## Balance the scale puzzle

How many blue faces are needed to balance scale #3?

*[Brain teaser created for smartkit site; click picture to jump to hi-res image] *

Note: Even though this puzzle is very hard, the comments below are actually quite insightful. If you can follow them, your brain is almost certain to get a decent workout!

For a different type of scale balancing puzzle (one that’s been asked on job interviews!), click here.

Joe| Guest May 16th, 2007 - 6:05 pm2 green and 2 red equal 2 purple.

On the next scale you can take away 2 green and 2 red and 2 purple on the other side to get 4 green being the equivalent of 2 purple and 1 orange.

For the last scale 4 of the greens will equal 2 purple and 1 orange, and the remaining 1 green and 1 red will equal 1 purple.

So the other side of the scale should have 3 purple and 1 orange.

arvin| Guest May 17th, 2007 - 1:25 ambut it asks how many blue’s you need, i assume it means that you need to only use blue faces? and you cannot use it in conjunction with another colored face?

jusdiver| Guest May 17th, 2007 - 10:23 amI worked it out exactly a Joe did before and came to the same conclusion (3 blues and 1 yellow). I can’t see how you can use only blues unless you adjust the position of the greens and red to occupy a position closer to the pivot so that they balanced perfectly with 2 blues. I await someones answer with interest.

max| Guest May 17th, 2007 - 10:48 amIf it were asking which combination of colored faces would balance the scale, then the complete answer should be 3 blue plus 1 yellow but since the question specifically asks about how many blue are needed to blance the scale the answer should be 3.

Riyad| Guest May 17th, 2007 - 12:43 pmYes it asks for how many Blues

Milo Zachary| Guest May 17th, 2007 - 2:58 pmScale 1 tells us that 1 blue is more than 1 green and that one green + one red is equal to one blue.

Scale 2 tells us that 1 blue is less than 2 green (and one of the green + the red equal a blue), therefore the number of blue must be less than the the 4 remaining green, but greater than one-half of the remainder, so it must be 3 (plus the one for the red + green).

The scale will balance with 4 blues faces on the right.

Vishvas Vasuki| Guest May 18th, 2007 - 1:32 pmVery good reasoning!

I agree: Strict reasoning indeed reveals the scale may be balanced by n blues, where 5>n>3. If n has to be a whole number, it has to be 4.

jusdiver| Guest May 18th, 2007 - 1:55 pmbut with 4 blues it would not balance – it would tilt towards the side on which the blues are perched, just as it would tilt the other way if only 3 were used. YES, the question only asks for the number of blues, but does it actually forbid the use of another colour for an exact balance?

vik| Guest May 19th, 2007 - 7:40 amto me, its 3 blue and 1 orange….

RK| Profile May 20th, 2007 - 2:30 pmMilo’s observation that 1 blue > 1green, and that 1 blue < 2 green- while not readily apparent- is actually very good (took awhile for me to see this!), and so Vishvas’s conclusion that the scale may be balanced by n blues, where 5>n>3 is correct.

However, if you are allowed to use other faces, you can also balance the scale exactly with 3 blue and 1 orange.

The puzzle as it stands is pretty difficult; probably could’ve made it more accessible by putting an orange face on scale 3 (right side) and then ask how many blue faces are need to balance…

fantabulous| Guest May 28th, 2007 - 9:15 pmfrom 1st scale you get1 blue = 1 red + 1 green

the difference between 2nd 3rd scale is exactly 1 red + 1 green therefore take away 1 blue from the 2nd scale and you get the answer for 3 scale = 3 blue + 1 orange ;D

Abhijit| Guest June 16th, 2007 - 12:19 pm3 blue faces are required .

skeletalmonkey| Guest June 20th, 2007 - 1:42 amBlue faces.

using calculus, lets make, green=g red=r blue=b orange=o

from first scale, 2g + 2r = 2b, therefore b = g + r

from second scale, 6g + 2r = 4b + o, that is 4g + (2g + 2r) = 4b + o, therefore 4g = 2b + o. subbing in result from first scale, b = r + g we get 4g = 2g + 2r + o. thats means o = 2g + 2r, which means o = 2b.

now on third scale we get 5g + r = ?, break that down to 4g + (g + r) = ?, and we get 4g + b = ?, from scale 2 we know that 4g = (2g + 2r) + o, and from that we get 4g = 2b + 2b. therefore to balance the last scale we would need 5 blue faces.

i hope i didn’t stuff up the maths

Riyad| Guest June 20th, 2007 - 11:19 amYou did stuff the math..

