1994 AIME Problema 3

Question

La función

f

tiene la propiedad de que, para todo número real

x,

f(x)+f(x-1)=x^2.

Si

f(19)=94,

¿cuál es el residuo al dividir

f(94)

entre

1000

?

The function

f

has the property that, for each real number

x,

f(x)+f(x-1)=x^2.

If

f(19)=94,

what is the remainder when

f(94)

is divided by

1000?

Respuesta

Primero usa

x=20

para hallar

f(20)

First use

x=20

to find

f(20)

Elimina los términos alternados deduciendo una recurrencia de

f(x)

a

f(x-2)

Eliminate alternating terms by deriving a recurrence from

f(x)

to

f(x-2)

Primero,

f(20)=20^2-f(19)=306.

Al avanzar dos pasos se obtiene

\begin{aligned}f(x)&=x^2-(x-1)^2+f(x-2)\\&=2x-1+f(x-2).\end{aligned}

Por lo tanto,

\begin{aligned}f(94)&=306+\sum_{j=11}^{47}(4j-1)\\&=306+4(1073)-37\\&=4561.\end{aligned}

El residuo pedido es

561.

First

f(20)=20^2-f(19)=306.

Subtracting through two steps gives

\begin{aligned}f(x)&=x^2-(x-1)^2+f(x-2)\\&=2x-1+f(x-2).\end{aligned}

Therefore

\begin{aligned}f(94)&=306+\sum_{j=11}^{47}(4j-1)\\&=306+4(1073)-37\\&=4561.\end{aligned}

The requested remainder is

561.

El Problema 3 en otros años