Answer to December 10, 2001 Problem |
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The Weight Problem Five young girls in Winchester want to weigh themselves on a coin-operated scale for the price of a single coin. Two of the girls climb on the scale at the same time, drop in the coin, and note their combined weight. Then one girl steps off, another steps on, they note the weight again, and so forth, until they have accumulated the following sequence of readings: 183, 186, 187, 190, 191, 192, 193, 194, 196, and 200 pounds. What are their individual weights?
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Answer:
The girls weigh 91, 92, 95, 99, and 101 pounds. Rick Jones sent in the following explanation: Solution:
Since 5 items taken two at a time is 5!/(2!3!) = 10 and since the ten readings are
distinct:
a+b = 183, d+e = 200
By adding the ten weighings, we obtain 4a + 4b + 4c + 4d + 4e = 1912 Consider the final weighing. Assuming integer weights for the moment, the possible values of e are restricted to 104, 103, 102 and 101. Adding these to c would give 199, 198, 197 and 196--only the last of which was actually recorded. Hence it must be that e = 101, d = 99 For the first weighing, possible values of b are 94, 93 and 92. If b = 94, then b+e = 195, which is not a recorded weighing. If b = 93, then b+c = 188, which again is not a recorded weighing. Hence, b = 92, a = 91 This gives us 91 < 92 < 95 < 99 < 101 and
a+b = 183 a+c = 186
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1. Richard K. Johnson * | La Jolla, California |
2. Rick Jones | Kennett Square, Pennsylvania |
3. Keith Mealy | Cincinnati, Ohio |
4. David Powell | Winchester, Virginia |
5. Tony Wu | Fort Collins, Colorado |
6. James Alarie | University of Michigan -- Flint |
7. Janine Oliver | Winchester, Virginia |
8. George Gaither | Winchester, Virginia |
9. Renata Sommerville | Austin, Texas |
10. Walt Arrison + | Philadelphia, Pennsylvania |
* Wrote a computer program to solve the problem
+ Pointed that there was an error in the sequence of weights |