The mistake is here:

If 4g = 2g + 2r + o

Then o= 2g – 2r NOT o = 2g + 2r

And that screws up the rest.

Marc| Guest June 21st, 2007 - 2:35 pmOne of the many laws of algeabra states that you need as many equations as you have variables to figure out the value of said variables.

We have 4 colors of marbles (i.e.: 4 variables).

We have 3 scales (i.e.: 3 equations).

It is therefore impossible to properly solve this puzzle.

However…

The question is: “How many blue faces are needed to balance scale 3?”

While this does imply ONLY blue faces, it does NOT exclude the yellow ones.

Hence, with the information we have, and as many people have already figured out, the answer is 3 blue faces. Yes, there will also be a yellow face, but that’s not what the question asks now, is it?

This is a puzzle; it’s not solid, concrete, exact math.

Sometimes you have to make the question to fit the answer rather than make the answer fit the question. ;-)

Marc| Guest June 21st, 2007 - 2:41 pmPS: The answer could even be 0 (zero) blue faces, because to balance out scale 3, you could just use 1 red and 5 green faces. Nowhere does it say you HAVE TO use blue faces.

MJ| Guest June 24th, 2007 - 2:26 amMy view on it… On the first scale, it shows that 2r + 2g = 2b. To split this up it could mean that 1r + 1g = 1b. Or it could mean that 2r = 1b and 2g = 1b. Either way you look at it, the bottom line is the same– one half of 1r = 1b; and one half of 1g = 1b. With that settled, we can move to the second scale. 2r + 6g = 8rg. We have established that 1/2r = 1b and that 1/2 g = 1b. So basically all we are doing is adding the halves. We have 8 halves of r/g which logically would = 4 wholes of b. The orange doesn’t fit. The only option in this strategy would be for o to = 0. If I am right and o = 0, then it would not be necessary to add any color other than blue to the third scale. We would simply count our halves again. We have 6 halves, which would make 3 wholes, therefore to balance the third scale we would need 3 blues.

I hope this makes sense, I am seriously sleep deprived right now…

syuen| Guest June 24th, 2007 - 10:30 pm3 blue faces are required

Balance the Scale Picture Puzzle - Smartkit Brain Enhancement News| Guest September 6th, 2007 - 12:49 am[…] hope this doesn’t turn out to be as difficult as the last scale balancing puzzle we posted… […]

Lara| Guest September 11th, 2007 - 3:42 amI get 3 blue faces

1 blue = 1 green + 1 red

or two red or two green

therefore 5 green + 1 red = 3 blue

it’s half

Nayan Shah| Guest September 11th, 2007 - 7:02 amIt should be 3 blue and 1 orange ball.

Soop| Guest October 6th, 2007 - 10:20 amIt’s not possible to get an exact answer.

From the first puzzle, we get b=r+g. So in the next puzzle, we can say

y+4r+4g=2r+6g. So we can also now say that y+2r=2g. Now unless y =0, we know that green is definitely greater than r, and continuing to assume that r > 0, that b>g. Unfortunately, there is no way of proving exact values. If y=0 and r=g, then the answer is 3b. But you could just as easily say that the values are thus:

r=1

g=2

y=2

b=3

In which case there is no correct answer. It’s also possible that y=4 and r=0, in which case there would be no real answer.

So technically, this puzzle is flawed.

