Senderov V.A.

1999 numbers stand in a row. The first number is 1. It is known that each number, except the first and the last, is equal to the sum of its two neighbors.

Find the last number.

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Let's denote our numbers as $a_{1}, a_{2}, \ldots, a_{1999}$. By adding the equations $a_{n+1}=a_{n+2}+a_{n}$ and $a_{n+2}=a_{n+3}+a_{n+1}$, we get $a_{n+3}+a_{n}=0$, or $a_{n+3}=-a_{n}$. From this, $a_{n+6}=-a_{n+3}=a_{n}$, which means the sequence has a period of 6. Therefore, $a_{1999}=a_{6 \cdot 333+1}=a_{1}=1$.

## Answer

1.