How many pairs of positive integers $a$ and $b$ satisfy the equation $\frac{a}{7}+\frac{2}{b}=1$ ?
(A) 4
(B) 1
(C) 0
(D) 5
(E) 2

Part C: Each correct answer is worth 8.
Since $a$ and $b$ are positive integers, then each of $\frac{a}{7}$ and $\frac{2}{b}$ is greater than 0 .

The sum of $\frac{a}{7}$ and $\frac{2}{b}$ is equal to 1 , and so each is less than 1 .

Since $\frac{a}{7}$ is greater than 0 and less than 1 , then the possible values of $a$ are $1,2,3,4,5,6$.

By substituting each of these values for $a$ into the equation one at a time, we can determine if there is a positive integer value of $b$ for which the equation is true.

Substituting $a=1$, we get $\frac{1}{7}+\frac{2}{b}=1$ or $\frac{2}{b}=1-\frac{1}{7}$, and so $\frac{2}{b}=\frac{6}{7}$.

Since $\frac{2}{b}=\frac{6}{7}$, we can multiply the numerator and denominator of the first fraction by 3 (which is $6 \div 2$ ) to get $\frac{6}{3 b}=\frac{6}{7}$.

This gives $3 b=7$ which does not have an integer solution $\left(b=\frac{7}{3}\right)$.

Thus when $a=1$, there is no positive integer value of $b$ that satisfies the equation.

Substituting $a=2$, we get $\frac{2}{7}+\frac{2}{b}=1$ or $\frac{2}{b}=1-\frac{2}{7}$, and so $\frac{2}{b}=\frac{5}{7}$.

Since $\frac{2}{b}=\frac{5}{7}$, we can multiply the numerator and denominator of the first fraction by 5 , and the numerator and denominator of the second fraction by 2 to get $\frac{10}{5 b}=\frac{10}{14}$.

This gives $5 b=14$ which does not have an integer solution.

Thus when $a=2$, there is no positive integer value of $b$ that satisfies the equation.

Substituting $a=3$, we get $\frac{3}{7}+\frac{2}{b}=1$ or $\frac{2}{b}=1-\frac{3}{7}$, and so $\frac{2}{b}=\frac{4}{7}$.

Since $\frac{2}{b}=\frac{4}{7}$, we can multiply the numerator and denominator of the first fraction by 2 to get $\frac{4}{2 b}=\frac{4}{7}$.

This gives $2 b=7$ which does not have an integer solution.

Thus when $a=3$, there is no positive integer value of $b$ that satisfies the equation.

Substituting $a=4$ and simplifying, we get $\frac{2}{b}=\frac{3}{7}$.

Since $\frac{2}{b}=\frac{3}{7}$, we can multiply the numerator and denominator of the first fraction by 3 , and the numerator and denominator of the second fraction by 2 to get $\frac{6}{3 b}=\frac{6}{14}$.

This gives $3 b=14$ which does not have an integer solution.

Thus when $a=4$, there is no positive integer value of $b$ that satisfies the equation.

Substituting $a=5$ and simplifying, we get $\frac{2}{b}=\frac{2}{7}$.

Since the numerators are equal, then the denominators must be equal, and so $b=7$ satisfies the equation.

Finally, substituting $a=6$ and simplifying, we get $\frac{2}{b}=\frac{1}{7}$.

Since $\frac{2}{b}=\frac{1}{7}$, we can multiply the numerator and denominator of the second fraction by 2 to get $\frac{2}{b}=\frac{2}{14}$, and so $b=14$.

Thus, there are two pairs of positive integers $a$ and $b$ that satisfy the given equation: $a=5, b=7$ and $a=6, b=14$.

ANSWER: (E)