Difference between revisions of "1991 AHSME Problems/Problem 28"

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== Solution ==
 
== Solution ==
<math>\fbox{B}</math>
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<math>\fbox{B}</math> The possible operations are <math>-3B+1B = -2B</math>, <math>-2B-W+W+B = -B</math>, <math>-B-2W+2W = -B</math>, or <math>-3W+B+W = -2W+B</math>. Notice that the only way the number of whites can change is from <math>-2W+B</math>, so it starts at 100 and only ever decreases by 2, so the final number of whites must be even, eliminating <math>D</math> and <math>E</math>. Now observe that we can keep repeating operation 1 (<math>-2B</math>) until we get to 2 blacks (and 100 whites) left, at which point we can't take out 3 blacks so we can't use this operation any more. We can now use operation 2 to get to 1 black and 100 whites, then operation 3 to get to 0 blacks and 100 whites, and then keep running operation 4 until we get to 2 whites and 49 blacks. Now run operation 2 to give 2 whites and 48 blacks, 2 whites and 47 blacks, and so on, and keep repeating until you reach 2 whites, which is <math>B.</math>
  
 
== See also ==
 
== See also ==

Revision as of 03:26, 24 February 2018

Problem

Initially an urn contains 100 white and 100 black marbles. Repeatedly 3 marbles are removed (at random) from the urn and replaced with some marbles from a pile outside the urn as follows: 3 blacks are replaced with 1 black, or 2 blacks and 1 white are replaced with a white and a black, or 1 black and 2 whites are replaced with 2 whites, or 3 whites are replaced with a black and a white. Which of the following could be the contents of the urn after repeated applications of this procedure?

(A) 2 black (B) 2 white (C) 1 black (D) 1 black and 1 white (E) 1 white

Solution

$\fbox{B}$ The possible operations are $-3B+1B = -2B$, $-2B-W+W+B = -B$, $-B-2W+2W = -B$, or $-3W+B+W = -2W+B$. Notice that the only way the number of whites can change is from $-2W+B$, so it starts at 100 and only ever decreases by 2, so the final number of whites must be even, eliminating $D$ and $E$. Now observe that we can keep repeating operation 1 ($-2B$) until we get to 2 blacks (and 100 whites) left, at which point we can't take out 3 blacks so we can't use this operation any more. We can now use operation 2 to get to 1 black and 100 whites, then operation 3 to get to 0 blacks and 100 whites, and then keep running operation 4 until we get to 2 whites and 49 blacks. Now run operation 2 to give 2 whites and 48 blacks, 2 whites and 47 blacks, and so on, and keep repeating until you reach 2 whites, which is $B.$

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 29
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
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