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2017 AMC 10A Problem 5

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2017 AMC 10A Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Algebraic Manipulation · difficulty 1 of 5 · about 1 min · position P1-10

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

A)
$1$
B)
$2$
C)
$4$
D)
$8$
E)
$12$

Answer

Key insight

The sum of reciprocals 1/x + 1/y equals (x+y)/(xy), and the given relation x+y = 4xy makes that quotient exactly 4.

Solution

Call the numbers $x$ and $y$ . The condition says $x + y = 4xy$ .

Combine the reciprocals over a common denominator:
$\frac{1}{x} + \frac{1}{y} = \frac{y + x}{xy} = \frac{x + y}{xy}.$
Substituting $x + y = 4xy$ gives
$\frac{4xy}{xy} = 4.$

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Why this works

The sum of reciprocals is always the sum divided by the product, so any relation between $x+y$ and $xy$ translates directly into a value for $\frac1x + \frac1y$ . There is no need to know the numbers themselves; the expression is determined by the relation alone.

Alternative approach

Take a concrete pair. With $x = y$ , the condition becomes $2x = 4x^2$ , so $x = \tfrac12$ . Then $\frac{1}{x} + \frac{1}{y} = 2 + 2 = 4$ .

The trap

Trying to solve for the two numbers individually instead of combining the reciprocals into a single fraction.

Common mistakes

Trying to solve for the two numbers individually instead of combining the reciprocals into a single fraction.
Inverting the relation the wrong way and answering $\tfrac14$ or $1$ .

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta