1971 AHSME Problems/Problem 27

Revision as of 21:14, 15 September 2020 by Yofro (talk | contribs) (Solution)

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

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