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n Part a of this problem, you found the values of integers that satisfied the system of equations. In this part of the problem you will write up a full solution that describes how to solve this problem. Four positive integers $a$, $b$, $c$, and $d$ satisfy \begin{align*} ab + a + b &= 524, \\ bc + b + c &= 146, \\ cd + c + d &= 104, \\ \end{align*} and $abcd=8!$. What are $a$, $b$, $c$, and $d?$

 Apr 20, 2018
 #1
avatar+972 
+4

First of all, we have the equations:

 

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

 

and abcd=8!.

 

From the first equation, we have: 

 

\(ab + a + b = 524.\)

 

After factoring out the b, we have:

 

\(b(1 + a)  =  524 - a.\)

 

Solving for b, we have:

 

\(b =  \frac{524 -a}{1 + a}\)

 

From the second equation, we are going to try to write c in the form of a.

 

\(bc + b + c =  146.\)

 

Factoring out the b, we have:

 

\(b(c+1)+c=146.\)

 

Solving for b, we have:

 

\(b=\frac{146-c}{1+c}.\)

 

Since b = b, we have:

 

\(\frac{146-c}{1+c}=\frac{524 -a}{1 + a}.\)After cross-multiplying, we end up with:

 

\( (1+c)(524-a)=(1+a)(146-c).\)

 

Using the distributive property, we have:

 

\((524-a)+c(524-a)=146(1+a)-(524-a).\)

 

Isolating the expressions with c's and simplifying the left side, we have:

 

\(c(524-a+1+a)=146 + 146a-524+a.\)

 

Finalizing, we have:

 

\(c=\frac{7a-18}{25}.\)

 

Using the same method, we solve for d, and we end up with:

 

\(d =   \frac{374 - a}{1+a}.\)

 

Since abcd=8!, we have:

 

\(a\cdot\frac{524 - a}{1 + a}\cdot\frac{7a-18}{25}\cdot\frac{374 - a}{1 +a}  = 40320.\)

 

After simplifying and factoring, we have:

 

\((a - 24) (7a^3 -6136a^2 +232732a + 42000)   = 0.\)

 

The only way a can be an integer is when a=24.

 

From this, we can plug in a to find the other values.

 

\(\boxed{a=24}. \boxed{b=20}. \boxed{c=6}. \boxed{d=14}.\)

 

I hope this helped! :)

 

Gavin.

 Apr 20, 2018
 #2
avatar+98005 
0

Nice, Gavin.....!!!

 

 

cool cool cool

CPhill  Apr 21, 2018

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