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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Revision as of 21:14, 15 September 2020
Solution
Let the number of white be 
. The number of blue is then 
for some constant 
. So we want 
. We take mod 3 to find y. 
, so blue is 1 more than half white. The number of whites is then 36, and the number of blues is 19. So 
~yofro
