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abid.uhscian
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Re: Solution of problem 1

Unread post by abid.uhscian » Wed Mar 24, 2021 4:55 pm

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I am noob in NT. if the answer is wrong please share me the right solution🙂

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Mehrab4226
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Re: Problem 7

Unread post by Mehrab4226 » Wed Mar 24, 2021 7:22 pm

Anindya Biswas wrote:
Wed Mar 24, 2021 3:26 pm
Let $a,b, c, d$ be integers. Show that the product \[(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)\] is divisible by $12$
Source :
Slovenia 1995
Let,
$(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)=p$
In $\text{mod 3}$ there are $3$ possible values of $a,b,c,d$. So by $PHP$ there must be $2$ integers from $\{a,b,c,d\}$ which have the same value in $\text{mod 3}$. So $p$ is divisible by $3$ as one of the factors of $p$ is $0\text{ (mod 3)}$.

For $\text{mod 4}$ if any two of the $4$ integers have the same value in $\text{mod 4}$ then $p$ is divisible by $4$ just like the last time. If they are no $2$ of the $4$ integers having the same value in $\text{mod 4}$ then without the loss of generaltiy,
$a \equiv 0\text{ (mod 4)}$
$b \equiv 1\text{ (mod 4)}$
$c \equiv 2\text{ (mod 4)}$
$d \equiv 3 \equiv -1\text{ (mod 4)}$. So,
$(a-c) \equiv 0-2 \equiv -2 \equiv 2\text{ (mod 4)}$
$(b-d) \equiv 1+1 \equiv 2\text{ (mod 4)}$
$\therefore (a-c)(b-d) \equiv 2\times 2 \equiv 4 \equiv 0 \text{ (mod 4)}$
So $p$ is divisible by 4.
So $p$ is divisible by 12. $\square$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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Mehrab4226
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Problem: 8

Unread post by Mehrab4226 » Wed Mar 24, 2021 7:44 pm

Prove that any integer greater than or equal to $7$ can be written as a sum of two relatively prime integers, both greater than $1$.
Source:
Bay Area Mathematical Olympiad
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Asif Hossain
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Re: Problem: 8

Unread post by Asif Hossain » Fri Mar 26, 2021 7:50 am

Mehrab4226 wrote:
Wed Mar 24, 2021 7:44 pm
Prove that any integer greater than or equal to $7$ can be written as a sum of two relatively prime integers, both greater than $1$.
Source:
Bay Area Mathematical Olympiad
wow this was a brain teaser (sorry saw from mathstacks cuz i am noob :()
if $n=2k+1$ then $k$ and $k+1$ are coprime.
if $n=4k$ then $2k-1$ and $2k+1$ are coprime
if $n=4k+2$ then $2k-1$ and $2k+3$ are coprime. $\square$
Hmm..Hammer...Treat everything as nail

Asif Hossain
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Problem 9

Unread post by Asif Hossain » Fri Mar 26, 2021 7:54 am

An old question nobody answered that so reposted
Problem 9
Find all $a,b \in \mathbb{N}$ such that $y=ax^2+6x+b$ and $y=ax+6$ intersect only once.
Hmm..Hammer...Treat everything as nail

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Mehrab4226
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Re: Problem 9

Unread post by Mehrab4226 » Fri Mar 26, 2021 2:23 pm

Asif Hossain wrote:
Fri Mar 26, 2021 7:54 am
An old question nobody answered that so reposted
Problem 9
Find all $a,b \in \mathbb{N}$ such that $y=ax^2+6x+b$ and $y=ax+6$ intersect only once.
There are too many calculations(so a lot of chances for mistakes by me. So I might have skipped some parts. If it is difficult to understand, feel free to ask.
Given,
$y=ax^2+6x+b \cdots (1)$
$y=ax+6 \cdots (2)$
Subtracting (2) from 1 we get,
$ax^2+(6-a)x+(b-6)=0\cdots (3)$

Here solution of equation (3) comprises the intersections of (1) and (2). And (1) and (2) intersect at 1 point if (3) has only 1 solution.
And we know the infamous quadratic formula, don't we?
For an equation $ax^2+b^x+c=0$ the solutions are,
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

And the equation will have only one solution if the discriminant $b^2-4ac=0$.
So in our equation (3) will have one solution(which is what we are looking for when,
$(6-a)^2-4(a)(b-6)=0$
Or,$36-12a+a^2-4ab+24a=0$
Or,$a^2+12a-4ab+36=0$
Or,$a^2+(12-4b)a+36=0$
We can use the quadratic formula again ,
$a=\frac{-12+4b \pm \sqrt{(12-4b)^2-4 \times 36}}{2}$
$a=\frac{-12+4b \pm \sqrt{(144-96b+16b^2-144}}{2}$
$a=\frac{-12+4b \pm \sqrt{(16b^2-96b}}{2}$
$a=\frac{-12+4b \pm 4\sqrt{b^2-6b}}{2}$
$a=-6+2b \pm 2\sqrt{b^2-6b}\cdots (3)$
For a to be integer the number inside the radical must be a perfect square.
Let,
$b^2-6b=k^2$
$b^2-6b-k^2=0$
The quadratic formula again :cry:

