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.

(In reply to re: ss-t! by nickson)

1's is position for the last digit.. and 100's is position for hundred... like 526.. 1's position is 6, then 100's position is 5! sorry my english is bad...

keep sMiLe

Posted by sragen on 2005-08-26 03:23:29