Question

Dado que $x_1 = 211,$ $x_2 = 375,$ $x_3 = 420,$ $x_4 = 523,$ y $\begin{aligned} x_n &= x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4} \\ &\quad \text{when } n \ge 5, \end{aligned}$ halla el valor de $x_{531} + x_{753} + x_{975}.$

Given that $x_1 = 211,$ $x_2 = 375,$ $x_3 = 420,$ $x_4 = 523,$ and $\begin{aligned} x_n &= x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4} \\ &\quad \text{when } n \ge 5, \end{aligned}$ find the value of $x_{531} + x_{753} + x_{975}.$

Respuesta

Solución:

Para $n \ge 6,$ sustituye la recurrencia de $x_{n-1}:$ $\begin{aligned} x_n &= \\ &(x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) \\ &\quad {}- x_{n-2} \\ &\quad {}+ x_{n-3} - x_{n-4} \\ &= -x_{n-5}. \end{aligned}$ Por tanto $x_{n+10} = -x_{n+5} = x_n,$ así que la sucesión tiene periodo $10.$

Como $531,$ $753,$ y $975$ dejan restos $1,$ $3,$ y $5$ al dividir entre $10,$ obtenemos $x_{531} = x_1 = 211,$ $x_{753} = x_3 = 420,$ y $x_{975} = x_5$ $= x_4 - x_3 + x_2 - x_1$ $= 523 - 420 + 375 - 211$ $= 267.$

La suma es $211 + 420 + 267 = 898.$

For $n \ge 6,$ substitute the recurrence for $x_{n-1}:$ $\begin{aligned} x_n &= \\ &(x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) \\ &\quad {}- x_{n-2} \\ &\quad {}+ x_{n-3} - x_{n-4} \\ &= -x_{n-5}. \end{aligned}$ Hence $x_{n+10} = -x_{n+5} = x_n,$ so the sequence has period $10.$

Since $531,$ $753,$ and $975$ leave remainders $1,$ $3,$ and $5$ upon division by $10,$ we get $x_{531} = x_1 = 211,$ $x_{753} = x_3 = 420,$ and $x_{975} = x_5$ $= x_4 - x_3 + x_2 - x_1$ $= 523 - 420 + 375 - 211$ $= 267.$

The sum is $211 + 420 + 267 = 898.$

2001 AIME II Problema 3

Solución:

El Problema 3 en otros años