1994 AIME Problems/Problem 3

Problem

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 $1000$ ?

Solution 1

$\begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\ &= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94 \\ &= 4561 \end{align*}$

So, the remainder is $\boxed{561}$ .

Solution 2

Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is, $T_{n-1} + T_n = n^2,$ where $T_n = 1+2+...+n = \frac{n(n+1)}{2}$ is the $n$ th triangular number.

Using this, as well as using the fact that the value of $f(x)$ directly determines the value of $f(x+1)$ and $f(x-1),$ we conclude that $f(n) = T_n + K$ for all odd $n$ and $f(n) = T_n - K$ for all even $n,$ where $K$ is a constant real number.

Since $f(19) = 94$ and $T_19 = 190,$ we see that $K = -96.$ It follows that $f(94) = T_94 - (-96) = \frac{94\cdot 95}{2} + 96 = 4561,$ so the answer is $561.$

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1994 AIME Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See also