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.
please write the way to reach the solution.
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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 $\dfrac{m(m-1)}{2}=666$, 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 $n = (38-1)\times 8+7=\boxed{303}$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 $\dfrac{m(m-1)}{2}=666$, which implies $m=37$. So, largest such $n$ is $(37-1)\times 8+7=\boxed{295}$
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)?