
Re: more "bales of hay" urgh!
Posted:
Nov 10, 1997 6:15 AM


well, here's the way I would do it, but i'm sure there are many variants:
> suppose five bales of hay are weighed two at a time in all > possible ways. The weights in lbs are 110, 112, 113,114,115, > 116,117,118,120 and 121. How much does each bale weigh?
as usual, call the five individual weights a,b,c,d, and e where a is less than or equal to b which is less than or equal to c which is less than or equal to d which is less than or equal to e
some things I know for sure  the lightest twobale combination is a+b and the heaviest is d+e. In fact, I even know that the SECOND lightest twobale combination must be a+c [do you see why?] and the SECOND heaviest must be c+e.
[unfortunately, I can't tell the thirdlightest, or thirdheaviest, without knowing the specific weights  though I do know that the thirdlightest must be EITHER a+d or b+c, etc.]
Still, I already know quite a lot: a+b=110 a+c=112 c+e=120 d+e=121
from the first two I can see that c=b+2; from the last two I can see that d=c+1 [and since c=b+2, I can say d=b+3] from the MIDDLE two I can see that e=a+8
putting this all together, I have a is the smallest weight, b comes next, then b+2 (i.e. "c"), b+3 ("d") and finally a+8 ("e").
Since I also know that a+b=110, this tells me a+d=a+b+3=113 and b+e=a+b+8=118.
I still don't know quite which pairs of bales yield the totals 114,115,116,117, but I do know that they must come from the four remaining pairs:
a+e=2a+8= one of {114,115,116,117} b+c=2b+2= one of {114,115,116,117} b+d=2b+3= one of {114,115,116,117} c+d=2b+5= one of {114,115,116,117}
but note that c+d must be 3 greater than b+c and, of the four numbers as yet unassigned, the only one that's 3 greater than another is 117. this quickly shows that b+c must be 114 and c+d must be 117, i.e. 2b+2=114, so b=56. a=11056, i.e. 54, and everything checks. the weights are 54,56,58,59,62
Ted Alper

