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Q27·CSAT · Prelims 2025

1/p + 1/q + 1/r = 1 natural number solutions

NumericalEquations in IntegersAlgebra & equations● Hard

Question

How many possible values of (p + q + r) are there satisfying 1/p + 1/q + 1/r = 1, where p, q and r are natural numbers (not necessarily distinct)?

Options

a

None

b

One

c

Three

Answer
d

More than three

Explanation

We need to find sets of natural numbers (p, q, r) satisfying 1/p + 1/q + 1/r = 1. Assume p \le q \le r. Clearly, p cannot be 1. Case 1: p = 3. The only way to reach 1 without exceeding it is q = 3, r = 3. Sum = 9. Case 2: p = 2. We have 1/q + 1/r = 1/2.

If q = 3, 1/r = 1/2 - 1/3 = 1/6 \implies r = 6. The set is {2, 3, 6}. Sum = 11.
If q = 4, 1/r = 1/2 - 1/4 = 1/4 \implies r = 4. The set is {2, 4, 4}. Sum = 10.

No other integer combinations work. The possible sums are 9, 10, and 11, yielding exactly three distinct values.

To solve 1/a + 1/b + 1/c = 1, enforce an order (a \le b \le c) and test the smallest possible integer for a to bound the remaining variables tightly.

Answer: (c).

Question details

Year

2025

Paper

CSAT

Question

Q27

Section

Numerical Ability

Sub-topic

Equations in Integers

Type

Algebra & equations

Difficulty

Hard

Source hint

Number theory

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