$b=\frac{6 \pm \sqrt{6^2-4\times k^2}}{2}$
$b=\frac{6 \pm \sqrt{36- 4k^2}}{2}$
$b=\frac{6 \pm 2\sqrt{9+k^2}}{2}$
$b=3 \pm \sqrt{9+k^2}$
For b to be an integer we need $9+k^2$ to be a square number,
Let,
$9+k^2=g^2$
$g^2-k^2=9$
$(g+k)(g-k)=9$
Factors of 9 are 1,3,9. So putting them on the equation we will get,
g=3,5
Now,
$b=3 \pm g$
$\therefore b=6,8$(Only natural numbers)
Now putting it in (3) we get,
$a=2,6,18$
So the requered solutions, $(a,b)=(6,6),(2,8),(18,8)$
Last edited by Mehrab4226 on Fri Mar 26, 2021 2:54 pm, edited 1 time in total.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: Problem 9

Unread post by Asif Hossain » Fri Mar 26, 2021 2:39 pm

Mehrab4226 wrote:
Fri Mar 26, 2021 2:23 pm
Asif Hossain wrote:
Fri Mar 26, 2021 7:54 am
An old question nobody answered that so reposted
Problem 9
Find all $a,b \in \mathbb{N}$ such that $y=ax^2+6x+b$ and $y=ax+6$ intersect only once.
There are too many calculations(so a lot of chances for mistakes by me. So I might have skipped some parts. If it is difficult to understand, feel free to ask.
Given,
$y=ax^2+6x+b \cdots (1)$
$y=ax+6 \cdots (2)$
Subtracting (2) from 1 we get,
$ax^2+(6-a)x+(b-6)=0\cdots (3)$

Here solution of equation (3) comprises the intersections of (1) and (2). And (1) and (2) intersect at 1 point if (3) has only 1 solution.
And we know the infamous quadratic formula, don't we?
For an equation $ax^2+b^x+c=0$ the solutions are,
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

And the equation will have only one solution if the discriminant $b^2-4ac=0$.
So in our equation (3) will have one solution(which is what we are looking for when,
$(6-a)^2-4(a)(b-6)=0$
Or,$36-12a+a^2-4ab+24a=0$
Or,$a^2+12a-4ab+36=0$
Or,$a^2+(12-4b)a+36=0$
We can use the quadratic formula again ,
$a=\frac{12-4b \pm \sqrt{(12-4b)^2-4 \times 36}}{2}$
$a=\frac{12-4b \pm \sqrt{(144-96b+16b^2-144}}{2}$
$a=\frac{12-4b \pm \sqrt{(16b^2-96b}}{2}$
$a=\frac{12-4b \pm 4\sqrt{b^2-6b}}{2}$
$a=6-2b \pm 2\sqrt{b^2-6b}\cdots (3)$
For a to be integer the number inside the radical must be a perfect square.
Let,
$b^2-6b=k^2$
$b^2-6b-k^2=0$
The quadratic formula again :cry:

$b=\frac{6 \pm \sqrt{6^2-4\times k^2}}{2}$
$b=\frac{6 \pm \sqrt{36- 4k^2}}{2}$
$b=\frac{6 \pm 2\sqrt{9+k^2}}{2}$
$b=3 \pm \sqrt{9+k^2}$
For b to be an integer we need $9+k^2$ to be a square number,
Let,
$9+k^2=g^2$
$g^2-k^2=9$
$(g+k)(g-k)=9$
Factors of 9 are 1,3,9. So putting them on the equation we will get,
g=3,5
Now,
$b=3 \pm g$
$\therefore b=6,8$(Only natural numbers)
Now putting it in (3) we get,
$a=\text{No positive integers}$
So there are no solutions for the problem.
Recheck your proof There are solutions for a,b one example is $(a,b)=(2,8)$
Hmm..Hammer...Treat everything as nail

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Mehrab4226
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Re: Problem 9

Unread post by Mehrab4226 » Fri Mar 26, 2021 3:00 pm

Asif Hossain wrote:
Fri Mar 26, 2021 2:39 pm
Mehrab4226 wrote:
Fri Mar 26, 2021 2:23 pm
Asif Hossain wrote:
Fri Mar 26, 2021 7:54 am
An old question nobody answered that so reposted
Problem 9
Find all $a,b \in \mathbb{N}$ such that $y=ax^2+6x+b$ and $y=ax+6$ intersect only once.
There are too many calculations(so a lot of chances for mistakes by me. So I might have skipped some parts. If it is difficult to understand, feel free to ask.
Given,
$y=ax^2+6x+b \cdots (1)$
$y=ax+6 \cdots (2)$
Subtracting (2) from 1 we get,
$ax^2+(6-a)x+(b-6)=0\cdots (3)$

Here solution of equation (3) comprises the intersections of (1) and (2). And (1) and (2) intersect at 1 point if (3) has only 1 solution.
And we know the infamous quadratic formula, don't we?
For an equation $ax^2+b^x+c=0$ the solutions are,
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