Derek Roll| Guest October 30th, 2007 - 12:32 pmi think i figured it out it takes 4 blues

green=3/4 blue

red=1/4blue

yellow=blue

scale #1 2g=1,1/2b 2r=1/2b so 1.5b+.5b=2b right

scale #2 6g=4,1/2b 2r=1/2b so 4.5b+.5b=5b but theres only 4b so i fighured yellow=blue not sure if them being equal is aginst the rules

scale #3 5g=3,3/4b r=1/4b so 3.75b+.25b=4b so it works i think but if this isnt right then yes the problem is faulthis was my math warm up and i wasted the hole period on it so i hope im right

Derek Roll| Guest October 30th, 2007 - 12:43 pmafter reading other answers it makes sense that yellow(or orange as some people see it) is equal to blue because 3blues and 1 yellow would be the same as 4 blues so either of them is the right answer because wether it is saying there can only be blue or just how many blues then they are both correct. i think they left out wether or not you can only uuse blues to confuse us

leosuth| Profile February 7th, 2008 - 7:13 amStrictly, speaking, if we read the puzzle literally, then any number of blues, or reds or greens or yellows, or combinations will balance #3

Why?

Well, nowhere does is actually assign any weight to any colour, so they could all be 0 – this mean that #3 balances quite happily as it is, indeed it is even shown in a balanced state!!

Think about it :-)

leosuth| Profile February 7th, 2008 - 7:21 amFollowing on from that bombshell :-)

Perhaps the balancing has nothing to do with weight, but with colour?

If colour, them remember that Y = G + R if light

Or G = B + Y if pigment

Here I leave it to someone else to prove/disprove mathematically!

michaelc| Profile February 8th, 2008 - 4:52 pmThere are 2 equations and 4 variables. One way of testing the 4 blue ball balance theory is substitute in values to check.

Let’s take the green = g = 2 lbs and the red = r = 1 lb. Now we have 2 equations and 2 variables which we can solve.

From first equation, 2g + 2r = 2b, g=2, and r = 1, implies b = 3

From 2nd equation, 6g+ 2r = 4b + 1o , and g=2, r=1, and b=3 implies o = 2.

Now 5g +1r = xb

5*2 + 1*1 = x3. implies x = 11/3. X does not equal 4, and yet the equations that we are given are all satisfied.

You can pick values for the weights of green and red where the weight of blue is equal to the weight of orange, but nothing we are given says that this is the case.

For example, g=6, r=2, b=o=8. In the cases the weight of blue does not equal the weight of orange, 3 blues and 1 orange is the only correct answer.

GreekMind| Profile March 25th, 2008 - 3:20 pm14/3 blue

The One Eyed Man| Profile March 25th, 2008 - 3:30 pmLet’s go back to the algebraic approach.

We know b = g + r, but we don’t know the relative weight of g vs. r (not enough info), which is the root of the problem.

We know that o = 2g – 2r from o+4b=2r+6g

If you look at the relative weight of o vs. b, you can see that b is much more massive than o (b = g + r, while o = 2g – 2r ), but even though we don’t know the relative weight of g vs. r, we know that they are relatively similar.

b is heavy

g and r are medium

o is very light

Again, since this is not looking for an exact value of the weight of each ball, we could scale the whole equation down until o approaches 0, keeping the relative weights proportional:

(let “~” mean approximately equal)

if o ~ 0, then

b ~ 2 * r or 2 * g,

and r ~ g ~ 1

This gives us:

o = 0

r = g = 1

b = 2

(if o = 2g – 2r, and o = 0, then g = r)

if g = r, then g vs. r is irrelevant.

The formula for blue is: b = (g + r)/2

Using this logic (that o is always insignificant at any scale), the answer is b = 3

michaelc| Profile March 26th, 2008 - 9:18 amThe One Eyed Man, there is nothing given that says o has to be weightless! Even it’s some light sythetic plastic material, it still has a weight.

To perfectly balance the scale the answer has to be 3 blues and 1 orange.

The orange ball could be 1/100th of an ounce or it could be 1,000 lbs. Once it is set however, the weights of the other balls are upwards proportionally folowing the equation “1 0range = 2 Green – 2 Red”.

The One Eyed Man| Profile March 26th, 2008 - 11:07 amI agree with you, michaelc:

o is twice the mass of the unbalanced difference between r and g.

In the equation b = r + g, the substitution can be made: b = g + g + 0.5o

Mine was a theoretical more than a practical exercise.

