A PROBLEM OF "BDMO PROSTUTI""

For students of class 9-10 (age 14-16)
SYED ASHFAQ TASIN
Posts: 28
Joined: Thu Jun 02, 2016 6:14 pm

A PROBLEM OF "BDMO PROSTUTI""

Unread post by SYED ASHFAQ TASIN » Tue Nov 29, 2016 12:59 pm

x+8y+8z=n it has 666 solutions. then,what could be the largest value of n?
please write the way to reach the solution.
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]

User avatar
Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

Re: A PROBLEM OF "BDMO PROSTUTI""

Unread post by Thanic Nur Samin » Tue Dec 06, 2016 1:07 am

Note that the number of positive integer solutions to $x+8y+8z=n$ is equal to the number of positive integer solutions of $8y+8z < n$.

However, the number of solutions to the second equation is equal to the number of positive integer solutions to $y+z < \left\lfloor \dfrac{n}{8}\right\rfloor+1$. Let $m=\left\lfloor \dfrac{n}{8}\right\rfloor+1$. So the number of solutions is equal to the number of positive integer solutions to $t+y+z=m$. By using stars and bars, we get that the number is simply $\dbinom{m}{2}$. So, we get that $\dfrac{m(m-1)}{2}=666$, which implies $m=37$. So, largest such $n$ is $(37-1)\times 8+7=\boxed{295}$
Last edited by Thanic Nur Samin on Wed Dec 07, 2016 4:54 pm, edited 1 time in total.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

SYED ASHFAQ TASIN
Posts: 28
Joined: Thu Jun 02, 2016 6:14 pm

Re: A PROBLEM OF "BDMO PROSTUTI""

Unread post by SYED ASHFAQ TASIN » Tue Dec 06, 2016 12:38 pm

STARS AND BARS? PLEASE! BUJHLAM NA....
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]

User avatar
Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

Re: A PROBLEM OF "BDMO PROSTUTI""

Unread post by Thanic Nur Samin » Tue Dec 06, 2016 6:06 pm

https://brilliant.org/wiki/integer-equa ... -and-bars/

https://www.artofproblemsolving.com/wik ... ll-and-urn

If you run into a word you don't understand, then please google it first. You might want to check brilliant.org articles, they are good for readability and example problems. Also, while using bangla over internet, please use avro. It is not very difficult to use.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

User avatar
samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

Re: A PROBLEM OF "BDMO PROSTUTI""

Unread post by samiul_samin » Tue Feb 20, 2018 7:06 am

Thanic Nur Samin wrote:
Tue Dec 06, 2016 1:07 am
Note that the number of positive integer solutions to $x+8y+8z=n$ is equal to the number of positive integer solutions of $8y+8z < n$.

However, the number of solutions to the second equation is equal to the number of positive integer solutions to $y+z < \left\lfloor \dfrac{n}{8}\right\rfloor+1$. Let $m=\left\lfloor \dfrac{n}{8}\right\rfloor+1$. So the number of solutions is equal to the number of positive integer solutions to $t+y+z=m$. By using stars and bars, we get that the number is simply $\dbinom{m}{2}$. So, we get that $\dfrac{m(m-1)}{2}=666$, which implies $m=37$. So, largest such $n$ is $(37-1)\times 8+7=\boxed{295}$
From where the $t$ comes from?

SYED ASHFAQ TASIN
Posts: 28
Joined: Thu Jun 02, 2016 6:14 pm

Re: A PROBLEM OF "BDMO PROSTUTI""

Unread post by SYED ASHFAQ TASIN » Sun Dec 16, 2018 11:25 pm

't' is just an integer.
I am not understanding why you added 1 with (n/8)?
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]

Shafin666
Posts: 1
Joined: Tue Dec 18, 2018 4:27 pm

Re: A PROBLEM OF "BDMO PROSTUTI""

Unread post by Shafin666 » Tue Dec 18, 2018 4:53 pm

Thanic Nur Samin wrote:
Tue Dec 06, 2016 1:07 am
Note that the number of positive integer solutions to $x+8y+8z=n$ is equal to the number of positive integer solutions of $8y+8z < n$.

However, the number of solutions to the second equation is equal to the number of positive integer solutions to $y+z < \left\lfloor \dfrac{n}{8}\right\rfloor+1$. Let $m=\left\lfloor \dfrac{n}{8}\right\rfloor+1$. So the number of solutions is equal to the number of positive integer solutions to $t+y+z=m$. By using stars and bars, we get that the number is simply $\dbinom{m}{2}$. So, we get that $\dfrac{m(m-1)}{2}=666$, which implies $m=37$. So, largest such $n$ is $(37-1)\times 8+7=\boxed{295}$
You made a slight mistake there. The number of positive integer solutions to $t+y+z=m$ should be $\dbinom{m-1}{2}$ . So $m = 38$, rather than $37$ which yields the answer $n = (38-1)\times 8+7=\boxed{303}$

SYED ASHFAQ TASIN
Posts: 28
Joined: Thu Jun 02, 2016 6:14 pm

Re: A PROBLEM OF "BDMO PROSTUTI""

Unread post by SYED ASHFAQ TASIN » Sun Aug 04, 2019 11:38 pm

If I get it right,according to Stars and Bars,shouldn't the result ((m+2)C2)?
As total value is m,and 2 bars needed?I'm a bit confused.
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]

SYED ASHFAQ TASIN
Posts: 28
Joined: Thu Jun 02, 2016 6:14 pm

Re: A PROBLEM OF "BDMO PROSTUTI""

Unread post by SYED ASHFAQ TASIN » Sun Aug 04, 2019 11:40 pm

SYED ASHFAQ TASIN wrote:
Sun Dec 16, 2018 11:25 pm
't' is just an integer.
I am not understanding why you added 1 with (n/8)?
Oh ,Got it.You used Floor function.
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]

Post Reply