+1836 Let $a$, $b$, $c$, and $n$ be positive integers. If $a + b + c = 19 \cdot 97$ and a+n=b−n=cn, compute the value of $a$.
Let$a$,$b$,$c$,and$n$bepositiveintegers.If$a+b+c=19⋅97$and\[a+n=b−n=cn,\]computethevalueof$a$.
+26397 \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} $}}
+26397 \small{\text{$ \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} $}}
