Difference between revisions of "1999 AHSME Problems/Problem 30"
(solution) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 13: | Line 13: | ||
<cmath>x^3 + y^3 + z^3 - 3xyz = \frac12\cdot(x + y + z)\cdot((x - y)^2 + (y - z)^2 + (z - x)^2)</cmath> | <cmath>x^3 + y^3 + z^3 - 3xyz = \frac12\cdot(x + y + z)\cdot((x - y)^2 + (y - z)^2 + (z - x)^2)</cmath> | ||
| − | Setting <math>x = m,y = n,z = - 33</math>, we have that either <math>m + n - 33 = 0</math> or <math>m = | + | Setting <math>x = m,y = n,z = - 33</math>, we have that either <math>m + n - 33 = 0</math> or <math>m = n = - 33</math> (by the [[Trivial Inequality]]). Thus, there are <math>35 \Longrightarrow \mathrm{(D)}</math> solutions satisfying <math>mn \ge 0</math>. |
== See also == | == See also == | ||
| Line 19: | Line 19: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:36, 5 July 2013
Problem
The number of ordered pairs of integers 
for which 
and
is equal to
Solution
We recall the factorization (see elementary symmetric sums)
![\[x^3 + y^3 + z^3 - 3xyz = \frac12\cdot(x + y + z)\cdot((x - y)^2 + (y - z)^2 + (z - x)^2)\]](http://latex.artofproblemsolving.com/f/1/4/f14a47fe73a060ead9d72bd72cf43adeecc72bbb.png)
Setting 
, we have that either 
or 
(by the Trivial Inequality). Thus, there are 
solutions satisfying .
See also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Last problem | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
