Roll a pair of ordinary dice.

1. Multiply the values of the two top faces.
2. Multiply the values of the two bottom faces.
3. Multiply the value on the top face of the first die by the value on the bottom face of the second die.
4. Multiply the value on the top face of the second die by the value on the bottom face of the first die.
5. Add the four numbers from the previous four steps.

If you multiplied and added correctly you will always get the same number.

Some questions:

1. What's the number?
2. Why do you always get that number? (Again, assuming that your arithmetic is correct.)

Mark

 

Hint: The spots (pips?) on a die follow certain rules. If I can see three faces of a die, I can tell you how many spots are on the remaining faces.

 

I haven't thrashed out a set of answers, but the first thing that springs to mind is the fact that opposite faces of a die always sum to 7. Is this a factor?

Dave

 

I haven't thrashed out a set of answers, but the first thing that springs to mind is the fact that opposite faces of a die always sum to 7. Is this a factor?
Dave

It most certainly is.

 

Greetings Mark44,

Before I blurt out my answer, I should ask if this is a homework assignment. I don't want to deprive you of the pleasure of deriving the answer yourself.

In either case, what do you believe the answers to be.

hgmjr

PS. Dave, I think you are on to the key.

 

Nope, hgmjr, it's not a homework assignment, but I did give out a lot of homework assignments in the 21 years I was a teacher. It's actually a fairly simple problem once you understand how the spots are arranged on a die.

 

It's half-past midnight here so I won't crack open my "Games-set". Pen and paper tomorrow lunchtime I think!

I've come across many-a-maths problem with dice, and the root is always the fact that the opposite sides sum to 7 - something I discovered by accident when a curious youngster!

Dave

 

Nope, hgmjr, it's not a homework assignment, but I did give out a lot of homework assignments in the 21 years I was a teacher. It's actually a fairly simple problem once you understand how the spots are arranged on a die.

Since that is the case then I attach my answer below.

hgmjr

 

Yep, a simple case in point example of factor by grouping...

Good example for an algebra 2 class.

Mark22, do you teach Algebra 2 or Pre-Calc, by any chance?

 

I taught all levels of high-school math for two years, and I taught all levels that we offered in the community college, near Seattle, where I worked for 18 years. There was also some student teaching before I got my job in the high school, plus a couple of quarters as a teaching assistant (I actually taught the class) in the university I went to.

While I was working at the community college I taught everything from just plain arithmetic to algebra and precalculus with trig, engineering calculus, linear algebra, and differential equations. I also taught classes in BASIC, Pascal, one class in Modula-2, many, many C classes, a couple of classes in C++, and several classes in FORTRAN. During that time I taught myself Java and x86 assembler.
Mark

 

Since that is the case then I attach my answer below.
hgmjr

Yep, it's 49. You can determine this with only two variables, though.

Let x = no. of spots on the top face of die 1
Let y = no. of spots on the top face of die 2

Carrying out the multiplications described in the problem gives:
xy + x (7 - y) + y (7 -x) + (7 - x)(7 - y)
= xy + 7x - xy + 7y -xy + 49 - 7x - 7y + xy
= 49

Mark

 

Wow, very impressive, that's all there is to say (to the reply two posts back, of course)

 

Since that is the case then I attach my answer below.

hgmjr

Variable designation is incorrect in solution paper...although the logic is clear.

Better to correct it....

 

Well,
The numbers on the first die are x and (7-x) and,
The numbers on the second die are y and (7-y).
Just go from there.....

drewlas

 

I haven't thrashed out a set of answers, but the first thing that springs to mind is the fact that opposite faces of a die always sum to 7. Is this a factor?
Dave

any idea why that is the case?
Of course the dice won't become biased if we change it a little.

 

any idea why that is the case?
Of course the dice won't become biased if we change it a little.

It is either a) convention (i.e. the way it is always done with a reason lost in annals of time), or b) there is some statistical reason which ensures that by having opposite faces summing to 7 statistically we have a "fair dice".

I honestly can't say which of the above it is, but a quick Google search doesn't shed any light on it.

Dave