Question 1164387: I already know the answer to this problem, but I need a better explanation as to how to get the answer, like a step-by-step kind of way.
Three circular cardboard disks have numbers written on the front and back sides.
The number written on the front of Disk 1: 6
The number written on the front of Disk 2: 7
The number written on the front of Disk 3: 8
By tossing all three disks and adding the numbers that show face up, we can obtain these totals: 15, 16, 17, 18, 19, 20, 21, and 22. What numbers are written on the back sides of these disks?
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52810)   (Show Source):
You can put this solution on YOUR website! .
See the solution at this link
https://answers.yahoo.com/question/index?qid=20120131193448AA5gFsG
Do not hesitate (and do not forget) to post your "THANKS" to me for pointing you this source.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
I'd rather reason something out all by myself when possible, rather than
go looking where someone else has already figured it out. That's just me.
Being able to reason it out yourself is the main thing, not to know the
answer. This student already knew the answer. Trying to do it yourself is
the only way to train your brain. When I studied math in school and
college, I would cover up the explanation in the book of how to solve each
problem, and first try to figure it out all by myself. Then only when I got
stuck, would I uncover the explanation. Then I'd only uncover a little at
a time.
Suppose the back side numbers are A, B and C, like this:
Disk Front Back
1 6 A
2 7 B
3 8 C
One thing to notice is that there are exactly 2 ways each disk can land,
so there are 2∙2∙2=8 ways they can all three land. And there are 8 numbers
15, 16, 17, 18, 19, 20, 21, and 22 numbers they must add to, so there are no
"extras", that is, there is only one way to get each of those sums.
If the first two disks fall 6 and 7, there must be a 9 on the back of the
third disk so that the largest sum, 6+7+9=22, can be made. So C=9.
If the first and third disks fall 6 and 8, there must be a 2 on the back of
the second disk so that the smallest sum, 6+8+2=16 can be made. So B=2.
We can now make these sums:
2+7+8=17, 2+7+9=18, 6+7+8=21, 6+7+9=22
If the second and third disks fall 7 and 8, there must be a 5 on the back of
the first disk so that the largest remaining sum 6+5+9=20 can be made. So
A=5.
Now we must check to see if we can make all the sums:
Disk Front Back
1 6 2
2 7 5
3 8 9
2+5+8=15, 2+5+9=16, 2+7+8=17, 2+7+9=18, 6+5+8=19, 6+5+9=20, 6+7+8=21,
6+7+9=22.
Ah, sweet success!
Edwin
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