# Difference between revisions of "1971 AHSME Problems/Problem 27"

## Problem

A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. The number which are white or blue is at least $55$. The minimum number of red chips is

$\textbf{(A) }24\qquad \textbf{(B) }33\qquad \textbf{(C) }45\qquad \textbf{(D) }54\qquad \textbf{(E) }57$

## Solution

Let the number of white be $2x$. The number of blue is then $x-y$ for some constant $y$. So we want $2x+x-y=55\rightarrow 3x-y=55$. We take mod 3 to find y. $55=1\pmod{3}$, so blue is 1 more than half white. The number of whites is then 36, and the number of blues is 19. So $19*3=\boxed{57}$

~yofro