A PROBLEM OF "BDMO PROSTUTI""

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A PROBLEM OF "BDMO PROSTUTI""
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.
please write the way to reach the solution.
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 Thanic Nur Samin
 Posts: 176
 Joined: Sun Dec 01, 2013 11:02 am
Re: A PROBLEM OF "BDMO PROSTUTI""
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 $\dfrac{m(m1)}{2}=666$, which implies $m=37$. So, largest such $n$ is $(371)\times 8+7=\boxed{295}$
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 $\dfrac{m(m1)}{2}=666$, which implies $m=37$. So, largest such $n$ is $(371)\times 8+7=\boxed{295}$
Last edited by Thanic Nur Samin on Wed Dec 07, 2016 4:54 pm, edited 1 time in total.
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Because destroying everything mindlessly isn't cool enough.

 Posts: 28
 Joined: Thu Jun 02, 2016 6:14 pm
Re: A PROBLEM OF "BDMO PROSTUTI""
STARS AND BARS? PLEASE! BUJHLAM NA....
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]
 Thanic Nur Samin
 Posts: 176
 Joined: Sun Dec 01, 2013 11:02 am
Re: A PROBLEM OF "BDMO PROSTUTI""
https://brilliant.org/wiki/integerequa ... andbars/
https://www.artofproblemsolving.com/wik ... llandurn
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.
https://www.artofproblemsolving.com/wik ... llandurn
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.
Because destroying everything mindlessly isn't cool enough.
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: A PROBLEM OF "BDMO PROSTUTI""
From where the $t$ comes from?Thanic Nur Samin wrote: ↑Tue Dec 06, 2016 1:07 amNote 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 $\dfrac{m(m1)}{2}=666$, which implies $m=37$. So, largest such $n$ is $(371)\times 8+7=\boxed{295}$

 Posts: 28
 Joined: Thu Jun 02, 2016 6:14 pm
Re: A PROBLEM OF "BDMO PROSTUTI""
't' is just an integer.
I am not understanding why you added 1 with (n/8)?
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]
Re: A PROBLEM OF "BDMO PROSTUTI""
You made a slight mistake there. The number of positive integer solutions to $t+y+z=m$ should be $\dbinom{m1}{2}$ . So $m = 38$, rather than $37$ which yields the answer $n = (381)\times 8+7=\boxed{303}$Thanic Nur Samin wrote: ↑Tue Dec 06, 2016 1:07 amNote 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 $\dfrac{m(m1)}{2}=666$, which implies $m=37$. So, largest such $n$ is $(371)\times 8+7=\boxed{295}$

 Posts: 28
 Joined: Thu Jun 02, 2016 6:14 pm
Re: A PROBLEM OF "BDMO PROSTUTI""
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.
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]

 Posts: 28
 Joined: Thu Jun 02, 2016 6:14 pm
Re: A PROBLEM OF "BDMO PROSTUTI""
Oh ,Got it.You used Floor function.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)?
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]