[wolfgang]

If you have an unlimited amount of suger,One standared 5 kilos wieght,and a balance scale.

I want you to weigh exactly 7 kilos suger using the balance scale ONLY.

you can use it as much as you want!

Any other method is NOT allowed.

[Guest]

take five kilos of sugar then weigh more until the scale is only almost balanced.

Thats a tough one but I heard something like it before. Could it be...

Edited March 11, 2011 by D.H.

[Guest]

take five kilos of sugar then weigh more until the scale is only almost balanced.

Thats a tough one but I heard something like it before. Could it be...

If you were to weigh 5 kilos of sugar then weigh another 5 kilos of sugar and divide it into 5 equal piles then remove 3 of the piles. You could further test the accuracy of your split by first weighing 2 of the piles against 2 other piles to ensure you were in fact accurate in your division.

[Guest]

Measure 5 kilos, split that in half two piles of 2.5 K split one pile again .625k do that 6 more times giving you .3125, .15625, .078125, .0390625, .01953125, .009765625.

combine 1.25+.625+.078125+.0390625+.009765625= 2.001953125 add that to another weighting of 5k

[Guest]

Sorry forgot spoliers

[fabpig]

Measure 5 kilos, split that in half two piles of 2.5 K split one pile again .625k do that 6 more times giving you .3125, .15625, .078125, .0390625, .01953125, .009765625.

combine 1.25+.625+.078125+.0390625+.009765625= 2.001953125 add that to another weighting of 5k

Yeah I was thinking along those lines.....unforunately, that's not "exact"

[Guest]

I bet if you drill down enough it will be

[curr3nt]

Get 5k of sugar

Divide into 5 piles and keep weighing and adjusting until all 5 are equal to each other => 5 1k piles.

Get 5k more of sugar and add two of the 1k piles.

Not terribly quick but it can be done.

[Guest]

It's somewhat similar to what the others said, but I was thinking it the other way around.

You weigh 5 kg and then you take another 5 and so on until you have taken 5 kg seven times. So you should have a total of 35 kg. now you split the 35 kg into 5 piles and a pile should weigh 7 kg.

However, I don't know how accurate this is and as I said before, it's quite the same thing in comparison to the previous solutions.

Anyway, I'm looking forward to the correct answer and great job, wolfgang, your puzzles are quite intriguing and exciting to think of.

[Guest]

1. Weigh 5 lbs of sugar. Divide this into two equal parts, giving 2 1/2 lbs. Divide once again, giving two piles of 1 1/4 lbs. Save in two separate piles.

2. Weigh another 5 lbs sugar. Place one of the piles of 1 1/4 lbs of sugar on one side of the scale and enough of the 5 lb pile to balance. This will leave 3 3/4 lbs.

3. Divide the 3 3/4 lbs into two equal parts, giving two piles 1 7/8 lbs each.

4. Weigh another 5 lbs sugar. Place one of the piles of 1 7/8 lbs of sugar on one side of the scale and enough of the 5 lb pile to balance. This will leave 3 1/8 lbs.

5. Combine a pile of the 1 1/4 lb pile of sugar with the pile of 3 1/8 lb of sugar, giving a pile weighing 4 3/8 lbs.

6. Weigh another 5 lbs sugar. Place the 4 3/8 lb of sugar on one side of the scale and enough of the 5 lb pile, leaving 7/8 lb.

7. Place the 7/8 lb pile of sugar on one side of the scale and enough of the sugar from the 1 7/8 lb pile of sugar from step 3, leaving 1 lb in the pile.

the rest is simple to complete.

[Guest]

1. Weigh 5 lbs of sugar. Divide this into two equal parts, giving 2 1/2 lbs. Divide once again, giving two piles of 1 1/4 lbs. Save in two separate piles.

2. Weigh another 5 lbs sugar. Place one of the piles of 1 1/4 lbs of sugar on one side of the scale and enough of the 5 lb pile to balance. This will leave 3 3/4 lbs.

3. Divide the 3 3/4 lbs into two equal parts, giving two piles 1 7/8 lbs each.

4. Weigh another 5 lbs sugar. Place one of the piles of 1 7/8 lbs of sugar on one side of the scale and enough of the 5 lb pile to balance. This will leave 3 1/8 lbs.

