Question

A sequence of integers $a_1,$ $a_2,$ $a_3,$ $\ldots$ is chosen so that $a_n=a_{n-1}-a_{n-2}$ for each $n\geq3.$ What is the sum of the first $2001$ terms of this sequence if the sum of the first $1492$ terms is $1985,$ and the sum of the first $1985$ terms is $1492?$

Answer

Answer: 986

Concepts:recursion pattern recognition system of equations

Difficulty rating: 2110

Small Hint:

Write the first six terms in terms of $a_1$ and $a_2$

Big Hint:

The sequence repeats every six terms, and each six-term block has sum $0$

Solution:

Put $a_1=x$ and $a_2=y.$ The first six terms are $x,\ y,\ y-x,\ -x,\ -y,\ x-y,$ after which the sequence repeats; these six terms sum to $0.$ Since $1492\equiv4\pmod6$ and $1985\equiv5\pmod6,$ the given equations are $\begin{aligned} 2y-x&=1985,\\ y-x&=1492. \end{aligned}$ Thus $y=493.$ Since $2001\equiv3\pmod6,$ the requested sum is $x+y+(y-x)=2y=986.$

1985 AIME Problem 5

Answer: 986

Difficulty rating: 2110

Small Hint:

Big Hint:

Solution:

Problem 5 in Other Years