Difference between revisions of "1991 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | <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 == | ||
| − | {{AHSME box|year=1991|num-b= | + | {{AHSME box|year=1991|num-b=27|num-a=29}} |
[[Category: Intermediate Combinatorics Problems]] | [[Category: Intermediate Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:12, 10 February 2019
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
The possible operations are 
, 
, 
, or 
. Notice that the only way the number of whites can change is from 
, so it starts at 100 and only ever decreases by 2, so the final number of whites must be even, eliminating 
and 
. Now observe that we can keep repeating operation 1 (
) 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 
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