As g and r converge, o approaches zero. So at some level, you could have values of g, r, and b where o is insignificant relative to the overall masses of b’s, g’s and r’s present in the solution. So a practical example of this would be that b is super-dense, g and r are made of something very dense (like lead), and o is made from a thin shell of synthetic material. With only one “o” in the solution (again — as r and g converge), the mass of “o” might not be enough to overcome the friction at the fulcrum.

The OTHER extreme is where 2g and o converge. In this scenario, o ~ r ~ 2g, and b ~ 3g. In this practical example, b might be lead, r and o might be light stone or metal, and g might be some kind of wood.

In the latter, the value of o is critical to the solution.

I was just taking a different path, illustrating how 3b (vs. 3b+o) could be a valid solution.

SolarFlare| Profile March 26th, 2008 - 3:49 pmANSWER:

There will be 4 blue balls on the right side of balance 3.

ASSUMPTIONS:

I use the letters R, G, B, G for the colours red, green, blue, yellow. I assume that no ball has zero or negative weight. I also assume that the weight of the balls can be expressed as integers.

GIVEN INFORMATION:

From the 2 complete balances we have 2 equations:

2G + 2R = 2B [Eqn 1]

6G + 2R = 4B + Y [Eqn 2]

Let n be the number of blue balls required on the right of the 3rd balance, then we have a 3rd equation:

5G + R = nB [Eqn 3]

We need to find the value of n (which is also a positive integer).

CONDITIONS WHICH MUST BE SATISFIED:

1st Condition:

Eqn 1 becomes:

G + R = B

Dividing Eqn 3 by this gives:

(5G + R) / (G + R) = n

Rearranging this gives:

5G + R = n(G + R)

(5 – n)G = (n – 1)R [Condition 1]

2nd Condition:

Rearranging Eqn 2 for Y and subsituting for B:

Y = 6G + 2R – 4B

= 6G + 2R – 4(G + R)

= 2G – 2R

As Y is a positive integer we know that:

2G – 2R > 0, and is an integer. As the difference between two even numbers is always an even number (so that we can divide this by 2 and still have an integer) this is the same as:

G – R > 0, and is an integer. [Condition 2]

FINDING POSSIBLE VALUES FOR N:

In Condition 1 it can be seen that n = 5 and n = 1 gives zero values for R and G, respectively. If n>5 then G is a negative multiple of R. If n<1 then n is not a positive integer. All of these cases fail to satisfy the assumptions that all colours and n are positive integers.

Hence, we have: 1 < n

3G = R

=>

R > G

=>

G – R

2G = 2 R

=>

G = R

=>

G – R = 0

So Condition 2 is not satisfied. Hence n=3 is NOT a solution.

When n = 4:

(5 – n)G = (n – 1)R

=>

G = 3R

=>

G > R

=>

G – R > 0

So Condition 2 IS satisfied. Hence n=4 IS a solution.

FINDING THE RELATIVE BALL WEIGHTS WHEN N=4:

We have G = 3R.

Let R = 1

Then G = 3

And B = R + G = 4.

And Y = 2G – 2R = 4

CHECKING FOR THE ORIGINAL BALANCES:

Using these values in balance 1: 2G + 2R = 2B gives:

2*3 + 2*1 = 2*4 which tallies.

And in balance 2: 6G + 2R = 4B + Y gives:

6*3 + 2*1 = 4*4 + 4 which is correct.

And in balance 3: 5G + R = nB gives:

5*3 + 1 = 4*4 which tallies.

FINAL COMMENT:

From the above it seems that this is the only solution for n which satisfies the initial assumptions.

SolarFlare| Profile March 26th, 2008 - 4:13 pmApologies. A section of my last message appears to have been corrupted by the angle brackets I used as part of the math. (Also, I hope my long answer is not unacceptably long. A shorter answer would have made the math hard to follow.)

FINDING POSSIBLE VALUES FOR N

…should have continued to state that n must lie between 1 and 5 exclusive, that is it is one of [2, 3, 4].

A new section followed with a heading which has disappeared:

IDENTIFYING THE CORRECT VALUE FOR N

… which took each of these possible values of n in turn, found values for R & G which satisfied them, and deduced that in the cases of n=2 and n=3 the condition that G – R greater than 0 is not satisfied, but that in the case of n=4 this was satisfied.

