Positive Integer Solutions to Reciprocal Equation

CAT 2017 Slot 2 · QA · Hard · Number Theory

How many different pairs (a,b)(a, b) of positive integers are there such that aba \le b and 1a+1b=19\frac{1}{a} + \frac{1}{b} = \frac{1}{9}?

Answer

3

Explanation

Equation: 1a+1b=19    9a+9b=ab    ab9a9b=0\frac{1}{a} + \frac{1}{b} = \frac{1}{9} \implies 9a + 9b = ab \implies ab - 9a - 9b = 0.

Add 81 to both sides: (a9)(b9)=81(a - 9)(b - 9) = 81

Since aba \le b, we look for factor pairs (x,y)(x, y) of 81 such that xyx \le y where x=a9,y=b9x = a-9, y = b-9:

  • 1×81    a9=1,b9=81    (10,90)1 \times 81 \implies a - 9 = 1, b - 9 = 81 \implies (10, 90)
  • 3×27    a9=3,b9=27    (12,36)3 \times 27 \implies a - 9 = 3, b - 9 = 27 \implies (12, 36)
  • 9×9    a9=9,b9=9    (18,18)9 \times 9 \implies a - 9 = 9, b - 9 = 9 \implies (18, 18)

There are 3 such pairs.

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