Let $n$ be a natural number such that $\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{n}$ is also a natural number. Which of the following statements is not true?

### Concept & Logic For a sum of fractions to result in a natural number (an integer $\ge 1$), the fractional parts must add up perfectly to a whole number. Finding a common denominator simplifies the arithmetic. ### Step-by-Step Solution 1. Add the known fractions: $\frac{1}{2} + \frac{1}{3} + \frac{1}{7}$. 2. The Least Common Multiple (LCM) of 2, 3, and 7 is 42. 3. Rewrite the fractions: $\frac{21}{42} + \frac{14}{42} + \frac{6}{42} = \frac{41}{42}$. 4. We are given that $\frac{41}{42} + \frac{1}{n} = k$, where $k$ is a natural number. 5. For the sum to be a natural number, the smallest possible value for $k$ is 1. 6. Set $k = 1$: $\frac{41}{42} + \frac{1}{n} = 1 \Rightarrow \frac{1}{n} = 1 - \frac{41}{42} = \frac{1}{42}$. 7. Thus, $n = 42$. 8. Evaluate the given options with $n = 42$: 2 divides 42 (True), 3 divides 42 (True), 7 divides 42 (True), $42 > 84$ (False). ### Exam Strategy & Shortcut Recognize that $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$. Adding $\frac{1}{7}$ brings it very close to 1. The deficit from 1 is exactly $\frac{1}{42}$. Therefore, $n$ must be 42 immediately. Testing the options against $n=42$ reveals the false statement instantly. ### Common Pitfall Assuming $k$ could be 2 or higher and getting stuck trying to solve for multiple values of $n$. If $k=2$, $\frac{1}{n} = \frac{43}{42}$, making $n = \frac{42}{43}$, which is not a natural number. $k$ must be 1. ### Final Answer **Therefore, the correct answer is $n > 84$.**