Using the given equations \( a + 5n = b - 3n = cn \), we can denote the common value as \( k \). Therefore, we have the equations:
\[
a + 5n = k
\]
\[
b - 3n = k
\]
\[
cn = k
\]
From these equations, we solve for \(a\), \(b\), and \(c\):
\[
a = k - 5n
\]
\[
b = k + 3n
\]
\[
c = \frac{k}{n}
\]
We substitute these expressions into the equation \(a + b + c = 793\):
\[
(k - 5n) + (k + 3n) + \frac{k}{n} = 793
\]
We simplify this equation:
\[
2k - 2n + \frac{k}{n} = 793
\]
To eliminate the fraction, multiply through by \(n\):
\[
2kn - 2n^2 + k = 793n
\]
Combine like terms:
\[
2kn + k - 2n^2 = 793n
\]
Factor out \(k\) from the terms involving \(k\):
\[
k(2n + 1) - 2n^2 = 793n
\]
Solve for \(k\):
\[
k(2n + 1) = 793n + 2n^2
\]
Then,
\[
k = \frac{793n + 2n^2}{2n + 1}
\]
Since \(k\) must be an integer, \(793n + 2n^2\) must be divisible by \(2n + 1\). We now check suitable values of \(n\) that satisfy this condition.
Testing \(n=23\):
\[
k = \frac{793 \cdot 23 + 2 \cdot 23^2}{2 \cdot 23 + 1} = \frac{18239 + 1058}{47} = \frac{19297}{47} = 411
\]
This quotient is an integer, so \(n = 23\) is a valid solution. Now calculate \(a\):
\[
a = k - 5n = 411 - 5 \cdot 23 = 411 - 115 = 296
\]
Therefore, the value of \(a\) is:
\[
\boxed{296}
\]
Let $a,$ $b,$ $c,$ and $n$ be positive integers. If a + b + c = 793 and
\[a + 5n = b - 3n = cn,\]
compute the value of $a.$
1 answer