5. Combine a pile of the 1 1/4 lb pile of sugar with the pile of 3 1/8 lb of sugar, giving a pile weighing 4 3/8 lbs.

6. Weigh another 5 lbs sugar. Place the 4 3/8 lb of sugar on one side of the scale and enough of the 5 lb pile, leaving 7/8 lb.

7. Place the 7/8 lb pile of sugar on one side of the scale and enough of the sugar from the 1 7/8 lb pile of sugar from step 3, leaving 1 lb in the pile.

the rest is simple to complete.

Step six is flawed, 5 - 4 3/8 = 5/8, not 7/8.

[fabpig]

1. Weigh 5 lbs of sugar. Divide this into two equal parts, giving 2 1/2 lbs. Divide once again, giving two piles of 1 1/4 lbs. Save in two separate piles.

2. Weigh another 5 lbs sugar. Place one of the piles of 1 1/4 lbs of sugar on one side of the scale and enough of the 5 lb pile to balance. This will leave 3 3/4 lbs.

3. Divide the 3 3/4 lbs into two equal parts, giving two piles 1 7/8 lbs each.

4. Weigh another 5 lbs sugar. Place one of the piles of 1 7/8 lbs of sugar on one side of the scale and enough of the 5 lb pile to balance. This will leave 3 1/8 lbs.

5. Combine a pile of the 1 1/4 lb pile of sugar with the pile of 3 1/8 lb of sugar, giving a pile weighing 4 3/8 lbs.

6. Weigh another 5 lbs sugar. Place the 4 3/8 lb of sugar on one side of the scale and enough of the 5 lb pile, leaving 7/8 lb.

7. Place the 7/8 lb pile of sugar on one side of the scale and enough of the sugar from the 1 7/8 lb pile of sugar from step 3, leaving 1 lb in the pile.

the rest is simple to complete.

but as I understand it this is assumin 5lb - 4 3/8lb = 7/8lb?

Told u I was slow MaestroOD beat me to it!

Edited March 12, 2011 by fabpig

[benjer3]

Wouldn't all of that handling of the sugar get it all moist and sticky, giving you more weight than just sugar?

Edited March 12, 2011 by benjer3

[Guest]

too late

Edited March 12, 2011 by solman

[wolfgang]

Deviding any amount of suger into piles without using a balance scale is not allowed.

Edited March 12, 2011 by wolfgang

[witzar]

6. Weigh another 5 lbs sugar. Place the 4 3/8 lb of sugar on one side of the scale and enough of the 5 lb pile, leaving 7/8 lb.

Wrong.

[witzar]

1. If you can weigh x and y then you can weigh x+y.

2. If you can weigh x and y then you can weigh x-y (for x>y).

3. If you can weigh x then you can weigh x/2.

And that's all.

Problem of finding 7 out of 5 is equivalent to finding 1 out of 5, because:

7 = (5+1)+1 (if you know how to weigh 1, you know how to weigh 7)

and

1 = (7-5)/2 (if you know how to weigh 7, you know how to weigh 1)

And problem of finding 1 out of 5 is equivalent to finding 1/5 out 1 (solution for one works for another, just with another starting pile).

Suppose that the last problem is solvable. Then there is a finite number of operations 1.-3., that produce 1/5 out of 1.

But how can finite number of additions, subtractions and divisions by 2 of number 1 produce a (irreducible) fraction with denominator 5?

It just can't. Clearly only powers of 2 are possible in the denominator of this fraction (in reduced form).

This shows that the original problem in also not solvable.

[Guest]

[spoiler=Well this may work ]5 x 5=25

25-2.5=22.5

22.5-1.25=21.25

21.25 x 3= 63.75

63.75 / 2 =31.875

31.875 / 2 =15.9375

15.9375 / 2 =7.96875

7.96875=7.9688=7.969=7.97=8

8 - 5 =3

3 x 2=6

6-5=1

1x2=2

2+5=7

[Guest]

but as I understand it this is assumin 5lb - 4 3/8lb = 7/8lb?

Told u I was slow MaestroOD beat me to it!

Sorry for the brain freeze.

