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

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== Solution 2 ==
 
== 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,
 
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,
−
<math></math>T_{n-1} + T_n = n^2,<math>
+
<cmath>T_{n-1} + T_n = n^2,</cmath>
−
where </math>T_n = 1+2+...+n = \frac{n(n+1)}{2}<math> is the </math>n$th triangular number.
+
where <math>T_n = 1+2+...+n = \frac{n(n+1)}{2}</math> is the <math>n</math>th triangular number.
  
 
== See also ==
 
== See also ==

Revision as of 01:43, 6 July 2016

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.

See also

1994 AIME (Problems • Answer Key • Resources)
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

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