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.

well, first i think we should see that ten's position of each three-digit number at one cube is same.. example at cube4 all have 2 at ten's position.. if we sum all of six cube we get 300. that for ten's position.

then sum of 1's and 100's of each faces of one cube is the same, example at cube 3, the sum is 7 (1+6=2+5). if we calcute all of six cube, the sum is 52 (9+9+7+9+8+10)... so we can get 100's by only sum 1's.. because 100's had add 3 (from sum of 6cube of 10's). so 100's can get from 55 minus sum of 1's.

for example if the 6 three-digit number is 564,623,384,247,773, and 236. then we just add 1's, it's 4+3+4+7+3+6=27, then 100's is 55-27=28. so the sum of 6 three-digit is 2827.

Sorry if my english is bad, and i had write too much secret here...It's cool pcbouhid.. i like it...keep sMiLe

Posted by sragen on 2005-08-25 06:58:51