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avatar+1760 

Let $a$, $b$, $c$, and $n$ be positive integers. If $a + b + c = 19 \cdot 97$ and \[a + n = b - n = \frac{c}{n},\] compute the value of $a$.

$$Let $a$, $b$, $c$, and $n$ be positive integers. If $a + b + c = 19 \cdot 97$ and
\[a + n = b - n = \frac{c}{n},\]
compute the value of $a$.$$

Mellie  Jul 2, 2015

Best Answer 

 #2
avatar+18712 
+15

$$\small{\text{$
\begin{array}{lcl}
$Let $ a, b, c, $ and $ n $ be positive integers. If $
a + b + c = 19 \cdot 97 $ and $ \\
\left[a + n = b - n = \dfrac{c}{n}\right],
$ compute the value of $ a .
\end{array}
$}}$$

 

$$\small{\text{$
\begin{array}{lrrrcl}
& a+b+c=19\cdot 97 \\
\\
\hline
\\
(1)& a+n=k \\
(2) & b-n=k &\qquad \qquad (1)+(2): & a+b&=& 2\cdot k\\
(3) & \dfrac{c}{n}=k &\qquad \qquad $so$ & c &=& k\cdot n\\\\
& & & a+b+c &=& 2\cdot k + k\cdot n=19\cdot 97
\\
\hline
\\
\end{array}
$}}\\\\
\small{\text{$
\begin{array}{rrclrcl}
& 2\cdot k + k\cdot n &=& 19\cdot 97\\\\
& k\cdot(2+n)&=& 19\cdot 97\\\\
I. & \textcolor[rgb]{1,0,0}{k}\cdot\textcolor[rgb]{0,0,1}{(2+n)}&=& \textcolor[rgb]{1,0,0}{19}\cdot \textcolor[rgb]{0,0,1}{97}\\\\
& \underline{k=19} && \underline{2+n = 97 }& \qquad \Rightarrow
\qquad n&=& 95\\
& && & a+n&=& k\\
& && & a+95&=& 19\\
& && & a&=& -76 ~$ negative! $\\ \\
II. & \textcolor[rgb]{1,0,0}{k}\cdot\textcolor[rgb]{0,0,1}{(2+n)}&=& \textcolor[rgb]{1,0,0}{97}\cdot \textcolor[rgb]{0,0,1}{19}\\\\
& \underline{k=97} && \underline{2+n = 19 }& \qquad \Rightarrow
\qquad n&=& 17\\
& && & a+n&=& k\\
& && & a+17&=& 97\\
& && & a&=& 80 ~$ okay! $\\
\end{array}
$}}$$

 

heureka  Jul 3, 2015
Sort: 

3+0 Answers

 #1
avatar+26328 
+15

Integers:

.

Alan  Jul 2, 2015
 #2
avatar+18712 
+15
Best Answer

$$\small{\text{$
\begin{array}{lcl}
$Let $ a, b, c, $ and $ n $ be positive integers. If $
a + b + c = 19 \cdot 97 $ and $ \\
\left[a + n = b - n = \dfrac{c}{n}\right],
$ compute the value of $ a .
\end{array}
$}}$$

 

$$\small{\text{$
\begin{array}{lrrrcl}
& a+b+c=19\cdot 97 \\
\\
\hline
\\
(1)& a+n=k \\
(2) & b-n=k &\qquad \qquad (1)+(2): & a+b&=& 2\cdot k\\
(3) & \dfrac{c}{n}=k &\qquad \qquad $so$ & c &=& k\cdot n\\\\
& & & a+b+c &=& 2\cdot k + k\cdot n=19\cdot 97
\\
\hline
\\
\end{array}
$}}\\\\
\small{\text{$
\begin{array}{rrclrcl}
& 2\cdot k + k\cdot n &=& 19\cdot 97\\\\
& k\cdot(2+n)&=& 19\cdot 97\\\\
I. & \textcolor[rgb]{1,0,0}{k}\cdot\textcolor[rgb]{0,0,1}{(2+n)}&=& \textcolor[rgb]{1,0,0}{19}\cdot \textcolor[rgb]{0,0,1}{97}\\\\
& \underline{k=19} && \underline{2+n = 97 }& \qquad \Rightarrow
\qquad n&=& 95\\
& && & a+n&=& k\\
& && & a+95&=& 19\\
& && & a&=& -76 ~$ negative! $\\ \\
II. & \textcolor[rgb]{1,0,0}{k}\cdot\textcolor[rgb]{0,0,1}{(2+n)}&=& \textcolor[rgb]{1,0,0}{97}\cdot \textcolor[rgb]{0,0,1}{19}\\\\
& \underline{k=97} && \underline{2+n = 19 }& \qquad \Rightarrow
\qquad n&=& 17\\
& && & a+n&=& k\\
& && & a+17&=& 97\\
& && & a&=& 80 ~$ okay! $\\
\end{array}
$}}$$

 

heureka  Jul 3, 2015
 #3
avatar+91001 
+5

2 great answers - thanks Alan and Heureka   

Melody  Jul 3, 2015

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