Difference between revisions of "2015 AMC 12B Problems/Problem 23"
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When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>. | When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>. | ||
− | Thus, there are <math>\boxed{\textbf{(B)}\;10}</math> solutions | + | Thus, there are <math>\boxed{\textbf{(B)}\;10}</math> solutions. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}} | {{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:14, 5 March 2015
Problem
A rectangular box measures , where
,
, and
are integers and
. The volume and the surface area of the box are numerically equal. How many ordered triples
are possible?
Solution
The surface area is , the volumn is
, so
.
Divide both sides by , we get that
First consider the bound of the variable . Since
we have
, or
.
Also note that , we have
.
Thus,
, so
.
So we have or
.
Before the casework, let's consider the possible range for if
.
From , we have
. From
, we have
. Thus
When ,
, so
. The solutions we find are
, for a total of
solutions.
When ,
, so
. The solutions we find are
, for a total of
solutions.
When ,
, so
. The only solution in this case is
.
When ,
is forced to be
, and thus
.
Thus, there are solutions.
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.