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

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

Revision as of 22:14, 15 September 2020

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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