Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 12 Problems/Problem 21"

(→‎Solution 2)
(→‎Solution 2)
Line 43: Line 43:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
 
−
As above, we can write the equations as follows:
 
−
 
−
<cmath>
 
−
\begin{align*}
 
−
a - b = 2(c+d)
 
−
b = a - 2
 
−
d = c + 5
 
−
\end{align*}
 
−
</cmath>
 
−
 
 
From the second equation, we can conclude that <math>a-b=2</math>. Plugging this into the first equation yields that <math>c+d=1</math>. The last equation implies that <math>d-c=5</math>, so by inspection, we know that <math>d=3</math> and <math>c=\boxed{-2}</math>.
 
From the second equation, we can conclude that <math>a-b=2</math>. Plugging this into the first equation yields that <math>c+d=1</math>. The last equation implies that <math>d-c=5</math>, so by inspection, we know that <math>d=3</math> and <math>c=\boxed{-2}</math>.
 
<cmath></cmath>
 
<cmath></cmath>

Revision as of 12:30, 6 July 2019

Problem

Solve the following system of equations for $c$: \begin{align*} a - b &= 2 (c+d)\\ b &= a-2 \\ d &= c+5 \end{align*}

Solution 1

Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:

\begin{align*} (a+1)(b+1) & = 525 \\ (b+1)(c+1) & = 147 \\ (c+1)(d+1) & = 105 \end{align*}

Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$. We get:

\begin{align*} ef & = 3\cdot 5\cdot 5\cdot 7 \\ fg & = 3\cdot 7\cdot 7 \\ gh & = 3\cdot 5\cdot 7 \end{align*}

Clearly $7^2$ divides $fg$. On the other hand, $7^2$ can not divide $f$, as it then would divide $ef$. Similarly, $7^2$ can not divide $g$. Hence $7$ divides both $f$ and $g$. This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$.

The first case solves to $(e,f,g,h)=(75,7,21,5)$, which gives us $(a,b,c,d)=(74,6,20,4)$, but then $abcd \not= 8!$. We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$. (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.)

The second case solves to $(e,f,g,h)=(25,21,7,15)$, which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$, and we have $a-d=24-14 =\boxed{10}$.

Solution 2

From the second equation, we can conclude that $a-b=2$. Plugging this into the first equation yields that $c+d=1$. The last equation implies that $d-c=5$, so by inspection, we know that $d=3$ and $c=\boxed{-2}$. \[\] ~AopsUser101

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_21&oldid=107401"