2010 AMC 10A Problem 6

P1-10~1 minPrint

Functions Set up the equation/formula and compute; no special trick needed

For positive numbers $x$ and $y$ the operation $x\spadesuit y$ is defined as

$x\spadesuit y = x -\dfrac{1}{y}$

What is $2\spadesuit (2\spadesuit 2)$ ?

Show solution without answering

Key insight

Do the inside first: 2 spade 2 = 3/2. Then 2 minus 1 divided by 3/2 is 2 - 2/3 = 4/3.

Solution

The rule says: take the first number and subtract $1$ divided by the second number. Work from the inside of the brackets outward.

1. Inside bracket. $2\spadesuit 2$ means $x = 2$ and $y = 2$ :
$2 - \frac{1}{2} = \frac{3}{2}.$
2. Outside. Now the problem is $2\spadesuit \tfrac32$ , so $x = 2$ and $y = \tfrac32$ :
$2 - \frac{1}{3/2}.$
3. Dividing $1$ by $\tfrac32$ flips the fraction: $\dfrac{1}{3/2} = \dfrac{2}{3}$ .
4. So the value is $2 - \dfrac23 = \dfrac63 - \dfrac23 = \dfrac43$ .

The answer is $\boxed{\textbf{(C)}\ \dfrac{4}{3}}$ .

Why this works

A made-up symbol is just a rule for combining two numbers. The brackets tell you which part to do first, and that result becomes the second number in the next step. The only tricky part here is $1$ divided by a fraction, which is that fraction turned upside down.

Concise solution

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

Work from the inside out. The inner expression is
$2\spadesuit 2 = 2 - \frac{1}{2} = \frac{3}{2}.$

Now substitute that as the second input:
$2\spadesuit \frac{3}{2} = 2 - \frac{1}{3/2} = 2 - \frac{2}{3} = \frac{4}{3}.$

The answer is $\boxed{\textbf{(C)}\ \dfrac{4}{3}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 16: Functions
Introduction to Algebra, Chapter 17: Graphing Functions (§17.2)
Intermediate Algebra, Chapter 19: Functional Equations (§19.1, §19.2)