2002 AMC 10B Problem 7

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

"The sum is an integer" is a strong condition when you can bound the sum: squeezing it between two consecutive integers leaves only one possible value. The size of $\frac1n$ is the lever. After $n$ is determined, the question is pure checking, and the "not true" wording means four statements will pass and one will fail.

• Trying to make the sum equal 2 (impossible since 1/n is at most 1) or assuming only that n is a multiple of 42 without pinning it down.
• Misadding the fractions (for example $\frac{40}{42}$), which makes $n = 21$ and wrongly marks (A) or (C) as false.