2001 AMC 12 Problems/Problem 21

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

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2001 AMC 12 Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

See Also