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Dear Mitch,

I saw you posted the answer to the first set of Christmas/Hanukkah balance scale puzzles, when are you going to post the answers to the second set, the ones you put up about a week ago?


Marc G.


Dear Marc G., 


Sometimes, when solving problems, a basic and standard step-by-step methodology is at least as easy as reinventing the wheel – after all, the procedures of math were developed for good reason! 


Recall, there were 3 scales presented, A, B, and C. 

Scale A showed:

2 Dreidels on the left side balanced with 2 Candles and one Present on the right side.

Scale B showed:

2 Dreidels and one Present on the left, balanced with 8 Candles on the right. 

Scale C showed:

An unbalanced scale with 2 Dreidels and one Candle on the left, and nothing on the right.

The question was: 

Based on what you see in scale A and scale B, how many Presents would you need to place on the empty, right-hand side of scale C in order to form a balance with the left-hand side of that scale (with 2 Dreidels and 1 Candle)?

So, using basic concepts from the area of math called "systems of equations" we substitute letters for the pictures of the different items... 

And, to keep things as clear as we can,

to represent a Dreidel we'll use the letter D,

to represent a Candle we'll use the letter C,


to represent a Present we'll use the letter P.


SCALE A shows: 2D = 2C + P

SCALE B shows: 2D + P = 8C

SCALE C shows: 2D + 1C = (?) P 

Well, there's a 'golden rule' in systems of equations which is this: "You need (at least) the same number of equations as variables."

And here, since only A + B count as complete equations (with C yet to be completed) we have 3 variables (D,C,P), but only 2 equations.

So we need to come up with a third equation.

One easy way to do that is to simply "add" equation A and equation B together.

If you think about it, since the left-hand side of A = its right-hand side, and the left of B = the right-hand side of B, then: 

The left of A + the left of B should = right of A + right of B,

(like this):

2 pounds = 2 pounds

3 pounds = 3 pounds


2 +3 pounds = 2 + 3 pounds

           (5)     =     (5)

By the way, this is a common approach, so, so far, we are not doing anything too strange.  BUT here we're going to add a little twist.  Rather than combining the left of A with the left of B, and right of A with right of B, we're going to do something just a little different. 

If you look again at this example:

2 = 2 and

3 = 3

Making 2 +3 = 2 + 3, you see that as long as the left of A = right of A, and left of B = right of B, we could simply turn either A or B around in "space" before we combine) which give us left A + right B = right A + left B.

So, we get: 2d + 2c + 2P = 2D +8C

Why do this? 

We're trying to make our way toward figuring out the value or weight of a P, so we can move things along a step or two faster by flipping scales around in space to get all P's on the same side.

NOW, going back ... 

2D + 2C + 2P  = 2D + 8C

Divide both sides by 2... 

D + C + P = D + 4C

Subtracting a C and a D from both sides, we get: 

P = 3C

So, C = 1/3 P

1 Candle = 1/3 Present 

Adding left A + left B = right A + right B

4D + P = 10C +P

4D = 10C

2D = 5C

D = 5/2C

But we want to know how many P each D equals...

So, scale A:  2D = 2C + P

2 (5/2C) = 2C + P

5C = 2C + P

3C = P

C = P/3 

Each Dreidel = 5/2 C, and each Candle = 1/3 P.

So (2 1/2)C = D.

And so we could replace each D with 2 1/2 C. so... 


Scale C:

... = 1/3P


5 Candles + 1 Candle = 6 Candles and each Candle = 1/3 P

So, left side of C = 6 x 1/3 P

6/1 x 1/3P = 2 Presents


For each D for Dreidel, substitute a T, for Tree,

For each C for Candle, substitute a C for Cane (in other words, leave the C's alone!),

For each P for Present, substitute an S for Snowman,

And that's all there is to it!


Hope this helps,

Happy New Year,