Suppose a, b and c are three distinct natural numbers such that a + b + c = abc. Consider the following statements: 1. The arithmetic mean of a, b and c is a natural number. 2. The harmonic mean of a, b and c lies between 1 and 2. Which of the statements given above is/are correct?

Correct — C, Both 1 and 2. The condition a + b + c = abc on three distinct natural numbers has exactly one solution, and once you find it both statements can be checked directly. Take a < b < c. Then abc = a + b + c < 3c, so ab < 3. Since the numbers are distinct naturals with a ≥ 1 and b ≥ 2, ab is at least 2, which forces ab = 2 and therefore a = 1, b = 2. Substituting, 1 + 2 + c = 2c gives c = 3. So the triple is {1, 2, 3}, and indeed 1 + 2 + 3 = 6 = 1 × 2 × 3. Now statement 1: the arithmetic mean is 6/3 = 2, a natural number, so it holds. Statement 2: the harmonic mean is 3 ÷ (1/1 + 1/2 + 1/3) = 3 ÷ (11/6) = 18/11 ≈ 1.64, which lies between 1 and 2, so it holds too.

• (a)1 only — Accepts the arithmetic mean but rejects the harmonic mean, though 18/11 is about 1.64 and sits squarely between 1 and 2. The reciprocals 1, 1/2 and 1/3 add to 11/6, and 3 divided by 11/6 is 18/11.
• (b)2 only — Accepts the harmonic mean but rejects the arithmetic mean, though the sum is 6 and 6 ÷ 3 = 2, as natural a number as could be asked for.
• (d)Neither 1 nor 2 — Rejects both. It would appeal only to a candidate who never found the triple and assumed the condition had many solutions with messy means; in fact {1, 2, 3} is the only one.

Three means are defined for positive numbers. The arithmetic mean is the sum divided by the count. The geometric mean is the nth root of the product. The harmonic mean is the count divided by the sum of the reciprocals, n ÷ Σ(1/x). For any set of positive numbers that are not all equal, AM > GM > HM strictly, which gives a free check on any computed value.

The bounding argument is the reusable part of this question. Whenever a symmetric condition ties a sum to a product, order the variables, replace each of them by the largest, and you get an inequality that boxes the smaller variables into a tiny range. Here a + b + c < 3c collapses abc < 3c to ab < 3 in one line, and there is only one way for two distinct naturals to multiply to 2. Once {1, 2, 3} is in hand, the three means are AM = 2, GM = 6^(1/3) ≈ 1.82 and HM = 18/11 ≈ 1.64, which respect AM > GM > HM as they must. It is worth remembering that a harmonic mean is always pulled towards the smallest member of the set, which is why it lands nearer 1 than 2 here; the arithmetic mean, by contrast, is pulled towards the largest.

• The only set of three distinct natural numbers with a + b + c = abc is {1, 2, 3}, since 1 + 2 + 3 = 6 = 1 × 2 × 3.
• The bounding step: with a < b < c, abc = a + b + c < 3c gives ab < 3, so a = 1 and b = 2.
• Arithmetic mean = 6/3 = 2 — a natural number.
• Harmonic mean = 3 ÷ (1 + 1/2 + 1/3) = 3 ÷ (11/6) = 18/11 ≈ 1.64.
• For unequal positive numbers, AM > GM > HM always: 2 > 6^(1/3) ≈ 1.82 > 1.64.

• Assuming the condition has many solutions and abandoning the question instead of bounding it.
• Computing the harmonic mean as the arithmetic mean of the reciprocals, forgetting to invert at the end.
• Reading 'natural number' as 'integer' and letting a negative or zero value in — natural numbers here start at 1.

Asked as a two-statement code item whose real work is a single hidden step — identifying the unique triple that satisfies the condition.

No directly related past PYQ was found.

• practice — not a real PYQ

  The harmonic mean of 2, 3 and 6 is

  • (a)2
  • (b)3
  • (c)11/3
  • (d)6

  Answer(b) 3 — the reciprocals 1/2 + 1/3 + 1/6 add to 1, and 3 ÷ 1 = 3.

• practice — not a real PYQ

  For two distinct positive numbers, which one of the following is always true?

  • (a)AM = GM = HM
  • (b)AM > GM > HM
  • (c)HM > GM > AM
  • (d)GM > AM > HM

  Answer(b) AM > GM > HM — equality holds only when the numbers are equal.