Difference between revisions of "1991 AJHSME Problems/Problem 12"
(→Solution 2) |
|||
| Line 12: | Line 12: | ||
==Solution 2== | ==Solution 2== | ||
| − | As we know that <math>\frac{1990+1991+1992}{n}</math> has to be some multiple of <math>\frac{2+3+4}{3}</math>, then we know that the first equation is <math>995</math>(1990/2) times bigger than the second one(in my solution), so the bottom must be <math>3\cdot995=\boxed{\text{(D)}1991}</math> | + | As we know that <math>\frac{1990+1991+1992}{n}</math> has to be some multiple of <math>\frac{2+3+4}{3}</math>, then we know that the first equation is <math>995</math>(1990/2) times bigger than the second one(in my solution), so the bottom must be <math>3\cdot995=\boxed{\text{(D)}1991}</math>-fn106068 |
==See Also== | ==See Also== | ||
Latest revision as of 11:29, 28 August 2021
Contents
Problem
If 
, then 
Solution 1
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 
.
Solution 2
As we know that 
has to be some multiple of 
, then we know that the first equation is 
(1990/2) times bigger than the second one(in my solution), so the bottom must be 
-fn106068
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
