# Difference between revisions of "2017 AMC 10A Problems/Problem 5"

## Problem

The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

## Solution

Let the two real numbers be $x,y$. We are given that $x+y=4xy,$ and dividing both sides by $xy$, $\frac{x}{xy}+\frac{y}{xy}=4.$

$$\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.$$

Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.

## Solution 2

Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction. See for yourself. And by looking into fractions, we immediately see that $\frac{1}{3}$ and $1$ would fit the rule. $\boxed{\textbf{(C)} 4}.$

## Solution 3

Notice that from the information given above, $x+y=4xy$

Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or $\frac{x+y}{xy}$

We can solve this by substituting $x+y\implies 4xy$.

Our answer is simply $\frac{4xy}{xy}\implies4$.

Therefore, the answer is $\boxed{\textbf{(C) } 4}$.

## Solution 4

Let the two numbers be $a$ and $b$, respectively. We wish to find $\frac{1}{a} + \frac{1}{b}$. Note that $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$. We are given that $a+b = 4ab$.

Subsituting, we have $\frac{a+b}{ab} = \frac{4ab}{ab} = \boxed{\textbf{(C) } 4}$