Difference between revisions of "2003 AMC 12A Problems/Problem 6"

Latest revision as of 14:38, 19 August 2023

The following problem is from both the 2003 AMC 12A #6 and 2003 AMC 10A #6, so both problems redirect to this page.

Define $x \heartsuit y$ to be $|x-y|$ for all real numbers $x$ and $y$ . Which of the following statements is not true?

$\mathrm{(A) \ } x \heartsuit y = y \heartsuit x$ for all $x$ and $y$

$\mathrm{(B) \ } 2(x \heartsuit y) = (2x) \heartsuit (2y)$ for all $x$ and $y$

$\mathrm{(C) \ } x \heartsuit 0 = x$ for all $x$

$\mathrm{(D) \ } x \heartsuit x = 0$ for all $x$

$\mathrm{(E) \ } x \heartsuit y > 0$ if $x \neq y$

We start by looking at the answers. Examining statement C, we notice:

$x \heartsuit 0 = |x-0| = |x|$

$|x| \neq x$ when $x<0$ , but statement C says that it does for all $x$ .

Therefore the statement that is not true is $\boxed{\mathrm{(C)}\ x\heartsuit 0=x\ \text{for all}\ x}$

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 14: / Line 14: @@
 == Solution ==
-Examining statement C:
+We start by looking at the answers. Examining statement C, we notice:
 <math> x \heartsuit 0 = |x-0| = |x| </math>
@@ Line 21: / Line 21: @@
 Therefore the statement that is not true is <math>\boxed{\mathrm{(C)}\ x\heartsuit 0=x\ \text{for all}\ x}</math>
+== Video Solution ==
+https://www.youtube.com/watch?v=d-IRsaIUdDA   ~David
 == See Also ==