Number Theory or Algebra?

$\begin{cases} a + \dfrac bc = 101 & ...(1) \\ \dfrac ac + b = 68 & ...(2) \\ \dfrac {a+b}c = k &...(3) \end{cases}$

$\begin{aligned} (1)+(2): \quad a + b + \frac {a+b}c & = 101+68 \\ (a + b)\left(1 + \frac 1c\right) & = 169 \\ a+b & = \frac {13^2c}{c+1} \end{aligned}$

Since the LHS is a positive integer, the RHS must also be an positive integer. This means that for $c \ge 1$ , $c+1$ can only be either 13 or $13^2$ . But if $c+1=13^2$ , then $c=168$ and $a+b = 168$ , $\implies a, b < c$ , then for $(1): \ a + \dfrac bc = 101$ , $a$ and $b$ are not integers. Therefore, $c+1 = 13$ , then $a+b = 13c$ , $\implies k = \dfrac {a+b}c = \boxed{13}$ .

Since $a+\dfrac{b}{c}$ is a positive integer, then $\dfrac{b}{c}$ is an positive integer, meaning $b$ must be a multiple of $c$ . Similarly, since $\dfrac{a}{c}+b$ is a positive integer, then $a$ must also be a multiple of $c$ ; we can rewrite $a$ and $b$ as $Ac$ and $Bc$ respectively, for some positive integers $A$ and $B$ .

Ergo, $a+\dfrac{b}{c}=101$ becomes $Ac+B=101$ and $\dfrac{a}{c}+b=68$ becomes $A+Bc=68$ . Adding these new equations gives us $Ac+B+A+Bc=169$ or $A(c+1)+B(c+1)=169$ , and so $(A+B)(c+1)=169$ , meaning $c+1$ must be a factor of $169$ .

$169=13^2$ , and therefore has positive divisors $1, 13,$ and $169$ . Since $A, B,$ and $c$ are positive integers, $A+B≥2$ and $c+1≥2$ . Neither $A+B$ nor $c+1$ can equal $1$ , meaning $A+B=c+1=13$ .

Finally, $\dfrac{a+b}{c}=\dfrac{Ac+Bc}{c}=A+B=13,$ and so, $\boxed{k=13}$ .

Proving integers $a, b, c$ exist

We have

$\begin{cases} a+\dfrac{b}{c}=101& \cdots(1) \\ \dfrac{a}{c}+b=68& \cdots(2) \\ \dfrac{a+b}{c}=13& \cdots(3)\end{cases}$

$(1)+(2) : a+b+\dfrac{a+b}{c}=169$

$a+b+13=169\Rightarrow a+b=156 \cdots (4)$

Substituting $(4)$ into $(3): \dfrac{156}{c}=13$

$\Rightarrow \boxed{c=12}$

$(4): a+b=156\Rightarrow a=156-b$

Substituting the above into $(1): 156-b+\dfrac{b}{12}=101$

$\Rightarrow \boxed{b=60}$

Substituting $b=60$ into $(2)$ we get: $\dfrac{a}{12}+60=68$

$\Rightarrow \boxed{a=96}$

Thus $a=96, b=60,$ and $c=12.$

Let a = cx and b = cy.

cx + y = 101

cy + x = 68

(c + 1)(y + x) = 13^2

(x + y) can't equal 1, which means c = 12 and x + y = 13.

(a + b)/c = x + y = 13

a = 96, b = 60

Add the first two expressions to get $\frac {ac + b} {c} + \frac {ac + b} {c} = 169$

Combine and factor to get $\frac {(a + b) } {c} (1 + c) = 169$

Since both factors are integers and 169= $13^2$ it follows that

$\frac {(a + b) } {c} = 13$ and $1 + c = 13$ is the only possible solution in integers.

The equations are satisfied by a = 96, b = 60, c = 12.