Difference between revisions of "1990 AJHSME Problems/Problem 11"
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The only possibilities for the numbers are <math>11,12,13,14,15,16</math> and <math>10,11,12,13,14,15</math>. | The only possibilities for the numbers are <math>11,12,13,14,15,16</math> and <math>10,11,12,13,14,15</math>. | ||
| − | In the second case, the common sum would be <math>(10+11+12+13+14+15)/ | + | In the second case, the common sum would be <math>(10+11+12+13+14+15)/3=25</math>, so <math>11</math> must be paired with <math>14</math>, which |
it isn't. | it isn't. | ||
Revision as of 12:38, 14 July 2016
Problem
The numbers on the faces of this cube are consecutive whole numbers. The sums of the two numbers on each of the three pairs of opposite faces are equal. The sum of the six numbers on this cube is
![[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); label("$15$",(1.5,1.2),N); label("$11$",(4,2.3),N); label("$14$",(2.5,3.7),N); [/asy]](http://latex.artofproblemsolving.com/1/b/1/1b1a85e70cc87bd44b6976f6eb76b21a969c7729.png)
Solution
The only possibilities for the numbers are 
and 
.
In the second case, the common sum would be 
, so 
must be paired with 
, which
it isn't.
Thus, the only possibility is the first case and the sum of the six numbers is 
.
See Also
| 1990 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
