Difference between revisions of "2001 AMC 12 Problems/Problem 21"
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Solve the following system of equations for <math>c</math>: | Solve the following system of equations for <math>c</math>: | ||
| + | <cmath> | ||
\begin{align*} | \begin{align*} | ||
a - b &= 2 (c+d)\\ | a - b &= 2 (c+d)\\ | ||
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d &= c+5 | d &= c+5 | ||
\end{align*} | \end{align*} | ||
| + | </cmath> | ||
== Solution 1 == | == Solution 1 == | ||
Revision as of 14:06, 19 May 2019
Contents
Problem
Solve the following system of equations for 
:
Solution 1
Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:
Let 
. We get:
Clearly 
divides 
. On the other hand, can not divide 
, as it then would divide 
. Similarly, can not divide 
. Hence 
divides both and . This leaves us with only two cases: 
and 
.
The first case solves to 
, which gives us 
, but then 
. We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by . (Also, a - d equals 
in this case, which is way too large to fit the answer choices.)
The second case solves to 
, which gives us a valid quadruple 
, and we have 
.
Solution 2
As above, we can write the equations as follows:
Looking at the first two equations, we know that 
but not 
is a multiple of 5, and looking at the last two equations, we know that 
but not 
must be a multiple of 5 (since if or 
was a multiple of 5, then 
would also be a multiple of 5).
Thus, 
, and 
. The only answer choice where this is true is 
.
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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. 
