A PROBLEM OF "BDMO PROSTUTI""

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SYED ASHFAQ TASIN
Posts: 21
Joined: Thu Jun 02, 2016 6:14 pm

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.
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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 $\dfrac{m(m-1)}{2}=666$, 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.
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SYED ASHFAQ TASIN
Posts: 21
Joined: Thu Jun 02, 2016 6:14 pm

Re: A PROBLEM OF "BDMO PROSTUTI""

STARS AND BARS? PLEASE! BUJHLAM NA....
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Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

Re: A PROBLEM OF "BDMO PROSTUTI""

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.

samiul_samin
Posts: 999
Joined: Sat Dec 09, 2017 1:32 pm

Re: A PROBLEM OF "BDMO PROSTUTI""

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: 21
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)?
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""

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