Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 12A Problems/Problem 6"
m (→‎Solution 2)
 
(6 intermediate revisions by 4 users not shown)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
<math>U-V=2004^{2005}</math>
+
<cmath>\begin{eqnarray*}
 +
U-V&=&2004*2004^{2004}\\
 +
V-W&=&1*2004^{2004}\\
 +
W-X&=&2001*2004^{2004}\\
 +
X-Y&=&1*2004^{2004}\\
 +
Y-Z&=&2003*2004^{2003}
 +
\end{eqnarray*}</cmath>
  
−
<math>V-W=2004^{2004}</math>
+
After comparison, <math>U-V</math> is the largest. <math>\mathrm {(A)}</math>
−
 
 
−
<math>W-X=2001*2004^{2004}</math>
 
−
 
 
−
<math>X-Y=2004^{2004}</math>
 
  
−
<math>Y-Z=2003*2004^{2003}</math>
 
  
−
After comparison, <math>U-V</math> is the largest. <math>\mathrm {(A)}</math>
+
==Solution 2==
 +
A quick check reveals the positive integers are in decreasing order. Then note <math>V = 2004^{2005}</math>. <math>\newline</math>
 +
<math>U - V = 2004^{2005} = V</math>, and any of the other differences cannot be greater than or equal to <math>V</math>, hence choose <math>\boxed{A}</math> as the answer.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2004|ab=A|num-b=5|num-a=7}}
 
{{AMC12 box|year=2004|ab=A|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 01:10, 30 December 2020

Problem

Let $U=2\cdot 2004^{2005}$, $V=2004^{2005}$, $W=2003\cdot 2004^{2004}$, $X=2\cdot 2004^{2004}$, $Y=2004^{2004}$ and $Z=2004^{2003}$. Which of the following is the largest?

$\mathrm {(A)} U-V \qquad \mathrm {(B)} V-W \qquad \mathrm {(C)} W-X \qquad \mathrm {(D)} X-Y \qquad \mathrm {(E)} Y-Z \qquad$

Solution

\begin{eqnarray*} U-V&=&2004*2004^{2004}\\ V-W&=&1*2004^{2004}\\ W-X&=&2001*2004^{2004}\\ X-Y&=&1*2004^{2004}\\ Y-Z&=&2003*2004^{2003} \end{eqnarray*}

After comparison, $U-V$ is the largest. $\mathrm {(A)}$


Solution 2

A quick check reveals the positive integers are in decreasing order. Then note $V = 2004^{2005}$. $\newline$ $U - V = 2004^{2005} = V$, and any of the other differences cannot be greater than or equal to $V$, hence choose $\boxed{A}$ as the answer.

See Also

2004 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_6&oldid=141017"