Suppose a, b and c are three distinct natural numbers such that a + b

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?

1 only

2 only

Both 1 and 2

Neither 1 nor 2

This question was previously asked in

UPSC CAPF – 2023

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Let the three distinct natural numbers be a, b, and c. The given condition is a + b + c = abc. We need to find the distinct natural numbers that satisfy this equation.

Assuming a <= b <= c, and since a, b, c are natural numbers (>= 1):
If a=1, the equation becomes 1 + b + c = bc. Rearranging gives bc – b – c = 1. Adding 1 to both sides to factor: bc – b – c + 1 = 2, which is (b-1)(c-1) = 2. Since b <= c, we have b-1 <= c-1. As b and c are natural numbers, b-1 >= 0 and c-1 >= 0. The only factors of 2 are (1, 2). So, b-1=1 and c-1=2, which gives b=2 and c=3. The set of distinct natural numbers is {1, 2, 3}. Checking: 1 + 2 + 3 = 6 and 1 * 2 * 3 = 6. This solution is valid.
We can show that there are no other solutions by considering a >= 2. If a >= 2, then b >= 2 and c >= 2. If a=2, 2+b+c = 2bc. Dividing by bc gives 2/bc + 1/c + 1/b = 2. If b=2, 2/4c + 1/c + 1/2 = 2 => 1/2c + 1/c + 1/2 = 2 => 3/2c = 3/2 => c=1. This contradicts c>=b=2. If b>=3, c>=b>=3, then 1/b <= 1/3 and 1/c <= 1/3. 1/b + 1/c <= 2/3. But 1/b + 1/c = 2 - 2/bc. So 2 - 2/bc <= 2/3 => 4/3 <= 2/bc => bc <= 1.5. This contradicts bc >= 3*3=9. If a>=3, then b>=3, c>=3, leading to the same contradiction bc <= 1.5 while bc >= 9.
Thus, the only set of distinct natural numbers satisfying the equation is {1, 2, 3}.

Statement 1: The arithmetic mean of a, b and c is a natural number.
AM = (1+2+3)/3 = 6/3 = 2. 2 is a natural number. Statement 1 is correct.

Statement 2: The harmonic mean of a, b and c lies between 1 and 2.
HM = 3 / (1/a + 1/b + 1/c) = 3 / (1/1 + 1/2 + 1/3) = 3 / (6/6 + 3/6 + 2/6) = 3 / (11/6) = 18/11.
18/11 is approximately 1.636. 1 < 18/11 < 2. Statement 2 is correct.

The equation a + b + c = abc is a Diophantine equation. The set {1, 2, 3} is the only solution in distinct natural numbers. If distinctness is not required, other solutions in natural numbers exist, e.g., {1, 1, 2} is not a solution (1+1+2=4, 1*1*2=2), {1, 1, 1} is not a solution (1+1+1=3, 1*1*1=1). In positive integers (which includes natural numbers), the only solution sets for a+b+c=abc are {1,2,3}.

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