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

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:

\[
\boxed{296}
\]