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

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<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
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(a+1)(b+1) & = 525
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a - b = 2(c+d)
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\\
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b = a - 2
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(b+1)(c+1) & = 147
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d = c + 5
−
\\
 
−
(c+1)(d+1) & = 105
 
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
−
Looking at the first two equations, we know that <math>a+1</math> but not <math>b+1</math> is a multiple of 5, and looking at the last two equations, we know that <math>(d+1)</math> but not <math>(c+1)</math> must be a multiple of 5 (since if <math>b+1</math> or <math>c+1</math> was a multiple of 5, then <math>(b+1)(c+1)</math> would also be a multiple of 5).
+
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>.
−
 
+
<math></math>
−
Thus, <math>(a+1) \equiv (d+1) \equiv 0 \hspace{1 mm} \text{(mod 5)}</math>, and <math>a - d \equiv 0 \hspace{1 mm}  \text{(mod 5)}</math>. The only answer choice where this is true is <math>\boxed{\text{D) 10}}</math>.
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~AopsUser101
  
 
== See Also ==
 
== See Also ==

Revision as of 11:29, 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

As above, we can write the equations as follows:

\begin{align*} a - b = 2(c+d) b = a - 2 d = c + 5 \end{align*}

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 $c=\boxed{-2}$. $$ (Error compiling LaTeX. Unknown error_msg) ~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

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