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 "2014 AMC 10B Problems/Problem 17"

(→‎Solution)
(→‎Solution: FInsihed Solution)
Line 6: Line 6:
 
==Solution==
 
==Solution==
  
−
We begin by factoring the <math>2^{1002}</math> out. This leaves us with <math>5^{1002} - 1</math> This gives us the final answer of <math>\textbf{(D) } 2^{1005}</math>.
+
We begin by factoring the <math>2^{1002}</math> out. This leaves us with <math>5^{1002} - 1</math>.
 +
We factor the difference of squares, leaving us with <math>(5^{501} - 1)(5^{501} + 1)</math>. We note that for all powers of 5 more than two, it ends in ...<math>25</math> Thus, <math>(5^{501} + 1)</math> would end in ...<math>26</math> and thus would contribute one power of two to the answer, but not more.
 +
We can continue to factor <math>(5^{501} - 1)</math> as a difference of cubes, leaving us with <math>(5^{167} - 1)</math> times an odd number. <math>(5^{167} - 1)</math> ends in ...<math>24</math>, contributing two powers of two to the final result.
 +
Adding these extra 3 power of two to the original 1002 factored out, we obtain the final answer of <math>\textbf{(D) } 2^{1005}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2014|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:01, 20 February 2014

Problem 17

What is the greatest power of $2$ that is a factor of $10^{1002} - 4^{501}$?

$\textbf{(A) } 2^{1002} \qquad\textbf{(B) } 2^{1003} \qquad\textbf{(C) } 2^{1004} \qquad\textbf{(D) } 2^{1005} \qquad\textbf{(E) }2^{1006}$

Solution

We begin by factoring the $2^{1002}$ out. This leaves us with $5^{1002} - 1$. We factor the difference of squares, leaving us with $(5^{501} - 1)(5^{501} + 1)$. We note that for all powers of 5 more than two, it ends in ...$25$ Thus, $(5^{501} + 1)$ would end in ...$26$ and thus would contribute one power of two to the answer, but not more. We can continue to factor $(5^{501} - 1)$ as a difference of cubes, leaving us with $(5^{167} - 1)$ times an odd number. $(5^{167} - 1)$ ends in ...$24$, contributing two powers of two to the final result. Adding these extra 3 power of two to the original 1002 factored out, we obtain the final answer of $\textbf{(D) } 2^{1005}$.

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

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=2014_AMC_10B_Problems/Problem_17&oldid=60073"