Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1990 AJHSME Problems/Problem 11"
m (→‎Problem)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==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
+
The numbers on the faces of this cube are consecutive whole numbers.  The sum 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>
 
<asy>
Line 16: Line 16:
 
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)/6=25</math>, so <math>11</math> must be paired with <math>14</math>, which
+
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.   
  

Latest revision as of 14:42, 23 May 2021

Problem

The numbers on the faces of this cube are consecutive whole numbers. The sum 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]

$\text{(A)}\ 75 \qquad \text{(B)}\ 76 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 81$

Solution

The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$.

In the second case, the common sum would be $(10+11+12+13+14+15)/3=25$, so $11$ must be paired with $14$, which it isn't.

Thus, the only possibility is the first case and the sum of the six numbers is $81\rightarrow \boxed{\text{E}}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_AJHSME_Problems/Problem_11&oldid=154140"