| From the introduction, the turkeys weigh even numbers of lbs., with no
two turkeys weighing the same amount; while from clue 1, the five birds
total 100 lbs. with the smallest weighing 15 lbs. By clue 2, the turkey
the Mathers got weighs 5 lbs. more than Bob the turkey, who tips the scales
5 lbs. higher than the Standishes' bird. If the Standishes' turkey weighed
19 lbs., then Bob would weigh 24 and the Mathers' turkey 29, a total with
the 15 lb. bird included of 87 lbs., leaving the fifth turkey at 13, a
conflict with clue 1; any weight higher than 19 would make the fifth bird
even smaller. So, if the Standishes' turkey doesn't weigh the least, then
it would weigh 16, 17, or 18 lbs. If the Standishes' turkey weighed
16 lbs., Bob would weigh 21 and the Mathers' turkey would weigh 26 (2).
The smallest turkey weighs 15; so the total of 78 would give the fifth
bird a weight of 22 lbs (clue 1). However, there is no way for clue 3 to
work, since the Bradfords turkey weighs 2 lbs. more than Zeke. If the
Standishes' turkey weighed 18 lbs. and Bob then 23 and the Mathers' turkey
28, adding the 15-lb. bird would give 84 lbs. and leave the fifth turkey
at 16 (1). However, there is no way for clue 4, where Earl weighs 4 lbs.
more than the Aldens' turkey, to work given these numbers. If the
Standishes' turkey weighed 17 lbs., Bob 22, and the Mathers' 27, adding
the 15-lb.'er would total 81 lbs., making the fifth turkey a 19 lb. one.
By clue 3, the 19 lb. turkey would be the Bradfords, and Zeke would be a
17 lb. bird. By clue 4, the Bradfords' gift turkey would be Earl, with
the Aldens then getting the smallest turkey. However, Tom then would have
to go to the Mathers or Aldens, which he doesn't (5). Therefore, none of
the alternatives to the Standishes' turkey weighing the least can work;
the Standishes' birds weighs 15 lbs., with Bob then a 20-lb. turkey and
the Mathers' gift weighing 25 lbs. (2). Since these three turkeys total
60 of the 100 lbs. (1), the other two birds must weigh 40 lbs. between
them. Given the weights of the three recovered so far--15, 20, 25--the
fourth bird must weigh 21-24 lbs. and the fifth then 19-16 respectively
to the fourth's weight. By clue 3, the Bradfords' turkey weighs
2 lbs. more than Zeke. If the Bradfords' turkey were the fourth bird
weighing 21-24 lbs., Zeke would have to be the fifth bird weighing 19-16
lbs. The only weights that can work are that the Bradfords' bird weighs
21 lbs. and Zeke 19. However, there is no way for clue 4, where Earl is
4 lbs. heavier than the Aldens' turkey, to fit. If the Bradfords' turkey
were the fifth one, weighing 19-16 lbs., Zeke would have to be the turkey
given to the Standish family, with the Bradfords' turkey then a 17-lb.
bird. The fourth turkey would then weigh 23 lbs. to make up the 100-lb.
total. However, there is now no 4-lb. difference as required by clue 4.
So, the Bradfords got 20-lb. Bob, with Zeke weighing in at 18 lbs. (3).
The last turkey must weigh 22 lbs. to bring the total to 100. By clue 4,
the bird weighing 22 lbs. is Earl, and the Aldens received 18-lb. Zeke.
By elimination, Earl will be the Williamses Thanksgiving guest. Finally,
Tom went to the Standishes and Max to the Mathers (5). In sum, Ben Butterball
gave away turkeys for Thanksgiving as follows:
- Max, 25 lbs., to the Mather family
- Earl, 22 lbs., to the Williams family
- Bob, 20 lbs., to the Bradford family
- Zeke, 18 lbs., to the Alden family
- Tom, 15 lbs., to the Standish family
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