2002 AMC 10B Problem 7

The question asks which statement is false, so first find out what $n$ actually is.

1. Add the three known fractions using the common denominator $42$:
$$\frac12 + \frac13 + \frac17 = \frac{21}{42} + \frac{14}{42} + \frac{6}{42} = \frac{41}{42}.$$
That is just a little less than $1$.
2. Now think about how big $\frac1n$ can be. For $n = 1$ it is $1$; for bigger $n$ it is smaller, but it is always more than $0$.
3. So the whole sum is more than $\frac{41}{42}$ and at most $\frac{41}{42} + 1$, which is less than $2$. The only whole number in that range is $1$.
4. So the sum must equal $1$, which means
$$\frac1n = 1 - \frac{41}{42} = \frac1{42}, \qquad n = 42.$$
5. Check the statements with $n = 42 = 2 \cdot 3 \cdot 7$:
- $2$, $3$, $6$ and $7$ all divide $42$, so (A) through (D) are true.
- $42 > 84$ is false.

The answer is $\boxed{\textbf{(E)}\ n > 84}$.

"The sum is a whole number" sounds weak, but if you can trap the sum between two neighboring whole numbers, only one value is possible. The size of $\frac1n$ was what set the trap. After that, the question is just checking four true statements and spotting the false one.

Combine the three known fractions over $42$:
$$\frac12 + \frac13 + \frac17 = \frac{21 + 14 + 6}{42} = \frac{41}{42}.$$

Since $n$ is a positive integer, $0 < \frac1n \le 1$. Therefore the full sum lies strictly between $\frac{41}{42}$ and $\frac{41}{42} + 1 < 2$. The only integer in that window is $1$, so
$$\frac{41}{42} + \frac1n = 1 \quad\Longrightarrow\quad \frac1n = \frac1{42} \quad\Longrightarrow\quad n = 42.$$

Now test the statements: $42 = 2\cdot 3\cdot 7$ is divisible by $2$, $3$, $6$ and $7$, so (A) through (D) are true, while $42 > 84$ is false.

The answer is $\boxed{\textbf{(E)}\ n > 84}$.