### FactFinders Cannonballs-Here's How To Solve

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Anybody that can do this without pencil and paper has my awe. Was posed to me in '69 while in VN. Took me best part of 2 days spare time.

Number the balls. Balance 1,2,3,4 vs 5,6,7,8. Assume an imbalance (we'll cover the other possible later). Next do 1,2,5 vs 3,6,9. Note we've left 4,7& 8 out that's a manageable threesome. We're weighing 1,2 6 another manageable threesome. 9 is a known now. The *trick* is with balls 5 & 3. By swapping them we make them tattle on each other if they're guilty. Got it?

If the balance is even @ the 2nd weighing then the unweighed 4,7,8 get our attention. If the balance goes down the opposite way then either 5 or 3 are guilty (the only possible allowing a 2 ball third weighing). If the balance does as before then we're down to 1,2 & 6.

Won't detail the 3rd arrangement cause if you've got it this far you've got it.

Should 1,2,3,4 vs 5,6,7,8 balance. Simple now huh? 9,10 vs 11,1

Again, thanks FactFinder. Never thought this courious bit from the past would ever come up again. Lets me look smart.

-- Carlos (riffraff1@cybertime.net), July 13, 2000

In the original quiz, did anybody answer the "day" problem with "yesterday"? The question DID just say day, not what day of the week......

-- Sam (Wtrmkr52@aol.com), July 13, 2000.

Thanks, Carlos, I think that anyone who solved this puzzle in three weighs more than just "looks" smart.

I solved it in four rather easily, but trying to do it in three gave me a good lesson in humility. I first weight three balls each side (which is the way I solved it in four), since by doing allows you to eliminate 6 balls on the first weigh. Weighing 4 on each side can eliminate 8 if the balls balance, but if they don't, then you've only eliminated 4. So I went with three on each side, and eventually solved it in three weights ....or so I thought! The next day, I took another (fresh) look and found a case that would not allow me to solve it in three weighs....so I worked it again, this time weighing four balls each side and solved it.

This was a bit counterintuitive to me, since you could have a situation where the balls didn't balance, and you would have eliminated only four, yet still solve the problem in three weighs! The key is that you learn information about ALL the balls in the case of the unbalanced first weigh: * You know the four unweighed balls are "good" balls and can thus eliminate them from further consideration * You know that one group of four balls are heavier than the second group of four balls (And vice versa). So, each ball on the heavy side can be labeled as "heavy or good" and each ball on the light side can be labeled as "light or good".

Of course, if the balls balance, the "oddball" is in the pile of four unweighed balls. In this case, it's rather straightforward to solve in two more weights.

All in all, a fun and challenging puzzle. Like Carlos, two days to solve! I really liked NGs possible solution "I know when I picked it up" (speaking of the oddball), this is way easier, lol.

Sam, regarding the "yesterday" quiz, no one did present "yesterday" as the answer until you (unless I missed it), and it appears to be technically correct to me, but not as much fun as working out the day of week ;)

-- FactFinder (FactFinder@bzn.com), July 13, 2000.

lol...... Try this one :-)

Subject: Missing Dollar

THE MISSING DOLLAR!

3 MEN GO INTO A MOTEL. THE MAN BEHIND THE DESK SAID THE ROOM IS \$30, SO EACH MAN PAID \$10 AND WENT TO THE ROOM.

A WHILE LATER THE MAN BEHIND THE DESK REALIZED THE ROOM WAS ONLY \$25, SO HE SENT THE BELLBOY TO THE 3 GUYS' ROOM WITH \$5.

ON THE WAY THE BELLBOY COULDN'T FIGURE OUT HOW TO SPLIT \$5 EVENLY BETWEEN 3 MEN, SO HE GAVE EACH MAN \$1 AND KEPT THE OTHER \$2 FOR HIMSELF.

THIS MEANT THAT EACH MAN PAID \$9 FOR THE ROOM, WHICH IS A TOTAL OF \$27, ADD THE \$2 THAT THE BELLBOY KEPT = \$29...

WHERE IS THE OTHER DOLLAR???????????

-- Netghost (ng@no.yr), July 13, 2000.

If seen that one, NG, it makes you think a bit the first time! Of course there is no "other dollar", the men paid \$27, the hotel got \$25 of it and the bellboy \$2.

Or mathatically, \$30 - \$5 = \$27 - \$2

-- FactFinder (FactFinder@bzn.com), July 14, 2000.

Doh! It wasn't a matter of a cat-skinning problem (As I said, it was obvious to me how to get down to three balls after the second weighing). What I didn't realize at the time is that the second weighing also tells you whether the oddball is heavy or light! In other words, you do it the same way as the four-weighing solution, just pay a bit more attention to the results.

This is one of the best logic problems I've come across in a long time. It almost makes all the annoyed lurking I've done in these forums worthwhile. Thanks factfinder!

-- paul (anonymous@coward.com), July 11, 2000.

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-- anon (anon@anon.anon), July 17, 2000.