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A PROBLEM OF "BDMO PROSTUTI""

Posted: 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.

Re: A PROBLEM OF "BDMO PROSTUTI""

Posted: 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 $(37-1)\times 8+7=\boxed{295}$

Re: A PROBLEM OF "BDMO PROSTUTI""

Posted: Tue Dec 06, 2016 12:38 pm
STARS AND BARS? PLEASE! BUJHLAM NA....

Re: A PROBLEM OF "BDMO PROSTUTI""

Posted: 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.

Re: A PROBLEM OF "BDMO PROSTUTI""

Posted: 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?

Re: A PROBLEM OF "BDMO PROSTUTI""

Posted: Sun Dec 16, 2018 11:25 pm
't' is just an integer.
I am not understanding why you added 1 with (n/8)?

Re: A PROBLEM OF "BDMO PROSTUTI""

Posted: 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 $n = (38-1)\times 8+7=\boxed{303}$

Re: A PROBLEM OF "BDMO PROSTUTI""

Posted: 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.

Re: A PROBLEM OF "BDMO PROSTUTI""

Posted: 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.