Let $a,$ $b,$ $c,$ and $n$ be positive integers. If a + b + c = 793 and

Question

$$a+5n=b-3n=cn,$$
compute the value of $a.$

Answer 1

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:

$$\boxed{{\displaystyle 296}}$$