And the equation will have only one solution if the discriminant $b^2-4ac=0$.
So in our equation (3) will have one solution(which is what we are looking for when,
$(6-a)^2-4(a)(b-6)=0$
Or,$36-12a+a^2-4ab+24a=0$
Or,$a^2+12a-4ab+36=0$
Or,$a^2+(12-4b)a+36=0$
We can use the quadratic formula again ,
$a=\frac{12-4b \pm \sqrt{(12-4b)^2-4 \times 36}}{2}$
$a=\frac{12-4b \pm \sqrt{(144-96b+16b^2-144}}{2}$
$a=\frac{12-4b \pm \sqrt{(16b^2-96b}}{2}$
$a=\frac{12-4b \pm 4\sqrt{b^2-6b}}{2}$
$a=6-2b \pm 2\sqrt{b^2-6b}\cdots (3)$
For a to be integer the number inside the radical must be a perfect square.
Let,
$b^2-6b=k^2$
$b^2-6b-k^2=0$
The quadratic formula again :cry:

$b=\frac{6 \pm \sqrt{6^2-4\times k^2}}{2}$
$b=\frac{6 \pm \sqrt{36- 4k^2}}{2}$
$b=\frac{6 \pm 2\sqrt{9+k^2}}{2}$
$b=3 \pm \sqrt{9+k^2}$
For b to be an integer we need $9+k^2$ to be a square number,
Let,
$9+k^2=g^2$
$g^2-k^2=9$
$(g+k)(g-k)=9$
Factors of 9 are 1,3,9. So putting them on the equation we will get,
g=3,5
Now,
$b=3 \pm g$
$\therefore b=6,8$(Only natural numbers)
Now putting it in (3) we get,
$a=\text{No positive integers}$
So there are no solutions for the problem.
Recheck your proof There are solutions for a,b one example is $(a,b)=(2,8)$
Thank you for pointing that out. I used so the quadratic formula so many times that I accidentally forgot to give the minus before b in a line. I updated my solution.
Here is a meme as a token of appreciation,
Maybe Schools are right after all.
137585712_239247837637156_3546407512043365392_n.jpg
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The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: Problem 10

Unread post by Asif Hossain » Fri Mar 26, 2021 9:32 pm

Asif Hossain wrote:
Fri Mar 26, 2021 7:54 am
An old question nobody answered that so reposted
Problem 9
Find all $a,b \in \mathbb{N}$ such that $y=ax^2+6x+b$ and $y=ax+6$ intersect only once.
Problem 10
Extend this to $\mathbb{Z}$ :D
Hmm..Hammer...Treat everything as nail

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Mehrab4226
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Solution of Problem 10

Unread post by Mehrab4226 » Fri Mar 26, 2021 9:59 pm

Given,
$y=ax^2+6x+b \cdots (1)$
$y=ax+6 \cdots (2)$
Subtracting (2) from 1 we get,
$ax^2+(6-a)x+(b-6)=0\cdots (3)$

Here solution of equation (3) comprises the intersections of (1) and (2). And (1) and (2) intersect at 1 point if (3) has only 1 solution.
And we know the infamous quadratic formula, don't we?
For an equation $ax^2+b^x+c=0$ the solutions are,
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

And the equation will have only one solution if the discriminant $b^2-4ac=0$.
So in our equation (3) will have one solution(which is what we are looking for when,
$(6-a)^2-4(a)(b-6)=0$
Or,$36-12a+a^2-4ab+24a=0$
Or,$a^2+12a-4ab+36=0$
Or,$a^2+(12-4b)a+36=0$
We can use the quadratic formula again ,
$a=\frac{-12+4b \pm \sqrt{(12-4b)^2-4 \times 36}}{2}$
$a=\frac{-12+4b \pm \sqrt{(144-96b+16b^2-144}}{2}$
$a=\frac{-12+4b \pm \sqrt{(16b^2-96b}}{2}$
$a=\frac{-12+4b \pm 4\sqrt{b^2-6b}}{2}$
$a=-6+2b \pm 2\sqrt{b^2-6b}\cdots (3)$
For a to be integer the number inside the radical must be a perfect square.
Let,
$b^2-6b=k^2$
$b^2-6b-k^2=0$
The quadratic formula again :cry:

$b=\frac{6 \pm \sqrt{6^2-4\times k^2}}{2}$
$b=\frac{6 \pm \sqrt{36- 4k^2}}{2}$
$b=\frac{6 \pm 2\sqrt{9+k^2}}{2}$
$b=3 \pm \sqrt{9+k^2}$
For b to be an integer we need $9+k^2$ to be a square number,
Let,
$9+k^2=g^2$
$g^2-k^2=9$
$(g+k)(g-k)=9$
Factors of 9 are 1,3,9. So putting them on the equation we will get,
g=3,5
Now,
$b=3 \pm g$
$\therefore b=-2,0,6,8$
Now putting it in (3) we get,[I am not listing them in order.]
$a=-18,-2,-6,2,6,18$
So the requered solutions, $(a,b)=(-2,-2),(-18,-2),(-6,0),(6,6),(2,8),(18,8)$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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