This was followed by the heading:

FINDING THE RELATIVE BALL WEIGHTS WHEN N=4

… after which everything was ok.

SolarFlare| Profile March 27th, 2008 - 12:00 pmTHIS REPLACES MY ABOVE POSTS.

My sincere, very humble apologies for my earlier corrupted & way too long post (RK: please delete that with my blessing!). Here is the same solution stated much more briefly. The letters R, G, B, Y represent red, green, blue, yellow. I assume that no face has zero or negative weight. I also assume that the weight of the faces can be expressed as integers. Below I show that 4 blue faces are required on the right side of balance 3.

G + R = B (from balance 1)

6G + 2R – 4B = Y (from balance 2)

Hence,

6G + 2R – 4(G + R) = Y

2G – 2R = Y

So, 2G – 2R is a non-zero, positive integer. As this is even we can divide by 2 and still have an integer: G – R is a non-zero, positive integer. (This fact is used below.)

We want to find n in:

5G + R = nB (from balance 3)

We divide the equation for balance 3 by that for balance 1:

(5G + R) / (G + R) = n (as the B’s cancel)

And rearrange:

5G + R = n(G + R)

(5 – n)G = (n – 1)R

As G & R are positive & non-zero n must be one of [2, 3, 4] (otherwise we have negative or zero weights). If n=2, then 3G = R, which means R is greater than G, so (G – R) is not a non-zero, positive integer. So n=2 cannot be the solution. If n=3, then 2G = 2R, so (G – R) is zero. So n=3 cannot be the solution. If n=4, then G = 3R, which means G is greater than R, so (G – R) is a non-zero, positive integer and is a solution (the only solution given my assumptions).

Hence, we need 4 blue balls on the right side of balance 3. Integer values can be applied to R, G, B, Y to check this solution using:

G = 3R

G + R = B

Y = 2G – 2R

Let R = 1, then G = 3, B = 4, Y = 4. Applying these numbers to the 3 balances (using 4 blue faces on the right of balance 3) tallies with the diagrams.

(Note above that if we allow Y to be zero then n=3 becomes a valid solution with G=R. Then: R = 1, G = 1, B = 2, Y = 0 which tallies with the diagrams, this time with 3 blue faces on the right of balance 3.)

michaelc| Profile March 27th, 2008 - 2:40 pmSolarFlare,

That is an interesting, impressive and correct arguement.

I still disagree with your results.

Mainly your assumption, “I also assume that the weight of the balls can be expressed as integers.”

Where you take the solution after that is pretty cool!

SolarFlare| Profile March 28th, 2008 - 10:41 ammichaelc,

Thanks for the “pretty cool” compliment.

Everything I’ve written will work just as well without the assumption that “the weight of the faces can be expressed as integers”. None of my equations & none of my logic requires it. I could have left that sentence out.

If no face has zero weight, and if all faces have positive weight relative to each other, and if there are only whole blue balls on the right side of balance 3, then the only solution is 4 blue balls, with (R, G, B, Y) = (1, 3, 4, 4).

There are many sets of values for (R, G, B, Y) which satisfy the equations for balances 1 & 2 as you pointed out in your original post, but we also know that the number of blue balls on the right of balance 3 cannot be fractional. This narrows the result down to one solution.

…though I’ll bow to any convincing proof to the contrary

michaelc| Profile March 28th, 2008 - 9:46 pmSolarFlare,

Your arguement is taking the assumption that there is an exact balance of 5 greens and 1 red to x blue balls with x being a counting number. 1,2,3…

Since the puzzle asks for this (how many blue balls), I guess you could look at it that way and assume that’s part of it that is given to you.

However, just judging from the pictures alone, you can’t logically deduce that 4 blue balls will balance the scale.

If you assume there is a number of blue balls somewhere in (1,2,3…, you can (and you did quite well) conclude it can only be 4.

I still stand by my original answer, 3b + 1o. or 3b + 1y

SolarFlare| Profile April 1st, 2008 - 10:46 ammichaelc,

Aha! Now I understand the discrepancy between our solutions, and I agree completely. The question can be read in two ways, one giving the extra piece of information that I used in my solution (that there are an integer number of blue faces on the rhs of balance 3) and the other not giving this.

