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

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== Problem ==
 
== Problem ==
  
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Four positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> have a product of <math>8!</math> and satisfy:
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Solve the following system of equations for <math>c</math>:
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<cmath>
 
 
\begin{align*}
 
\begin{align*}
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ab + a + b & = 524
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a - b &= 2 (c+d)\\
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\\  
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b &= a-2 \\
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bc + b + c & = 146
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d &= c+5
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\\  
 
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cd + c + d & = 104
 
 
\end{align*}
 
\end{align*}
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</cmath>
 
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What is <math>a-d</math>?
 
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<math>
 
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\text{(A) }4
 
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\qquad
 
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\text{(B) }6
 
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\qquad
 
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\text{(C) }8
 
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\qquad
 
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\text{(D) }10
 
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\qquad
 
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\text{(E) }12
 
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</math>
 
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 13:10, 17 May 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+1)(b+1) & = 525 \\ (b+1)(c+1) & = 147 \\ (c+1)(d+1) & = 105 \end{align*}

Looking at the first two equations, we know that $a+1$ but not $b+1$ is a multiple of 5, and looking at the last two equations, we know that $(d+1)$ but not $(c+1)$ must be a multiple of 5 (since if $b+1$ or $c+1$ was a multiple of 5, then $(b+1)(c+1)$ would also be a multiple of 5).

Thus, $(a+1) \equiv (d+1) \equiv 0 \hspace{1 mm} \text{(mod 5)}$, and $a - d \equiv 0 \hspace{1 mm} \text{(mod 5)}$. The only answer choice where this is true is $\boxed{\text{D) 10}}$.

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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