# 1961 AHSME Problems/Problem 24

## Problem 24

Thirty-one books are arranged from left to right in order of increasing prices. The price of each book differs by $\textdollar{2}$ from that of each adjacent book. For the price of the book at the extreme right a customer can buy the middle book and the adjacent one. Then: $\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad \\ \textbf{(B)}\ \text{The middle book sells for \textdollar 36} \qquad \\ \textbf{(C)}\ \text{The cheapest book sells for \textdollar4} \qquad \\ \textbf{(D)}\ \text{The most expensive book sells for \textdollar64 } \qquad \\ \textbf{(E)}\ \text{None of these is correct }$

## Solution

Let the leftmost book cost $c$ dollars. The 16th book (the middle book) costs $c+30$ dollars, and the 31st book costs $c+60$ dollars.

If the book to the left of the middle book is used, the equation to solve for the price of the leftmost book is $c+30+c+29=c+60$. Solving yields $c=1$, so the price of the middle book is $\textdollar{31}$ and the price of the rightmost book is $\textdollar{61}$.

If the book to the right of the middle book is used, the equation to solve for the price of the leftmost book is $c+30+c+31=c+60$. Solving yields $c=-1$, which can not happen.

Thus, the answer is $\boxed{\textbf{(A)}}$.

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