In your interpretation the question is: given only the balances 1 & 2 how many blue faces are needed on the rhs of balance 3 for balance to be achieved? And in this case it cannot be guaranteed that there will be an integer number of blue faces which will balance. We do know that (3B + Y) will definitely balance, as will any rearrangement of this solution which expands some or all of the B’s into (G + R), and also the trivial solution in which the rhs is identical to the lhs. (There may be more ways of organising these 3 variables to give further solutions.)

Interesting! Thanks for the insight.

Scale Balance | Smartkit Puzzles and Brain Teasers| Guest April 1st, 2008 - 9:55 pm[…] some comments show up recently on our first scale balance puzzle (hard!), which reminded me how long its been since we put one of these up. With that in […]

Wayne| Profile February 18th, 2009 - 3:06 pmGreen = Red and Orange = 0.

i used excel sheet to change the values and multiplier.

JunzJeahnz| Profile June 6th, 2011 - 7:03 am3 and 3/4 blues

…

MichelleSam.| Profile October 27th, 2011 - 2:44 amI got 3 Blue Balls and 1 Orange.

To get this, I had 4 equations figured out:

B= blue

R= red

G= green

E= orange.

1. 1B = 1R + 1G

(2 blues and 2 greens make 2 blues. Using that ratio, you would get 1 red and 1 green make 1 blue.)

2. 4G = 1E + 2B

(According to scale #2, one orange and 4 blues make 2 blues and 4 greens:

1E + 4B = 2B + 4G

Subtract 2B from both sides and you have:

1E + 2B = 4G)

3. 2G = 2R + 1E

(Here, instead of using the 4 blue balls from scale #2, I converted the 4 blue balls into 4 green ones and 4 red ones, because 1 green and 1 red make 1 blue, so 4 red ones and 4 greens ones would make 4 blues.

So according to scale #2, once we converted the blues to reds and greens, we would have this equation:

4R + 4G + E1 = 2R + 6G

Subtract 2R from both sides and subtract 4G from both sides, you get:

2R + 1E = 2G)

4. 2B = 1E + 4R

(To figure this out, I used the ratios I had already figured out to make the right side of scale #2 equal itself to find the ratio between orange, red, and blue.

So there are 6 greens and 2 reds.

6G + 2R = 2B + 4G

The 6G = 3E +6R, according to equation #3, and using the same ratio.

And 4G = 4R + 2E, according to equation #3, and using the same ratio.

So the equations turns into:

3E + 6R + 2R = 2B + 4R + 2E

Simplify that, and it turns into:

1E + 4R = 2B)

Knowing that,

One l blue ball is a given, because there’s a red and a green in balance #3. That equals one blue ball.

Now we have to figure out how many blue balls will make 4 green balls.

1B + 4G —> 1B + 1E + 2B —–> 3B + 1E

3 Blue Balls and 1 Orange.

Barks| Profile December 7th, 2014 - 7:47 pmSuper easy to do algebraically.

g = green ball

r = red ball

b = blue ball

o = orange ball

Assuming weight remains constant:

(1st scale) 2g + 2r = 2b

? g + r = b

(2nd scale) 6g + 2r = 4b + o

? 4g = 2b + o (remember b = r + g)

(3rd scale) 5g + r = ?? (will come back to this!)

So if we substitute our own numbers for g and r we get b, which we can then use to get o.

so, for (1): g + r = b

if

g = 7

r = 3

?

b = 10

So now that we have a value for b:

(2) 4g = 2b + o

: 4 x 7 = (2 x 10) + o

28 = 20 + o

: 28-20 = 20 – 20 + o

? 8 = o

So now we have 4 numbers which fit into the equation.

Green (g) = 7

Red (r) = 3

Blue (b) = 10

Orange (o) = 8

So,

(3) 5g + r = 5 x 7 + 1 x 3

5g + r = 38

38/b = 38/10 = 3 remainder 8 **

8/o = 8/8 = 1

? the answer is 3 blue balls and 1 orange ball.

** if only blue balls may be used, the answer is 3.80 blue balls (38/10 = 3 8/10)