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.

All you do is add the right most digits of the showing faces.  The sum of those digits can be anywhere from 8 to 44. When you add those six faces together the sum is always a four digit number. If you think of the 4 digit number in the form XXYY  then you will see that for any combination XX + YY where both are 2 digit numbers = 55.  For example 3817.  38 + 17 = 55.  So you count the right most digit and simply subtract that from 55 to obtain the two left digits.  If you count 20 for example then you know the number must be 3520. Neat problem!

Posted by Thomas Harris on 2005-08-27 19:33:18