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/sci/ - Science & Math

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2542301 No.2542301 [Reply] [Original]

You are given 8 identical-looking balls, one of which is heavier than the other 7 (all of which weigh the same). Using an old-fashioned mechanical set of scales you must identify the heavier ball using the scale as few times as possible. The scale is constructed using two bowls and an arm enabling the bowls to either balance or have one bowl rising while the other (and heavier bowl) falling. You can't just add one ball at a time thinking its one weighing, however, you may put any number of balls in each bowl. How do you find the heavier ball in only 2 weighings?

>> No.2542310

Uhhhhh

>> No.2542319

put 4 balls on each side of the scale
then split the heavier side into 2 groups of 2
put those 2 groups on the scale
then split that heavier side into 1 and 1
bingo bango I fucked you mom

>> No.2542324

>>2542319
clearly 3 weighings

>> No.2542329

Place 3 on one side and 3 on other side. One of these things will happen;

1) One side dips down indicating it's heavier, if so, place 2 random balls from that side on the scale again, to compare one to the other. If they're the same weight that means the ball you excludd is the heaviest, if one side is more heavy then you found your ball.

2) Both sides are equal, meaning that one of the two balls left is the heavy one, just weigh them against eachother,

>> No.2542336
File: 32 KB, 400x330, Near happy puzzle.jpg [View same] [iqdb] [saucenao] [google]
2542336

Simple logic is simple...

let's see...take 6 balls and put three in each side.

If one side is heavier, then the heavier ball is one of the three that fell.

Take two of those three balls and place one on either side of the scale. If the two balls you weighed are equal, then the third ball you didn't weigh is the heavier one.

Also, for the two balls you didn't measure at the beginning, if each of the three balls weigh the same then you know the heavier ball is one of the two you left out.

Simply weigh those two and you'll find out which one it is.

/thread

btw that was kind of fun; do you have any other similar riddles? :3

>> No.2542339

>>2542329
*excluded

>> No.2542340

1. 3 on one side, 3 on the other
IF THEY BALANCE:
2. Weigh the other 2
IF 1. DOESN'T BALANCE
2. Weigh 2 of the heavier 3
IF THEY BALANCE
The third is heavier
IF THEY DO NOT BALANCE
The heavier ball is heaviest

2 weighings.

>> No.2542343

You measure 3 and 3 - there are two possibilities
1) It's balanced
-You measure the two balls that haven't been weighed
2) It's imbalanced
-You measure 2 of the 3 balls in the group
--If it's equal, then the unweighed of the 3 is the heavier ball
--If it's imbalanced, the heavier side contains the heavier ball

>> No.2542346

3 balls in each side.

scenario A) the sides balance, the heavy ball is one of the two you didn't put in, weigh them, find the heavy one


Scenario B) one side is heavier. Take the three balls from this side, put all other balls away, arbitrarily pick two of these 3 balls to weigh. If they balance, the heavy one is the one you didn't weigh, if they don't balance, the heavy one is the one thats lower.

I enjoyed this very much ty op

>> No.2542359

Op here, it was a question on an application, I'm just not the smartest guy around. I appreciate the help :)

>> No.2542368
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2542368

>>2542359

What was the application for?

Good luck! :3

>> No.2542364

So I'm just a ball chillen with 6 of my homies and that huge fuck fat albert when this niggah yanks 2 and me of my hommies up and puts us on a

>> No.2542372

Weighing 3v3 on the first weighing, you get two situations: Either one is heavier or they are the same. If they are the same, measure the other two and find the heavier.
In the case that they are different, let us label the three heavier balls 1 2 and 3, and the lighter ones 4 5 and 6. Weigh 1v2. If they are different, the heavier one is the heaviest. If they are the same, the 3 is the heaviest.
HAHAHAHAHAH

>> No.2542382

>>2542301
weight 3 and 3. If weigh the same, the last ball is heavier

If one is heavier discard the other 4 balls. take any 2 of the 3 left and weight against each other. If one is heavier that is the ball. If same the 3rd unweighted ball is the heavy one.

>> No.2542384
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2542384

btw guys if you like these types of problems you should start studying problem solving in depth.

There's this national contest for high schoolers called USAMTS, which features 6 math/logic puzzles that are highly complex but also very fun.

Requires tremendous problem solving skills, creativity and knowledge of high school level mathematical theory.

>> No.2542385

Similarly, suppose that you have 12 balls that weigh the same except for an odd ball which is either heavier OR lighter than the rest. Using a relative scale that is similar to OP's question, and only 3 measurements, how do you find the odd ball?

