Question

How many different pairs (a, b) of positive integers are there such that a ≤ b and 1/a+1/b=1/9=?

• A. 1
• B. 3
• C. 5
• D. 4

Answer 1

The Correct Option is B

To find the number of pairs \((a, b)\) where \(a \leq b\) and \(\frac{1}{a}+\frac{1}{b}=\frac{1}{9}\), let's begin by manipulating the equation:

\(\frac{1}{a}+\frac{1}{b}=\frac{1}{9}\)

Combine the fractions: 

\(\frac{a+b}{ab}=\frac{1}{9}\)

Cross-multiply to eliminate the fractions:

\(9(a+b)=ab\)

Rearrange the equation:

\(ab - 9a - 9b = 0\)

To factor it, add 81 to both sides:

\((a-9)(b-9)=81\)

Now, solve for integer pairs \((a-9, b-9)\) that multiply to 81. Consider the factor pairs of 81: \(1\times81\), \(3\times27\), \(9\times9\), \((-1)\times(-81)\), \((-3)\times(-27)\), \((-9)\times(-9)\). Since \(a\) and \(b\) are positive, we only consider positive factors:

1. \((1, 81)\) results in: \(a=10\), \(b=90\) (since \(a \leq b\))
2. \((3, 27)\) results in: \(a=12\), \(b=36\)
3. \((9, 9)\) results in: \(a=18\), \(b=18\)

Therefore, the valid pairs \((a, b)\) that satisfy the conditions are \((10, 90)\), \((12, 36)\), and \((18, 18)\), totaling to 3 different pairs.

Pair	\(a\)	\(b\)
\(1\)	10	90
\(2\)	12	36
\(3\)	18	18

Thus, the correct answer is 3.