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.

I am interpreting this as all 6 dice are rolled and we want the sum of each face of the six dice.  If so, then it is easy.

D1: Each face has a sum of 13

D2: Each face has a sum of 15

D3: Each face has a sum of 15

D4: Each face has a sum of 11

D5: Each face has a sum of 11

D6: Each face has a sum of 17

Therefore,

D1+D2+D3+D4+D5+D6 = 13+15+15+11+11+17 = 82

Am I missing something here?

Posted by Matt on 2005-08-29 01:03:47