Difference between revisions of "2015 AMC 12B Problems/Problem 23"
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Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Find that 2ab+2bc+2ac = abc | ||
+ | |||
+ | I'm not going to go through the bashing process of this solution, I will just give the outline | ||
+ | |||
+ | Split it up into cases: | ||
+ | |||
+ | Case 1: a=b=c | ||
+ | You should get a=6 | ||
+ | |||
+ | Case 2: a=b | ||
+ | You should get a^2(2-c)+4ac = 0. The easiest way to bash this is to set c equal to an integer, starting from 3 (so that a^2 is negative). You should get a=12, 8, 6, 5, but cut off 6 as we have already counted that in Case 1. Thus, three cases here. However, we can also do b=c and a=c on top of this - thus, we get 3*3 = 9 cases. | ||
+ | |||
+ | Case 3: Can we have a/b/c independent of each other? | ||
+ | Simply put, no. It's just not possible for there to be no overlap. You can verify this by setting persay a equal to some integer, plugging it in to the equation, and then using SFFT, but you will be to no avail. | ||
+ | |||
+ | A practical approach here is to just look at the answer choices. If a/b/c were independent of each other, and a/b/c have nothing differentiating them, then for each independent ordered triple {a,b,c}, we should have 6 solutions. That would mean that the solution has to be 10+6n, where n is some integer. Obviously the only answer choice here that works is B, as the the other answer choices we cannot plug in an integral n to get them. Thus, either by POE or complicated induction, we find the answer to be 10. | ||
+ | |||
+ | iron | ||
==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 20:58, 10 February 2019
Contents
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 , and the volume is
, so equating the two yields
Divide both sides by to obtain
First consider the bound of the variable . Since
we have
, or
.
Also note that , hence
.
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 , we get
, so
. We find the solutions
,
,
,
,
, for a total of
solutions.
When , we get
, so
. We find the solutions
,
,
, for a total of
solutions.
When , we get
, so
. The only solution in this case is
.
When ,
is forced to be
, and thus
.
Thus, there are solutions.
Solution 2
Find that 2ab+2bc+2ac = abc
I'm not going to go through the bashing process of this solution, I will just give the outline
Split it up into cases:
Case 1: a=b=c You should get a=6
Case 2: a=b You should get a^2(2-c)+4ac = 0. The easiest way to bash this is to set c equal to an integer, starting from 3 (so that a^2 is negative). You should get a=12, 8, 6, 5, but cut off 6 as we have already counted that in Case 1. Thus, three cases here. However, we can also do b=c and a=c on top of this - thus, we get 3*3 = 9 cases.
Case 3: Can we have a/b/c independent of each other? Simply put, no. It's just not possible for there to be no overlap. You can verify this by setting persay a equal to some integer, plugging it in to the equation, and then using SFFT, but you will be to no avail.
A practical approach here is to just look at the answer choices. If a/b/c were independent of each other, and a/b/c have nothing differentiating them, then for each independent ordered triple {a,b,c}, we should have 6 solutions. That would mean that the solution has to be 10+6n, where n is some integer. Obviously the only answer choice here that works is B, as the the other answer choices we cannot plug in an integral n to get them. Thus, either by POE or complicated induction, we find the answer to be 10.
iron
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.