SAT1000 - P900

Algebra Level pending

A group of members from brilliant.org decide to plant trees on the playground defined by coordinate plane, and the schedule is as follows:

The k t h k\ th tree is planted at P k ( x k , y k ) P_k(x_k,y_k) , where x 1 = 1 , y 1 = 1 x_1=1, y_1=1 , and when k 2 k \geq 2 :

{ x k = x k 1 + 1 5 ( k 1 5 k 2 5 ) y k = y k 1 + k 1 5 k 2 5 \begin{cases} \begin{aligned} x_k & = x_{k-1}+1-5(\lfloor \dfrac{k-1}{5} \rfloor - \lfloor \dfrac{k-2}{5} \rfloor) \\ y_k & = y_{k-1} + \lfloor \dfrac{k-1}{5} \rfloor - \lfloor \dfrac{k-2}{5} \rfloor \end{aligned} \end{cases}

What's the coordinate of the 2008 t h 2008\ th tree?

Let the coordinate be x 0 , y 0 x_0,y_0 . Submit 2 y 0 x 0 2y_0-x_0 .


Have a look at my problem set: SAT 1000 problems


The answer is 801.

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1 solution

Chris Lewis
Jun 30, 2020

k 1 5 k 2 5 = { 1 if k 1 ( m o d 5 ) 0 otherwise \left \lfloor \frac{k-1}{5} \right \rfloor - \left \lfloor \frac{k-2}{5} \right \rfloor = \begin{cases} 1 &\mbox{if }k\equiv 1 \pmod{5} \\ 0 &\mbox{otherwise} \end{cases}

From this we see that x k 1 x_k-1 is the remainder when k 1 k-1 is divided by 5 5 and y k 1 y_k-1 is the corresponding quotient. Hence the coordinates of the 2008 2008 th tree are 3 , 402 3,402 and the answer is 801 \boxed{801} .

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