I give you 6 cubes with 3-digit numbers in their faces:

cube 1: 643 / 445 / 742 / 247 / 346 / 544
cube 2: 465 / 564 / 267 / 366 / 762 / 663
cube 3: 186 / 285 / 384 / 483 / 681 / 582
cube 4: 821 / 227 / 722 / 623 / 326 / 128
cube 5: 533 / 137 / 236 / 335 / 632 / 731
cube 6: 278 / 377 / 179 / 872 / 773 / 971

I bet you that every time you throw them, I can evaluate (mentally) the sum of the 6 three-digit numbers that appear in the top faces faster than you, even if you use a calculator (about 6 or 7 seconds, and I´m not too good at mental calculations).

Explain how I can do this.

To summarize sragen's solution:
Add up the right most digits from the six cubes--that will give you the last two digits of the sum.  Subtract this result from 55; that gives you the first 2 digits.

Example: 445+267+483+326+137+179
5+7+3+6+7+9 = 37.
55-37 = 18.
So the sum is 1837.

The reason it works is: First, every middle digit (the 10s) is identical on all 6 faces of each cube.  So the 10's position will always contribute 300 to the sum.  Second, the sum of the units and 100s position on every cube is constant.  For cube 1 the sum is 9, cube 2 is 9, cube 3 is 7, cube 4 is 9, cube 5 is 8, and cube 6 is 10.  Adding these up you get 52. 

So, in our example, if I know the units add up to 37, then the hundredss must add up to 52-37 = 15 (or 1500).  But there's another 300 from the middle digit making a total of 18 hundreds (or 1800).  So the sum is 1837.  By subtracting from 55 instead of 52 we automatically take the 300 into account, saving a step in the mental calculation.

Posted by Ken Haley on 2005-08-26 05:48:03