# 1974 AHSME Problems/Problem 9

## Problem

The integers greater than one are arranged in five columns as follows: $$\begin{tabular}{c c c c c}\ & 2 & 3 & 4 & 5\\ 9 & 8 & 7 & 6 &\ \\ \ & 10 & 11 & 12 & 13\\ 17 & 16 & 15 & 14 &\ \\ \ & . & . & . & .\\ \end{tabular}$$

(Four consecutive integers appear in each row; in the first, third and other odd numbered rows, the integers appear in the last four columns and increase from left to right; in the second, fourth and other even numbered rows, the integers appear in the first four columns and increase from right to left.)

In which column will the number $1,000$ fall? $\mathrm{(A)\ } \text{first} \qquad \mathrm{(B) \ }\text{second} \qquad \mathrm{(C) \ } \text{third} \qquad \mathrm{(D) \ } \text{fourth} \qquad \mathrm{(E) \ }\text{fifth}$

## Solution

We try pairing numbers with the column number they're in. $$\begin{tabular}{c c}\ \text{Number} & \text{Column Number} \\ 2 & 2\\ 3 & 3\\ 4 & 4\\ 5 & 5\\ 6 & 4\\ 7 & 3\\ 8 & 2\\ 9 & 1\\ 10 & 2\\ 11 & 3\\ 12 & 4\\ \end{tabular}$$

Now we can see a pattern. The column number starts at $2$, goes up to $5$, then goes back down to $1$, and repeats. Each of these blocks has a length of $7$, so the column number has a period of $8$. Since $1000\equiv8(\bmod\,8)$, $1000$ is in the same column as $8$, which is the second column. $\boxed{\text{B}}$

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