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?
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:
$1 + b + c = 1 \cdot b \cdot c$ $1 + b + c = bc$Rearrange the terms to solve for $b$ and $c$:
$bc - b - c = 1$Add 1 to both sides to factor the expression:
$bc - b - c + 1 = 1 + 1$ $(b-1)(c-1) = 2$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 $\{1, 2, 3\}$:
$AM = \frac{1+2+3}{3} = \frac{6}{3} = 2$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 $\{1, 2, 3\}$:
$HM = \frac{3}{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}}$First, calculate the sum of the reciprocals:
$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{6+3+2}{6} = \frac{11}{6}$Now, calculate the harmonic mean:
$HM = \frac{3}{\frac{11}{6}} = 3 \times \frac{6}{11} = \frac{18}{11}$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.
Important Questions from Integers
- The average of eleven consecutive positive integers is d. If the last two numbers are excluded, by how much will the average increase or decrease?
The numerator of fraction is 3 more than the denominator. When 5 is added to the numerator and 2 is subtracted from the denominator, the fraction becomes 8/3, When the original fraction is divided by \(5 \frac{1}{2}\) , the fraction so obtained is:
The sum of a non - zero number and twenty times its reciprocal is 9. What is the number?
If \(\frac{{45}}{{53}} = \frac{1}{{a + \frac{1}{{b + \frac{1}{{c - \frac{2}{5}}}}}}},\) where a, b and c are positive integers, then what is the value of (4a - b + 3c)
The denominator of a fraction is 4 more than the double of its numerator. When 3 is added to the numerator and 3 is subtracted from denominator the fraction becomes 2/3. Then find the difference between denominator and numerator of the original fration.
