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2001 AMC 12 Problems/Problem 21: Difference between revisions

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Basically, the problem displayed on the page was the incorrect problem from the AMC, and the solution didn't make sense in context of the incorrect problem. I have edited to replace the incorrect problem with the correct problem.
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== Problem ==
== Problem ==


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


== Solution 1 ==
== Solution 1 ==

Revision as of 13:10, 17 October 2019

Problem

Four positive integers $a$, $b$, $c$, and $d$ have a product of $8!$ and satisfy:

\begin{align*} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{align*}

What is $a-d$?

$\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$

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 $a-d=24-14 =\boxed{10}$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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