Mental Nuts-- Can You Crack 'em?

[footfootfoot]

sold 49, 28, and 7 bushels @7/$1
1,2, and 3 bushels @1/$3

Nuts.

[ZenGum]

:

I liked Bigv's answer of zero better.

[glatt]

When I was a lad, I remember the older generation deriding what we were being taught as the "new math." Like math could somehow change from one generation to the next. But obviously it did.

[Sundae]

Originally Posted by BigV

The corn was worthless, and they're taking it to the town corn dump.

We might have a butter mountain. But every town has its own corn dump?!
Wow, you guys know how to live

[ZenGum]

If the 2012 asteroid hits just right, we could get some awesome popcorn!

[Clodfobble]

Originally Posted by footfootfoot

sold 49, 28, and 7 bushels @7/$1
1,2, and 3 bushels @1/$3

I have to assume that this answer involves some inherent cultural knowledge of the time, like bushels of corn are always sold either singly or in groups of 7. Because otherwise, it's just retarded.

[footfootfoot]

I think it is a bit of both. and now for the next nut:

(Chrome crashed when I was loading yesterday's nut, so here are two)

[ZenGum]

#17 started with 15 cents.

[HungLikeJesus]

That's what I get also.

[HungLikeJesus]

For #18, >> you have to rent a sheep to get twenty; the first guy gets 10, second 5, third 4, then return the rental sheep.

[ZenGum]

#18Get one of the sheep knocked up?


With the ram, ya pervert

[Clodfobble]

#18

1/2 plus 1/4 plus 1/5 does not equal 1, it equals .95. So 19 is the actual number he left them, not the number he started with. The total number he had was (.95)(x) = 19, so x=20.

[BigV]

A man left nineteen sheep to three heirs. One to get one-half, one one-quarter, and one one-fifth. No sheep were to be killed, and all to be dealt fairly with.

I see how the math works out when you add one sheep to make the total twenty. But why just add one to do that, why not add eighty-one and distribute 50 and 25 and 20 then return 5? This better not be the answer, adding one...

Still, very creative friends I have here!

****

A man left nineteen sheep to three heirs.

Ok, fine. This seems very straightforward. 19 sheep, heirs. Pretty unambiguous.

One to get one-half, one one-quarter, and one one-fifth.

Here's the nut that needs cracking. Half of what? A quarter of what?, a fifth of what? As y'all have pointed out, this doesn't "add up". What's missing? Or what is to be left out? How can these work out... What about half of ... the weight? Regardless, those fractions will never add up to 1, the presumptive "whole". As long as we're talking about only one kind of thing (a group of 19 sheep for example) those pieces will leave some left over.

Nothing in this nut says that 1/2 and 1/4 and 1/5 is the complete list of what was "to get".

Nothing says that the "one" who gets is a different heir. I can imagine that the same "one gets" 1/2, and 1/4, and 1/5 (of the sheep??) so this heir gets 19 sheep. Still leaves 1/20th of 19 sheep unallocated. Grrrrrrr...

Maybe these "fractions" aren't talking about fractions of the flock of sheep. What about one heir getting half the sheep, another heir getting a quarter (a twenty five cent piece or a $2.50 coin, take your pick) and another gets a jug of shine (a fifth). ... ... What am I gonna do about the other "half of nineteen sheep" (ha fucking ha)??

A man left nineteen sheep to three heirs. One to get one-half, one one-quarter, and one one-fifth. No sheep were to be killed, and all to be dealt fairly with.

still thinking...

[footfootfoot]

15 cents

Borrow or rent a sheep

Moving on, a very easy one!

[BigV]

pour three gallons from 8gal cask into 3gal jug leaving five gallons in the 8gal cask

now pour three gallons from the 3gal jug into the 5gal carboy

now pour three gallons from the 8gal cask into the 3gal jug leaving 2 gallons in the 8

shit


three columns, representing three vessels, 8gal cask, 5gal carboy, 3gal jug
each row will always add up to 8 gallons, and the changes, the pours should be obvious.

8--5--3
--------
8--0--0 (pour 3 from cask to jug, filling jug, to get to the next line)
5--0--3 (pour 3 from jug to carboy, emptying jug, to get to the next line)
5--3--0 (pour 3 from cask to jug, filling jug, to get to the next line)
2--3--3 (pour 2 from jug to carboy, filling carboy, leaving 1 in jug, to get to the next line)
2--5--1 (pour 5 from carboy to cask, emptying carboy, to get to the next line)
7--0--1 (pour 1 from jug to carboy, emptying jug, to get to the next line)
7--1--0 (pour 3 from cask to jug, filling jug, to get to the next line)
4--1--3 (pour 3 from jug to carboy, emptying jug, to get to the next line)
4--4--0 (ta-da!)