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{\textbf{(E) }57}</math> |
~yofro | ~yofro | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 35p box|year=1971|num-b=26|num-a=28}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:00, 6 August 2024
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 
. The minimum number of red chips is
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
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
