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 "2003 AMC 10A Problems/Problem 16"

(→‎Solution 2 (Patterns))
(→‎Solution 2 (Patterns))
 
Line 17: Line 17:
  
 
By testing the first few values or through previous knowledge, you might see that the units digit of exponents with base 3 follow this pattern:  
 
By testing the first few values or through previous knowledge, you might see that the units digit of exponents with base 3 follow this pattern:  
−
<cmath>3^1=3
+
<cmath>3^1=3</cmath>
−
3^2=9  
+
<cmath>3^2=9</cmath>
−
3^3=27  
+
<cmath>3^3=27</cmath>
−
3^4=81,</cmath>
+
<cmath>3^4=81,</cmath>
 
giving us the rotation <math>3-9-7-1.</math>
 
giving us the rotation <math>3-9-7-1.</math>
  

Latest revision as of 14:52, 19 August 2023

Problem

What is the units digit of $13^{2003}$?

$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9$

Solution 1

$13^{2003}\equiv 3^{2003}\pmod{10}$

Since $3^4=81\equiv1\pmod{10}$:

$3^{2003}=(3^{4})^{500}\cdot3^{3}\equiv1^{500}\cdot27\equiv7\pmod{10}$

Therefore, the units digit is $7 \Rightarrow\boxed{\mathrm{(C)}\ 7}$

Solution 2 (Patterns)

Since we are looking for the units digit of $13^{2003}$, we only have to focus on the units digit of the base (13) as none of the other digits of the base affect the units digit of the resulting value.

By testing the first few values or through previous knowledge, you might see that the units digit of exponents with base 3 follow this pattern: \[3^1=3\] \[3^2=9\] \[3^3=27\] \[3^4=81,\] giving us the rotation $3-9-7-1.$

As this cycle resets every time the index increases by 4, we know that this cycle ends on 2000, and starts once again on 2001. As our expression is raised to the power of 2003, we know that the units digit of our expression must end with the third term of our pattern: $7$.

Therefore, the units digit of our expression is $7 \Rightarrow\boxed{\mathrm{(C)}\ 7}$

~ JinhoK

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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=2003_AMC_10A_Problems/Problem_16&oldid=197199"