Difference between revisions of "2003 AMC 10A Problems/Problem 16"

m (Solution)
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<math>3^{2003}=(3^{4})^{500}\cdot3^{3}\equiv1^{500}\cdot27\equiv7\pmod{10}</math>
 
<math>3^{2003}=(3^{4})^{500}\cdot3^{3}\equiv1^{500}\cdot27\equiv7\pmod{10}</math>
  
Therefore, the units digit is <math>7 \Rightarrow C</math>
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Therefore, the units digit is <math>7 \Rightarrow\boxed{\mathrm{(C)}\ 7}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 22:42, 16 July 2014

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

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

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 $7 \Rightarrow\boxed{\mathrm{(C)}\ 7}$

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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

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