The problem asks us to evaluate two statements about three distinct natural numbers $a, b, c$ that satisfy the equation $a + b + c = abc$.

Finding Distinct Natural Numbers for $a + b + c = abc$

Let the three distinct natural numbers be $a, b, c$. Without loss of generality, assume $a < b < c$. Since they are natural numbers, the smallest possible values are $a=1, b=2, c=3$.

If we set $a=1$, the equation becomes:

Rearrange the terms to solve for $b$ and $c$:

Add 1 to both sides to factor the expression:

Since $a, b, c$ are distinct natural numbers and $a < b < c$, we have $1 < b < c$. This means $b-1$ and $c-1$ must be positive integers.

The integer pairs that multiply to 2 are (1, 2).

Therefore, we must have:

• $b-1 = 1 \implies b = 2$
• $c-1 = 2 \implies c = 3$

So, the only set of distinct natural numbers satisfying the condition is $\{1, 2, 3\}$. Let's verify: $1 + 2 + 3 = 6$ and $1 \times 2 \times 3 = 6$. The equation holds.

If $a \ge 2$, then $a \ge 2, b \ge 3, c \ge 4$. This would imply $abc \ge 2 \times 3 \times 4 = 24$. However, $a+b+c \le c+c+c = 3c$. If $a=2, b=3, c=4$, $a+b+c = 9$ and $abc = 24$. If $a, b, c$ increase, $abc$ grows much faster than $a+b+c$. If we divide the original equation by $abc$: $1/bc + 1/ac + 1/ab = 1$. If $a \ge 2$, the maximum value of the left side is $1/(2 \times 3) + 1/(2 \times 4) + 1/(3 \times 4) = 1/6 + 1/8 + 1/12 = (4+3+2)/24 = 9/24 < 1$. Thus, $a$ must be 1.

Statement 1: Arithmetic Mean (AM) Analysis

The statement says the arithmetic mean of $a, b, c$ is a natural number.

The arithmetic mean is calculated as $AM = \frac{a+b+c}{3}$.

Using the numbers $\{1, 2, 3\}$:

Since 2 is a natural number, Statement 1 is correct.

Statement 2: Harmonic Mean (HM) Analysis

The statement says the harmonic mean of $a, b, c$ lies between 1 and 2.

The harmonic mean is calculated as $HM = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$.

Using the numbers $\{1, 2, 3\}$:

First, calculate the sum of the reciprocals:

Now, calculate the harmonic mean:

To check if $HM$ lies between 1 and 2:

• $1 = \frac{11}{11}$
• $2 = \frac{22}{11}$

Since $\frac{11}{11} < \frac{18}{11} < \frac{22}{11}$, we have $1 < HM < 2$.

Therefore, Statement 2 is correct.

Conclusion

Both Statement 1 and Statement 2 are correct.