1991 AJHSME Problems/Problem 12: Difference between revisions
5849206328x (talk | contribs) Created page with '==Problem== If <math>\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}</math>, then <math>N=</math> <math>\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1…' |
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{{AJHSME box|year=1991|num-b=11|num-a=13}} | {{AJHSME box|year=1991|num-b=11|num-a=13}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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Revision as of 23:07, 4 July 2013
Problem
If 
, then 
Solution
Note that for all integers 
,
![\[\frac{(n-1)+n+(n+1)}{n}=3.\]](http://latex.artofproblemsolving.com/2/8/9/289d6817c50adb030fb467e9ee666513d2a67d5f.png)
Thus, we must have 
.
See Also
| 1991 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
