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Math Challenge II-A Number Theory

 
 
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Math Challenge II-A Number Theory
by Neo Liang - Friday, 2 August 2019, 2:29 PM
 

Lecture 9, Problem 9.25

The solution to this problem is incomplete.

 
Picture of David Reynoso
Re: Math Challenge II-A Number Theory
by David Reynoso - Wednesday, 7 August 2019, 5:44 PM
 

Apparently there was a problem when uploading the solution to this problem to the system.

The solution should say:

Let $\displaystyle  t=\left\lfloor\frac{x-2}{3} \right\rfloor$.  Then $\displaystyle x=\frac{4t+3}{2}$.  By definition, we have $\displaystyle t\leq\frac{x-2}{3} < t+1$, thus $\displaystyle t\leq\frac{4t-1}{6} < t+1$.  Solving for $t$, we get $\displaystyle -\dfrac{7}{2} < t \leq - \dfrac{1}{2}$.  So $t=-1,-2,-3$. Finally, plug $t$ back in to solve for $x$.

The solutions are $-\dfrac{1}{2}$, $-\dfrac{5}{2}$, and $-\dfrac{9}{2}$.

The solution has been updated now. Thanks for letting us know!