Problem Statement
Sylvia chose positive integers $a$, $b$ and $c$. Peter determined the value of $a + \dfrac{b}{c}$ and got an answer of 101. Paul determined the value of $\dfrac{a}{c} + b$ and got an answer of 68. Mary determined the value of $\dfrac{a + b}{c}$ and got an answer of $k$. The value of $k$ is
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Solution
$$a + \dfrac{b}{c} = 101 \implies ac + b = 101c \; \text{(1)}$$ $$\dfrac{a}{c} + b = 68 \implies a + bc = 68c \; \text{(2)}$$ $$\dfrac{a + b}{c} = k \implies a + b = kc$$ $$\text{(1) + (2): }(a + b)(c + 1) = 169c \implies (kc)(c + 1) = 169c \implies k(c+1) = 169$$ $$\text{Since } c \in \mathbb{N} \implies c+1 \in \mathbb{N} \implies k \in \mathbb{N}$$ $$169 = 1 \times 169 \text{ or } 169 = 13 \times 13 \text{ or } 169 = 169 \times 1$$ $$\text{Case 1: let } k = 1, c+1 = 169 \implies c = 168 \implies a + b = 168$$ $$\text{From (1): } a(168) + b = 101(168) \implies 167a + 168 = 101(168) \implies a = \dfrac{100(168)}{167}$$ $$\text{Since } a \in \mathbb{N} \implies a \ne \dfrac{100(168)}{167}$$ $$\text{Case 2: let } k = 13, c+1 = 13 \implies c = 12 \implies a + b = 156$$ $$\text{From (1): } a(12) + b = 101(12) \implies 11a + 156 = 101(12) \implies a = \dfrac{101(12) -156}{11} = 96$$ $$\text{Since } a \in \mathbb{N} \implies a = 96 \text{ is possible}$$ $$\therefore k = 13$$ 