here4fun
01-12-2011, 01:22 AM
The two dice in each pair are identical.
As usual, the numbers on opposite sides add up to seven.
On each pair, what is the total of the two hidden sides that face each other?
8005
This isn't hard. Just logic. I know with all the really smart people that are part of this site this should be solved by now.
OK here's the solution for the first pair of dice.
Pair A
First die.
The top is 4. Therefore the bottom is 3. Because the sum =7
The front is 6. Therefore the back is 1. As above.
Two faces left that you can't see must be 2 & 5 because you've got 4,3,6,1.
Second die.
The face opposite the 4 is 3 because the sum =7 so the first amount to add is 3.
Now turn the die counter-clockwise so it has the same orientation as die 1 which will bring the 5 which is opposite the 2 to the end (sum=7) and because the two dice are identical the inside face of die 1 must also be 5 so the second amount to add is 5.
So the sum is 5+3=8.
Now can anyone else get Pair B?
As usual, the numbers on opposite sides add up to seven.
On each pair, what is the total of the two hidden sides that face each other?
8005
This isn't hard. Just logic. I know with all the really smart people that are part of this site this should be solved by now.
OK here's the solution for the first pair of dice.
Pair A
First die.
The top is 4. Therefore the bottom is 3. Because the sum =7
The front is 6. Therefore the back is 1. As above.
Two faces left that you can't see must be 2 & 5 because you've got 4,3,6,1.
Second die.
The face opposite the 4 is 3 because the sum =7 so the first amount to add is 3.
Now turn the die counter-clockwise so it has the same orientation as die 1 which will bring the 5 which is opposite the 2 to the end (sum=7) and because the two dice are identical the inside face of die 1 must also be 5 so the second amount to add is 5.
So the sum is 5+3=8.
Now can anyone else get Pair B?