>> No.2542388

>>2542368
Application for a guild on WoW.
I'm a huge WoWfag.

>> No.2542393
File: 35 KB, 210x358, near 4rvfa4.jpg [View same] [iqdb] [saucenao] [google]
2542393

>>2542385

You can use literally the same logic for the last problem and apply it to this one, so why ask the question twice? haha

You guys are so cute :3

>> No.2542398

-put 7 balls on one side and one ball on the other.
-calculate the difference of in-equal distribution
-find the off balance, if their is no off balance then that means one of 2 things.
1)the heavy ball is the one by itself
2)the heavy ball is the same weight as the others

-put 4 balls on one side
-carefully place the 4 other on the other side, looking for the in-equal distribution.
-go fuck your mother.

OR. electro magnetic vibrations

>> No.2542402
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2542402

>>2542388

So the people in that guild are so pathetic they're trying to turn the application process for their club into some sort of university app thing?

Sounds like you shouldn't socialize with those people at all.

There's way better ways to spend your time OP. Learn about cosmology, develop problem solving skills, go outside and make friends, etc.

>> No.2542400

>>2542385
are you sure? without knowing whether it's lighter of hravier?

>> No.2542415

>>2542385
IMPOSSIBIRU!!!

>> No.2542430

OPs pic is more interesting than his post

>>2542340
This seems to work though

Good job

>> No.2542476

>>2542415

This anons' correct; it would literally be impossible to solve this problem in 3 measurements.

>> No.2542491

>>2542476
It's initially difficult to believe, but there is a solution. This puzzle is posted at least daily on /sci/.

>> No.2542523

>>2542393
>>2542415

[x] told.

>> No.2542529

>>2542402
dude, he only cares about making the approved impression necessary to participate in raids without retards
a lot of people in WoW value guilds without people who will screw you over
is basic logic!

>> No.2542551
File: 270 KB, 1440x900, Light_Yagami,_Death_Note,_Manga.jpg [View same] [iqdb] [saucenao] [google]
2542551

>>2542523

>> No.2542589

least amount of times i can think of is 3 simply by putting 4 balls on each side of the scale, taking the 4 from the side that was heavier and placing 2 on each side, then taking the 2 from the heavier side and placing them on either side to determine the heavier one.

Perhaps im not thinking cleverly enough, but 3 steps seems to be the most logical and straight-forward solution.

>> No.2542654
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2542654

>>2542393
not so fast..

>> No.2542656
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2542656

>>2542523

>> No.2542667

>>2542491
i'm actually pretty sure there isn't
you'd need two weighings alone just to figure the odd one out of 3 balls

>> No.2542686

Split into 3 groups. Let group A have balls 1,2,3, B have balls 4,5,6 and group C have balls 7,8. Weigh A vs B. If equal, weigh group C, easy. If not equal, then WLOG we can assume heavy ball is in group A (balls 1,2,3). Then weigh balls 1,7 against 2,8. If not equal, then ball 1 or 2 is heavy (depending which side fell). If not equal, then 3 is heavy.

Shit's easy.

>> No.2542698

>>2542686
why does an idiot like you feel the need to post the fucking solution when it's been answered a thousand fucking times in the same thread?

>> No.2542732

>>2542491
DO share

>> No.2543162

>>2542385

>Similarly, suppose that you have 12 balls that weigh the same except for an odd ball which is either heavier OR lighter than the rest. Using a relative scale that is similar to OP's question, and only 3 measurements, how do you find the odd ball?

got it!

okay first split to two groups, 8 and 4.

take 8 balls, split to two groups of four, measure.

if the group is balanced you have 4 balls, two masseurs, easy.

if it is unbalanced from this point on balls on side that went up are:
l1, l2, l3, l4

balls on side that went down are:
h1, h2, h3, h4

put h1, h2, l1, l2, l3 on left side.

put four of the balanced balls (the ones not in the original group of 8 balls) on the right side and l4.

now if the right side goes up, l4 is odd.

if left side goes up either l1, l2, l3 are odd.

measure l1 and l2, if they are balanced l3 is odd.
if l2 goes up it is odd.
else l1 is odd.

left side goes down h1 or h2 are odd.
measure, the one that goes down is odd.

if it is balanced non of the balls are odd, meaning either h3 or h4 are odd.

put both on scale and the one that goes down is the odd ball.

this was a fun riddle.

>> No.2543175

>>2543162

a crap, sorry if right side goes up it's either l4 or h1 h2.

it's not that hard either, way h1 and h2.

if they are balanced it's l4, if they are unbalanced the heavier one is odd.

fuck can't belive i missed that.