2002 AMC 10B Problem 7

P1-10~2 minPrint

Fractions & Decimals Divisibility & Factors Use the answer choices (mod checks, size, form) to eliminate or select Bound the quantity above/below or estimate to pin it down

Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:

Show solution without answering

Key insight

The known fractions add to 41/42, just under 1. Since 1/n is at most 1, the sum stays below 2, so it must equal 1 and n = 42.

Solution

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

Why this works

"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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 11: Decimals and Fractions
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Prealgebra, Chapter 6: Decimals (§6.3, §6.4)
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)