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.

For any cube given. Each face of the cube has a 3 digit number with sum of those three digits a constant.
For cube 1 :- 6+4+3=4+4+5=7+4+2=..=13
cube 2 :-15
cube 3:- 15
cube 4:-11
cube 5:-11
cube 6:-17
hence in any throw we get the sum of 6 three digit numbers that appear in the top faces =13+15+15+11+11+17=82.
hmm not much mental power is required when once we know the answer.

Posted by thanveer on 2005-08-31 09:31:36