A logic (Maybe...) problem

You have 9 marbles: 8 of them weigh 1 ounce each; 1 weighs 1.1 ounce. The 9 marbles are all uniform in size, appearance and shape. You have access to a balance scale containing 2 trays - you may use the balance 2 times. You must determine which of the 9 marbles is the heavier one using the balance only 2 times.

If there are really people who don't know that one yet, I'll leave it to them. 🙂


How about this one now:

You have 10 bags with 10 coins in each bag. Each coin in all the bags weighs 1 gramm, except for all coins in one (the odd) bag, which all weigh 1.1 gramm. Using a digital balance, you may do only one weighing to find the odd bag. How can you do that?

(One weighing = Put something on the balance, see what weight is displayed and then switch off the balance right away).

and the advanced problem is if you have 11 real coins and 1 counterfeit coin which is either lighter or heavier (you don't know which)
3 wayings on a two-tray scale. you must determine which coin is the counterfeight and whether it is lighter or heavier.

Originally posted by chess kid1
You have 9 marbles: 8 of them weigh 1 ounce each; 1 weighs 1.1 ounce. The 9 marbles are all uniform in size, appearance and shape. You have access to a balance scale containing 2 trays - you may use the balance 2 times. You must determine which of the 9 marbles is the heavier one using the balance only 2 times.

I think i have it.

If you weigh 6 first, 3 on each tray, then either one tray will be heavier or they will balance, meaning the remaining three marbles not first placed on the trays contains the heavier one. From whichever three the heavy marble is discovered to be in, two marbles can be removed and placed in the trays. Now either one marble will be heavier and therefore the heavy marble, or they will balance meaning the remaining third marble of the three is the heaviest.

Therefore you have found the heavy marble using the scales only twice.

Originally posted by crazyblue
If there are really people who don't know that one yet, I'll leave it to them. 🙂


How about this one now:

You have 10 bags with 10 coins in each bag. Each coin in all the bags weighs 1 gramm, except for all coins in one (the odd) bag, which all weigh 1.1 gramm. Using a digital balance, you may do only one weighing to find the odd bag. How can you d ...[text shortened]... hing on the balance, see what weight is displayed and then switch off the balance right away).

Could you put one coin from one bag, two coins from a second bag three coins from a third bag, and so on (all at the same time)? how many ever tenths of a gram it is over a full gram is the bag that had the 1.1 gram coins. Am i correct in my assumption?

Originally posted by hypothetical
Could you put one coin from one bag, two coins from a second bag three coins from a third bag, and so on (all at the same time)? how many ever tenths of a gram it is over a full gram is the bag that had the 1.1 gram coins. Am i correct in my assumption?

yes they are.
and for the advanced problem: i've expressed it as a flowchartish
1: ballence1 sidea 4 against sideb 4
2: if 1 yeilded a ballence goto 17
3: mark which coins were on the high side and which were on the low side and which were unweighed in 1 as h, l, and b (for blank) respectivly
4: ballence2 sidea 3b and l against sideb 3l and h
5: if 4 yeilded a ballence goto 14
6: mark which coins were low, high, and unballenced in 4 as i, m, and c respectivly in addition to their previous markings
7: if 4 yeilded sidea high goto 11
8: ballence3 sidea hi against sideb bc
9: if 8 yeilded a ballence then lm is conterfit and goto end
10: hi is conterfit and goto end
11: ballence3 sidea lm against sideb lm
12: if 11 yeilded a ballence then the lm which was not ballenced in 11 is conterfit and goto end
13: the lm which weighed low in 11 is counterfit and goto end
14: ballance3 sidea h against sideb h
15: if 14 yeilded a ballence then the h which was not ballenced in 14 is counterfit and goto end
16: the h which weighed high in 14 is conterfit and goto end
17: mark the coins which were ballenced and not ballenced in 1 as d and u respectivly
18: ballence2 sidea 3d agianst sideb 3u
19: if 18 yeilded a ballence then the u which was unballenced in 18 is counterfit and goto end
20: change the marking on the u which was unballenced in 18 to a d
21: ballence3 sidea u against sideb u
22: if 21 yeilded a ballence then the u which was unballenced in 21 is counterfit and goto end
23: if 18 yeilded sideb high then the u which weighed high in 21 is counterfit and goto end
24: if 18 yeilded sideb low then the u which weighed low in 21 is counterfit
end

in retrospect there must have been a better way of expressing that.
i'm pretty sure it's airtight, but it dont have any more paintence with which to check.

ok how about this one
you are standing in a large square room 100x100x10 ft
in the center of the room hang down to ropes each 10 ft long. the ropes are 16ft apart. your objective is to tie the two ropes together to form a triangle with the ceiling. your only tool is a large razor sharp machette.
GO!

Originally posted by aginis
ok how about this one
you are standing in a large square room 100x100x10 ft
in the center of the room hang down to ropes each 10 ft long. the ropes are 16ft apart. your objective is to tie the two ropes together to form a triangle with the ceiling. your only tool is a large razor sharp machette.
GO!

umm...

Good reply!

Originally posted by Gastel
Good reply!

Why?

It wasn't... but thanks for getting the verification. You continue to prove your ineptitude.

Originally posted by aginis
ok how about this one
you are standing in a large square room 100x100x10 ft
in the center of the room hang down to ropes each 10 ft long. the ropes are 16ft apart. your objective is to tie the two ropes together to form a triangle with the ceiling. your only tool is a large razor sharp machette.
GO!

What is the problem here? You grab one rope and swing on it to another rope and tie them together, you don't need a knife.
You just start some rope swinging, maybe tie the machete to on of them to make a pendulum and swing on another rope to catch the one with the knife. Big deal. You could ask what the angles of the resultant triangle are, assuming the ropes are 10 feet long at the end of the operation so each short leg is ten feet and the base 16 feet. At least you would have to do some trig.

Originally posted by chess kid1
You have 9 marbles: 8 of them weigh 1 ounce each; 1 weighs 1.1 ounce. The 9 marbles are all uniform in size, appearance and shape. You have access to a balance scale containing 2 trays - you may use the balance 2 times. You must determine which of the 9 marbles is the heavier one using the balance only 2 times.

I seem to have lost my marbles. Is it possible to use a substitute?

Originally posted by FreakyKBH
I seem to have lost my marbles. Is it possible to use a substitute?

Just use your balls.

Originally posted by sonhouse
Just use your balls.

I only have three of those. Could you send your mum and two of her sisters over?