Difference between revisions of "1994 AIME Problems/Problem 3"

(Solution 2)
 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
The function <math>f_{}^{}</math> has the property that, for each real number <math>x,\,</math>
 
The function <math>f_{}^{}</math> has the property that, for each real number <math>x,\,</math>
<center><math>f(x)+f(x-1) = x^2\,</math></center>.
+
<center><math>f(x)+f(x-1) = x^2.\,</math></center>
 
If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>?
 
If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>?
  

Latest revision as of 14:58, 13 June 2021

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 $561.$

See also

1994 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS