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I have two 6-sided dice which between them have 12 different numbers, but some of the numbers are hard to tell apart because one is the same as the other upside down (like $89$ and $68$).

I throw the dice once and they land like this:

throw 1

The numbers on the bases of the two dice now sum to $105$.

I throw them again and they land like this:

throw 2

The numbers on the bases of the two dice now sum to $149$.

I throw them a third time and they land like this:

enter image description here

The question is: what do the numbers on the bases of the dice now sum to?

(I got this puzzle from here, where you'll also be able to check your answer - but please don't brute-force it to find out which answer is right by clicking all the possibilities! I'll only accept an answer if you've solved it all by yourself.)

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  • $\begingroup$ From where we are observing these dice's....>,<,V,^ directions....since it will be dependent on that.? $\endgroup$ – user2408578 May 25 '15 at 11:51
  • $\begingroup$ @user2408578 The shadow is on the table. $\endgroup$ – LeppyR64 May 25 '15 at 12:08
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    $\begingroup$ I constructed the dice and finally... I happened to had messed up >__< It's really diabolical $\endgroup$ – Masclins May 25 '15 at 12:13
  • $\begingroup$ So this is what quantum computing looks like ;) $\endgroup$ – Mark N May 25 '15 at 12:15
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    $\begingroup$ Well I've worked out the answer, but I have a mess of paper scrawlings. Good luck to anyone that can provide a well written step by step solution! @MisterEman22 the dice can be on either side each roll $\endgroup$ – Pete May 25 '15 at 14:20
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The answer is

61 + 98 = 159

We are given that there are two six sided dice and that there are 12 unique numbers on them. Some of the numbers look the same as others but they still only occur once. For example we might see 66 and 99, we don't know which is which, but we know that they are not both 66.

The problem requires us to identify which die is which in the two images for each of the three rolls.

For the sake of clarity, we will call the die on the left in roll 1 "DIE 1", and the one on the right "DIE 2". We know nothing about the sums at this point.

enter image description here

Notice that DIE 1 has a 69 on it. Rotated 180 degrees it is still 69 so it is unique. Therefore the die on the left in roll 3 is DIE 1. We can add the (68/89) to DIE 1. Note the orientation of the (86/98) in DIE 2 of roll 1 and DIE 2 of roll 3. It would be impossible for these two numbers to be the same based on their orientations, so the DIE 2 in roll 3 is the opposite three faces that we couldn't see in roll 1.

enter image description here

Now looking at roll 2 we know that the die on the right must be DIE 1 because 19/61 cannot be on DIE 2. We can now fill in all of the numbers on DIE 1.

enter image description here

Now we can start looking at the sums. We know that the bottom number of DIE 1 in roll 1 is 19 or 61. 105 - 61 = 44 which is not possible so the bottom number of DIE 1 is 19. This identifies the 61 in DIE 1 and tells us that the bottom of DIE 2 is 86. This also identifies the 98.

enter image description here

The number on the bottom of DIE 1 in roll 2 is either 18 or 81. 149 - 18 = 131 which is not possible, so the number is 81. This identifies the 18 on DIE 2 and the 68 which we already know is opposite the 86 on DIE 2.

There is only one possible way to merge the two images of DIE 2 and that is like this:

enter image description here

The number on the bottom of DIE 1 in roll 3 is 61 and the number on the bottom of DIE 2 in roll 3 is 98. The sum is 159.

Proof that DIE 2 can only be built one way

enter image description here

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    $\begingroup$ The hardest part of this solution was making the step by step pictures :) $\endgroup$ – LeppyR64 May 25 '15 at 14:17
  • $\begingroup$ Nicely done with the pictures! $\endgroup$ – Pete May 25 '15 at 14:22
  • $\begingroup$ Thanks @Pete I had a bunch of paper scrwalings much like you :) $\endgroup$ – LeppyR64 May 25 '15 at 14:38
  • $\begingroup$ Beautiful answer! I was expecting a lot of typed explanation, rather than such nice pictures :-) $\endgroup$ – Rand al'Thor May 25 '15 at 16:30

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