
So, y < 5. And if y = 2, $5a_1=a_4$, which isn't possible. => y > 2. So, y = 3,4
If y = 3, $a_2=5a_1$, $a_3=7a_1$ and $a_4=11a_1$
If y = 4, $a_2=11a_1$, $a_3=19a_1$ and $a_4=29a_1$
So, A = $(a,5a,7a,11a)$, $(a,11a,19a,29a)$
Let $a_1 < a_2 < a_3 < a_4$. Then, $(a_3+a_4)$ can't divide $(a_1+a_2)$ and $(a_2+a_4)$ can't divide $(a_1+a_3)$.
So, $n_A$ < 4. So, to get the max value 4, $(a_1+a_3)(a_2+a_4)$, $(a_1+a_2)(a_3+a_4)$, $(a_1+a_4)(a_2+a_3)$ and $(a_2+a_3)(a_1+a_4)$. So, $a_1+a_4 = a_2+a_3$.
Since, $(a_1+a_3)(a_2+a_4)$, we can write, $w(a_1+a_3) = a_2+a_4 = a_2+(a_2+a_3a_1) = 2a_2+a_3a_1$.
=> $a_1(w+1) + a_3(w1) = 2a_2$. w > 1, else the numbers aren't distinct. If w > 2, $w(a_1+a_3)$ is larger than $2a_2+a_3a_1$. So, w = 2. And $a_3=2a_23a_1$.
Moving on, as $(a_1+a_2)(a_3+a_4)$, $y(a_1+a_2)=a_3+a_4=2a_23a_1+a_2+a_3a_1=3a_24a_1+2a_23a_1=5a_27a_1$. => $a_1(y+7) = a_2(5y)$.
So, y < 5. And if y = 2, $5a_1=a_4$, which isn't possible. => y > 2. So, y = 3,4
If y = 3, $a_2=5a_1$, $a_3=7a_1$ and $a_4=11a_1$
If y = 4, $a_2=11a_1$, $a_3=19a_1$ and $a_4=29a_1$
So, A = $(a,5a,7a,11a)$, $(a,11a,19a,29a)$
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: Beginner's Marathon
Problem $22$
The distance between Mirpur and Lalmatia is $24$ km. Two of three friends need to reach Lalmatia from Mirpur and another friend wants to reach Mirpur from Lalmatia. They have only one bike, which is initially in Mirpur. Each guy may go on foot (with velocity at most $6$ kmph) or on a bike (with velocity at most $18$ kmph). It is forbidden to leave a bike on a road. Prove that all of them may achieve their goals in $2$ hours $40$ minutes. (Only one guy may seat on the bike simultaneously).
The distance between Mirpur and Lalmatia is $24$ km. Two of three friends need to reach Lalmatia from Mirpur and another friend wants to reach Mirpur from Lalmatia. They have only one bike, which is initially in Mirpur. Each guy may go on foot (with velocity at most $6$ kmph) or on a bike (with velocity at most $18$ kmph). It is forbidden to leave a bike on a road. Prove that all of them may achieve their goals in $2$ hours $40$ minutes. (Only one guy may seat on the bike simultaneously).
Frankly, my dear, I don't give a damn.

 Posts: 57
 Joined: Sun Dec 11, 2016 2:01 pm
Re: Beginner's Marathon
What has been meant by 'after 2 hours 30 min'?
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm