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

(New page: == Problem == 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> \begin{align*} ab + a + b &...)
 
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Revision as of 21:04, 3 July 2013

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

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 $7$.)

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

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