Let's try this. If we can get a weight of 1 kilo of sugar, we can get 7 kilos. The idea I am using is to create 5 separate piles of sugar that weigh the same and adjust them until their total weight is 5 kilos.

1. Using the balance scale create 5 small but equal weighing piles of sugar.

2. Place the five piles of sugar onto one side of the balance scale, keeping them separate from each other, with the 5 kilo weight on the other.

3. If they balance, each pile of sugar will weigh 1 kilo and you are done. Otherwise,

4. Remove the 5 kilo weight and 4 of the piles of sugar, again keeping the sugar in separate piles. If the total weight of the sugar was greater than 5 kilos, remove a small amount of sugar from the pile on the balance scale. If the weight was less than 5 kilos, add a small amount of sugar.

5. Use the balance scale to equalize the weight of each of the piles of sugar one at a time.

5. Go back to step 2.

Eventually, the 5 individual piles of sugar will balance with the 5 kilo weight if you add or remove smaller and smaller amount of sugar as needed.

Sorry for the brain freeze.

Let's try this. If we can get a weight of 1 kilo of sugar, we can get 7 kilos. The idea I am using is to create 5 separate piles of sugar that weigh the same and adjust them until their total weight is 5 kilos.

1. Using the balance scale create 5 small but equal weighing piles of sugar.

2. Place the five piles of sugar onto one side of the balance scale, keeping them separate from each other, with the 5 kilo weight on the other.

3. If they balance, each pile of sugar will weigh 1 kilo and you are done. Otherwise,

4. Remove the 5 kilo weight and 4 of the piles of sugar, again keeping the sugar in separate piles. If the total weight of the sugar was greater than 5 kilos, remove a small amount of sugar from the pile on the balance scale. If the weight was less than 5 kilos, add a small amount of sugar.

5. Use the balance scale to equalize the weight of each of the piles of sugar one at a time.

5. Go back to step 2.

Eventually, the 5 individual piles of sugar will balance with the 5 kilo weight if you add or remove smaller and smaller amount of sugar as needed.

[Guest]

Since it seems impossible to measure exactly one kilo, maybe we could try to cheat the scale. if we know that we can measure exactly any number dividable by 5 we can use proportion

5 : 25 = 7 : 35

So, first we measure 10kg of sugar and put it aside, than we measure 25 kg of sugar, put it on one side of scales and put 5 kg weight on the other side. 25kg side goes down, another goes up, we mark on scale excat height where lighter side went up. Now we put 10kg on 25kg side which makes 35kg, on the other side we remove weight and add sugar untill it goes down to the line we marked. Voilla, 7kg of sugar measured! I'm not sure if this is possible though, and how precise it would be but i guess it can be very accurate if done carefully.

[Guest]

I can't see the spoilers on my mobile so sorry if this method is already posted and said to be wrong. But i'll give it a try.

Weight 5 times 5 kg and put each of them apart. U have now 5 piles of 5 kg each. Take an almost equal amount from each pile (using any reference like the grab of ur hand), a bit less than what u see as 1/5 of a pile, and weight the sugar, add to that again an equal amount from each pile, till the averall sugar u removed is exactely 5kg.

U will know now than by reducing 5kg of sugar, u reduced 1 kg of each pile, and there are 4kg remaining in each. Do that again, now u have 3kg in each pile, and once more, u'd have 2kg in each.

Weight 5kg of sugar again and add them to any pile of 2kg. Now it's 7kg

[wolfgang]

I can not solve if it was Flour instead of Suger!

[Guest]

I can not solve if it was Flour instead of Suger!

Hey!! What do you mean by SUGER

[Guest]

I know, you measure 10KG and bring kindergarten class in and tell to the kids to eat eaxtly 3kg. No more, no less. They are sugar experts, you know!

Edited March 13, 2011 by Ctpubop

[wolfgang]

I know, you measure 10KG and bring kindergarten class in and tell to the kids to eat eaxtly 3kg. No more, no less. They are sugar experts, you know!

[benjer3]

Wolfgang's hint made me think...

weighing 5kg of the sugar, counting the crystals, and adding another two-fifths. Of course, you would probably need tweezers to do that, and it would take forever, but it would work with very good accuracy.