# 2 brain teasers that I am having trouble with...
**URL:** https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165
**Category:** Factual Questions
**Created:** [October 3, 2001, 12:23pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165 "2001-10-03T12:23:24Z")
**Posts on this page:** 9
**Page:** 2
### Author:  [@Patricinus\_Scriblerus](https://boards.straightdope.com/u/Patricinus_Scriblerus)
#### Post date: [October 3, 2001, 2:26pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/21 "2001-10-03T14:26:09Z")
> [@](#):
>
> \*Originally posted by CurtC \*
> \*\*No, the fake statue can always be found with a maximum of **three** weighings.
>
> And **grimpixie** , one will never be enough, since even if you picked two trophies and got lucky, you still wouldn’t know whether the lighter or heavier one was the fake.
>
> \*\*
I know you can find the fake with 3 scalings but determining if it is heavier of lighter takes the 4th right?? If not please explain.
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### Author:  [@erislover](https://boards.straightdope.com/u/erislover)
#### Post date: [October 3, 2001, 2:54pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/22 "2001-10-03T14:54:55Z")
The algorithm used for the weighing problem is somewhat complicated. I spent some time figuring it out (heard it told differently, though—using pool balls, but the core riddle is the same).
The solution can only be accurately depicted as a sort of flow chart. First weighing is 4 and 4. You have two possible outcomes. Either the scale will tip or it won’t. From there each additional weighing has more outcomes.
If weighed correctly the answer is indeed three. Of course, given infinite weighs the answer is easy, but the smallest number of weighs to guarantee success _not_ knowing at the start whether the “fake” is heavier or lighter is certianly three.
If anyone is interested I know a webpage which will check the solution you propose for it. **I will not solve this riddle for someone** as it is one of my favorites and finding the solution can prove to be very difficult, even though many here have started anyone off in the right direction to solving it.
Rest assured the maximum numer of weighs needed is really three.
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### Author:  [@jcgmoi](https://boards.straightdope.com/u/jcgmoi)
#### Post date: [October 3, 2001, 3:49pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/23 "2001-10-03T15:49:00Z")
[Here](http://boards.straightdope.com/sdmb/showthread.php?threadid=44387) you will find a previous SD discussion of the 12 coins problem.
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### Author:  [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)
#### Post date: [October 3, 2001, 4:00pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/24 "2001-10-03T16:00:42Z")
Given that these are brainteasers I don’t feel too bad in saying I just don’t understand the answer to #2.
Total Mass = 1000
of that…
Water = 990
Solid = 10
Now, the water has lost 1% of its mass over the course of a month so…
990 = 100% (of the water mass)
990 \* 0.01 = 9.9 (or the weight of 1% of the water)
990 - 9.9 = 980.1 (weight of water after losing 1% of its mass)
New total weight of berries…
980.1 + 10 = 990.1 (or rounded to 990)
How does water losing 1% of its total mass nearly halve the final weight? What am I missing? More importantly, please explain the logic to me.
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### Author:  [@KeithT](https://boards.straightdope.com/u/KeithT)
#### Post date: [October 3, 2001, 4:20pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/25 "2001-10-03T16:20:36Z")
**Whack-a-mole** , you’re just misinterpreting what the change in percent means. The pie didn’t lose 1% of its water mass; its composition changed from 99% water and 1 percent solid to 98% water and 2% solid.
990 kg H[sub]2[/sub]O : 10 kg solid is 99% water
980 kg H[sub]2[/sub]O : 20 kg solid is 98% water
But, the mass of the solids can’t change. So, dividing each side by two:
490 kg H[sub]2[/sub]O : 10 kg solid is 98% water
The sum of these values is 500kg.
QED
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### Author:  [@duffman](https://boards.straightdope.com/u/duffman)
#### Post date: [October 3, 2001, 4:20pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/26 "2001-10-03T16:20:56Z")
whack-a-mole,
Your confusion lies in the statement “Now, the water has lost 1% of its mass over the course of a month so…”. Really the water has lost alot more than 1 percent of its mass. The only thing that stays constant in this problem is the mass of the solid material so you have to use the percentages that are given for it.
initially
0.01 \* 1000 = 10kg of solid
after a month the 10kg of solid comprise 2 percent of the total, so use proportion
10/2 = total/100
therefore total = 500
hope that helps
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### Author:  [@duffman](https://boards.straightdope.com/u/duffman)
#### Post date: [October 3, 2001, 4:23pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/27 "2001-10-03T16:23:13Z")
whack-a-mole,
Your confusion lies in the statement “Now, the water has lost 1% of its mass over the course of a month so…”. Really the water has lost alot more than 1 percent of its mass. The only thing that stays constant in this problem is the mass of the solid material so you have to use the percentages that are given for it.
initially
0.01 \* 1000 = 10kg of solid
after a month the 10kg of solids comprise 2 percent of the total, so use proportion
10/2 = total/100
therefore total = 500
hope that helps
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### Author:  [@ShibbOleth](https://boards.straightdope.com/u/ShibbOleth)
#### Post date: [October 3, 2001, 4:32pm UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/28 "2001-10-03T16:32:19Z")
> [@](#):
>
> \*Originally posted by Whack-a-Mole \*
> \*\*Given that these are brainteasers I don’t feel too bad in saying I just don’t understand the answer to #2.
>
> Total Mass = 1000
> of that…
> Water = 990
> Solid = 10
>
> Now, the water has lost 1% of its mass over the course of a month so…\*\*
The water hasn’t lost 1% of it’s mass, instead the ratio has been shifted. If you start thinking about it from the solid (constant) perspective it makes more sense.
Ah, I see **jwalk01** saying the same thing.
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### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [October 4, 2001, 12:40am UTC](https://boards.straightdope.com/t/2-brain-teasers-that-i-am-having-trouble-with/85165/29 "2001-10-04T00:40:40Z")
Some general material on the “counterfeit coin” problem (yes, it’s presented as statues, here, but coins are the traditional form): You need to find one statue out of 12, and you want to know whether it’s heavy or light. That means that you need to distinguish between 24 different possibilities. Now, each weighing can have one of three possible outcomes: Left heavy, right heavy, balanced. That means that one weighing can distinguish between as many as three possibilities, two weighings can distinguish between 9, and three weighings can distinguish between 27 possibilities. 27 is greater than 24, so we just barely have enough information to solve the problem.
The key to solving the problem is that third possibility, balance. If you set up a weighing that can’t balance, you’re probably wasting a weighing. For instance, if we do the first weighing 6 vs. 6, we know that the scale won’t balance, so we’ve lost that much information. In general, you’ll want to set up each weighing such that there’s approximately a 1/3 chance of each outcome: That’s why we start with 4 vs. 4, because there’s a 1 in 3 chance each of the counterfeit being on the left or right, or off the scale.
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